Practice · Perpendicular is half of a straight angle · Sensim Math

Variant 1 answer: 60 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $30^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 30 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 30 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 30 deg from MO toward the left, inside the upper-left region.

\angle MOE = 30^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (30 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 30^\circ = 60^\circ

The upward segment to OE uses 30 deg of the 90 deg quarter-turn, leaving 60 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 60^\circ

Crossing lines make equal angles straight across, so the 60 deg angle on the upper-left reappears as a on the lower-right.

Answer: 60 degrees

Review

OE is 60 deg above the horizontal line on the left, so its opposite ray OF dips 60 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 30 + 60 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 30 deg and angle MOD = 90 deg give angle EOD = 120 deg; since EOD and DOF are supplementary on line EF, a = 180 - 120 = 60 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 30 = 60 deg and matching it across as the vertical angle a.

💡 Take 30 deg out of the 90 deg right angle to get 60 deg, then bounce it straight across the crossing: angle a is 60 deg!

Variant 2 answer: 20 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $70^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 70 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 70 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 70 deg from MO toward the left, inside the upper-left region.

\angle MOE = 70^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (70 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 70^\circ = 20^\circ

The upward segment to OE uses 70 deg of the 90 deg quarter-turn, leaving 20 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 20^\circ

Crossing lines make equal angles straight across, so the 20 deg angle on the upper-left reappears as a on the lower-right.

Answer: 20 degrees

Review

OE is 20 deg above the horizontal line on the left, so its opposite ray OF dips 20 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 70 + 20 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 70 deg and angle MOD = 90 deg give angle EOD = 160 deg; since EOD and DOF are supplementary on line EF, a = 180 - 160 = 20 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 70 = 20 deg and matching it across as the vertical angle a.

💡 Take 70 deg out of the 90 deg right angle to get 20 deg, then bounce it straight across the crossing: angle a is 20 deg!

Variant 3 answer: 35 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $55^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 55 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 55 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 55 deg from MO toward the left, inside the upper-left region.

\angle MOE = 55^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (55 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 55^\circ = 35^\circ

The upward segment to OE uses 55 deg of the 90 deg quarter-turn, leaving 35 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 35^\circ

Crossing lines make equal angles straight across, so the 35 deg angle on the upper-left reappears as a on the lower-right.

Answer: 35 degrees

Review

OE is 35 deg above the horizontal line on the left, so its opposite ray OF dips 35 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 55 + 35 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 55 deg and angle MOD = 90 deg give angle EOD = 145 deg; since EOD and DOF are supplementary on line EF, a = 180 - 145 = 35 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 55 = 35 deg and matching it across as the vertical angle a.

💡 Take 55 deg out of the 90 deg right angle to get 35 deg, then bounce it straight across the crossing: angle a is 35 deg!

Variant 4 answer: 50 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $40^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 40 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 40 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 40 deg from MO toward the left, inside the upper-left region.

\angle MOE = 40^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (40 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 40^\circ = 50^\circ

The upward segment to OE uses 40 deg of the 90 deg quarter-turn, leaving 50 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 50^\circ

Crossing lines make equal angles straight across, so the 50 deg angle on the upper-left reappears as a on the lower-right.

Answer: 50 degrees

Review

OE is 50 deg above the horizontal line on the left, so its opposite ray OF dips 50 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 40 + 50 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 40 deg and angle MOD = 90 deg give angle EOD = 130 deg; since EOD and DOF are supplementary on line EF, a = 180 - 130 = 50 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 40 = 50 deg and matching it across as the vertical angle a.

💡 Take 40 deg out of the 90 deg right angle to get 50 deg, then bounce it straight across the crossing: angle a is 50 deg!

Variant 5 answer: 15 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $75^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 75 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 75 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 75 deg from MO toward the left, inside the upper-left region.

\angle MOE = 75^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (75 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 75^\circ = 15^\circ

The upward segment to OE uses 75 deg of the 90 deg quarter-turn, leaving 15 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 15^\circ

Crossing lines make equal angles straight across, so the 15 deg angle on the upper-left reappears as a on the lower-right.

Answer: 15 degrees

Review

OE is 15 deg above the horizontal line on the left, so its opposite ray OF dips 15 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 75 + 15 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 75 deg and angle MOD = 90 deg give angle EOD = 165 deg; since EOD and DOF are supplementary on line EF, a = 180 - 165 = 15 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 75 = 15 deg and matching it across as the vertical angle a.

💡 Take 75 deg out of the 90 deg right angle to get 15 deg, then bounce it straight across the crossing: angle a is 15 deg!

Variant 6 answer: 30 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $60^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 60 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 60 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 60 deg from MO toward the left, inside the upper-left region.

\angle MOE = 60^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (60 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 60^\circ = 30^\circ

The upward segment to OE uses 60 deg of the 90 deg quarter-turn, leaving 30 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 30^\circ

Crossing lines make equal angles straight across, so the 30 deg angle on the upper-left reappears as a on the lower-right.

Answer: 30 degrees

Review

OE is 30 deg above the horizontal line on the left, so its opposite ray OF dips 30 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 60 + 30 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 60 deg and angle MOD = 90 deg give angle EOD = 150 deg; since EOD and DOF are supplementary on line EF, a = 180 - 150 = 30 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 60 = 30 deg and matching it across as the vertical angle a.

💡 Take 60 deg out of the 90 deg right angle to get 30 deg, then bounce it straight across the crossing: angle a is 30 deg!

Variant 7 answer: 25 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $65^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 65 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 65 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 65 deg from MO toward the left, inside the upper-left region.

\angle MOE = 65^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (65 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 65^\circ = 25^\circ

The upward segment to OE uses 65 deg of the 90 deg quarter-turn, leaving 25 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 25^\circ

Crossing lines make equal angles straight across, so the 25 deg angle on the upper-left reappears as a on the lower-right.

Answer: 25 degrees

Review

OE is 25 deg above the horizontal line on the left, so its opposite ray OF dips 25 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 65 + 25 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 65 deg and angle MOD = 90 deg give angle EOD = 155 deg; since EOD and DOF are supplementary on line EF, a = 180 - 155 = 25 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 65 = 25 deg and matching it across as the vertical angle a.

💡 Take 65 deg out of the 90 deg right angle to get 25 deg, then bounce it straight across the crossing: angle a is 25 deg!

Variant 8 answer: 40 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $50^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 50 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 50 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 50 deg from MO toward the left, inside the upper-left region.

\angle MOE = 50^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (50 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 50^\circ = 40^\circ

The upward segment to OE uses 50 deg of the 90 deg quarter-turn, leaving 40 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 40^\circ

Crossing lines make equal angles straight across, so the 40 deg angle on the upper-left reappears as a on the lower-right.

Answer: 40 degrees

Review

OE is 40 deg above the horizontal line on the left, so its opposite ray OF dips 40 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 50 + 40 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 50 deg and angle MOD = 90 deg give angle EOD = 140 deg; since EOD and DOF are supplementary on line EF, a = 180 - 140 = 40 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 50 = 40 deg and matching it across as the vertical angle a.

💡 Take 50 deg out of the 90 deg right angle to get 40 deg, then bounce it straight across the crossing: angle a is 40 deg!

Variant 9 answer: 65 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $25^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 25 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 25 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 25 deg from MO toward the left, inside the upper-left region.

\angle MOE = 25^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (25 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 25^\circ = 65^\circ

The upward segment to OE uses 25 deg of the 90 deg quarter-turn, leaving 65 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 65^\circ

Crossing lines make equal angles straight across, so the 65 deg angle on the upper-left reappears as a on the lower-right.

Answer: 65 degrees

Review

OE is 65 deg above the horizontal line on the left, so its opposite ray OF dips 65 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 25 + 65 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 25 deg and angle MOD = 90 deg give angle EOD = 115 deg; since EOD and DOF are supplementary on line EF, a = 180 - 115 = 65 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 25 = 65 deg and matching it across as the vertical angle a.

💡 Take 25 deg out of the 90 deg right angle to get 65 deg, then bounce it straight across the crossing: angle a is 65 deg!

Variant 10 answer: 45 degrees

Segment $\overline{CO}$ and segment $\overline{MO}$ are perpendicular to each other. Find the measure of angle $a$ .

[Figure] A horizontal line $CD$ is drawn through point $O$ (left end $C$ , right end $D$ ). From point $O$ , segment $\overline{MO}$ is drawn straight up, perpendicular to segment $\overline{CO}$ . A second straight line $EF$ passes through point $O$ , with $E$ at the upper left and $F$ at the lower right. The angle between ray $\overline{OE}$ and the upward segment $\overline{MO}$ is marked as $45^\circ$ . The angle $a$ is the angle between ray $\overline{OD}$ (horizontal, to the right) and ray $\overline{OF}$ (down to the right).

Show solution

Understand

A horizontal line CD passes through point O. Segment MO points straight up from O and is perpendicular to CO. A second straight line EF also passes through O, with E up-left and F down-right. The angle between ray OE and the upward segment MO is 45 deg. I need angle a, between ray OD (horizontal, right) and ray OF (down-right).

Givens

CD is a straight horizontal line through O.
MO is perpendicular to CO, so angle MOD = 90 deg and angle MOC = 90 deg.
EF is a straight line through O (E up-left, F down-right), so OE and OF are opposite rays.
The angle MOE between MO (up) and OE (up-left) is 45 deg.
Angle a is between OD (right) and OF (down-right).

Unknowns

The measure of angle a (angle DOF).

Constraints

Angles on one side of a straight line at a point add to 180 deg.
Vertical (opposite) angles are equal.
Perpendicular lines make a 90 deg angle.

Plan

#1 Draw a Diagram · also uses: #7 Identify Subproblems

Use the angles around O on the upper-left side. The upward ray OM is 90 deg from the horizontal ray OC, and ray OE splits that right angle. Angle a (between OD and OF) is the vertical angle to angle EOC, giving a quick subtraction.

Execute

#1 Draw a Diagram 4.G.A.1

MO is straight up and perpendicular to the horizontal line, so angle MOD = 90 deg on the right and angle MOC = 90 deg on the left. Ray OE is 45 deg from MO toward the left, inside the upper-left region.

\angle MOE = 45^\circ, \quad \angle MOC = 90^\circ

Perpendicular means a clean 90 deg between the upward segment and the horizontal line, a fixed reference to measure OE from.

#7 Identify Subproblems 4.MD.C.7

On the upper-left, the 90 deg angle MOC is split by ray OE into angle MOE (45 deg) and angle EOC. So angle EOC is the leftover.

\angle EOC = 90^\circ - 45^\circ = 45^\circ

The upward segment to OE uses 45 deg of the 90 deg quarter-turn, leaving 45 deg from OE down to the horizontal line OC.

#7 Identify Subproblems 4.MD.C.7

OE and OF are opposite rays, and OC and OD are opposite rays, so angle DOF (which is a) is the vertical angle of angle EOC and therefore equal to it.

a = \angle DOF = \angle EOC = 45^\circ

Crossing lines make equal angles straight across, so the 45 deg angle on the upper-left reappears as a on the lower-right.

Answer: 45 degrees

Review

OE is 45 deg above the horizontal line on the left, so its opposite ray OF dips 45 deg below the horizontal on the right, making angle a match the slant of OF in the figure. Check: 45 + 45 = 90, exactly the right angle MOC.

Use the straight line idea (tool 7): angle MOE = 45 deg and angle MOD = 90 deg give angle EOD = 135 deg; since EOD and DOF are supplementary on line EF, a = 180 - 135 = 45 deg.

Standards · min grade 4

4.G.A.1 Draw points, lines, line segments, rays, angles, and identify in figures — Using the perpendicular MO to fix the 90 deg reference angles at O.
4.MD.C.7 Recognize angle measure as additive and solve addition and subtraction problems — Computing 90 - 45 = 45 deg and matching it across as the vertical angle a.

💡 Take 45 deg out of the 90 deg right angle to get 45 deg, then bounce it straight across the crossing: angle a is 45 deg!

Perpendicular is half of a straight angle · 10 practice problems

Generated variants — 10

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