Let’s solve each problem step by step. We’re looking for unknown angles using what we know about straight lines (180°) and right angles (90°).
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Problem 1:
We see a right angle marked at ∠yoz = 90°. The line xoy is straight, so total angle on that line is 180°.
∠xoz + ∠zoy = 180°
∠xoz + 90° = 180°
→ ∠xoz = 180° - 90° =
90°
Wait — actually, looking again: point O has rays going to X, Y, Z. Ray OZ is perpendicular to XY (marked with square), so ∠xoz and ∠yoz are both 90°? But the diagram shows only one 90° mark between z and y. Actually, since x-o-y is a straight line, and o-z is perpendicular to it, then both ∠xoz and ∠yoz are 90°. So yes, ∠xoz = 90°.
But let me double-check: if x-o-y is straight (180°), and o-z goes up forming 90° with o-y, then angle from o-x to o-z must also be 90°, because 180 - 90 = 90. Correct.
✔ Problem 1:
90°
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Problem 2:
Line p-o-q is straight → 180°.
Angle given: ∠por = 105°? Wait — label says “105°” near ray r, between p and r? Actually, diagram shows angle between p and r is 105°, and we need ∠roq.
Since p-o-q is straight, ∠por + ∠roq = 180°
So ∠roq = 180° - 105° =
75°
✔ Problem 2:
75°
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Problem 3:
Line a-c-b is straight → 180°.
Given: ∠dcb = 60° (angle between d and b).
We need ∠acd — which is the other part of the straight line.
So ∠acd + ∠dcb = 180°
∠acd + 60° = 180°
→ ∠acd = 120°
✔ Problem 3:
120°
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Problem 4:
Same setup: line a-c-b straight → 180°.
Given: ∠dcb = 30°
Need ∠acd.
∠acd + 30° = 180°
→ ∠acd = 150°
✔ Problem 4:
150°
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Problem 5:
Line x-q-y straight → 180°.
Given: ∠pqy = 135°? Wait — diagram shows angle between p and y is 135°, but labeled at q. Actually, angle shown is ∠p q y = 135°? No — look: ray qp and qy form 135°, and we need ∠xqp.
Points: x—q—y is straight line. Ray qp goes up-left. Angle between qp and qy is 135°. Then angle between xq and qp is what’s left to make 180°.
So ∠xqp + ∠pqy = 180°
∠xqp + 135° = 180°
→ ∠xqp = 45°
✔ Problem 5:
45°
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Problem 6:
Line p-t-q straight → 180°.
Given: ∠stq = 120°? Diagram shows angle between s and q is 120°, at point t. Need ∠pts.
So ∠pts + ∠stq = 180°
∠pts + 120° = 180°
→ ∠pts = 60°
✔ Problem 6:
60°
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Problem 7:
Line a-y-b straight → 180°.
Given: ∠xya = 15° (angle between x and a). Need ∠xyb.
Note: ∠xya and ∠xyb are adjacent angles making up the straight line at y.
Actually, ray yx is going up-left, ya is left, yb is right. So angle between yx and ya is 15°, and we want angle between yx and yb.
Since a-y-b is straight, angle from ya to yb is 180°.
So ∠xyb = ∠xya + ∠ayb? Wait — no.
Better: Points a-y-b straight. Ray yx makes 15° with ya. So angle between yx and yb is 180° - 15° = 165°.
Because from ya to yb is 180°, and yx is 15° away from ya toward the top, so from yx to yb is 180 - 15 = 165°.
Yes.
✔ Problem 7:
165°
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Problem 8:
Line p-o-q straight → 180°.
Given: ∠pos = 100°? Diagram shows angle between p and s is 100°, at o. Need ∠soq.
So ∠pos + ∠soq = 180°
100° + ∠soq = 180°
→ ∠soq = 80°
✔ Problem 8:
80°
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Final Answer:
1. 90°
2. 75°
3. 120°
4. 150°
5. 45°
6. 60°
7. 165°
8. 80°
Parent Tip: Review the logic above to help your child master the concept of angles worksheet geometry.
