Here is the step-by-step solution for each problem on the worksheet, using the properties of triangles and straight lines.
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Key Concepts Used:
1.
Sum of Interior Angles in a Triangle = 180°
2.
Exterior Angle Theorem: An exterior angle of a triangle equals the sum of the two opposite interior angles.
3.
Linear Pair (Straight Line): Angles on a straight line add up to 180°.
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Problem 1
Given: Triangle CDE with ∠D = 21°, ∠E = 31°
Find: ∠ECD (interior angle at C)
> Use the triangle angle sum:
> ∠ECD = 180° - ∠D - ∠E = 180° - 21° - 31° =
128°
✔ Answer: ∠ECD = 128°
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Problem 2
Given: Triangle TQR with ∠T = 55°, ∠R = 39°
Find: ∠TQR (interior angle at Q)
> ∠TQR = 180° - ∠T - ∠R = 180° - 55° - 39° =
86°
✔ Answer: ∠TQR = 86°
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Problem 3
Given: Triangle NOP with ∠O = 77°, ∠P = 77°
Find: ∠MNP (exterior angle at N)
> First, find interior ∠N:
> ∠N = 180° - 77° - 77° = 26°
> Since ∠MNP is an exterior angle at N, it forms a linear pair with ∠N:
> ∠MNP = 180° - ∠N = 180° - 26° =
154°
> *Alternatively*, use Exterior Angle Theorem:
> ∠MNP = ∠O + ∠P = 77° + 77° =
154°
✔ Answer: ∠MNP = 154°
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Problem 4
Given: Triangle OPQ with ∠P = 87°, ∠Q = 63°
Find: ∠MOQ (exterior angle at O)
> First, find interior ∠O:
> ∠O = 180° - 87° - 63° = 30°
> ∠MOQ is exterior at O → linear pair with ∠O:
> ∠MOQ = 180° - 30° =
150°
> *Or by Exterior Angle Theorem*:
> ∠MOQ = ∠P + ∠Q = 87° + 63° =
150°
✔ Answer: ∠MOQ = 150°
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Problem 5
Given: Triangle MNP with ∠M = 60°, ∠N = 60°
Find: ∠NPM (interior angle at P)
> ∠NPM = 180° - 60° - 60° =
60°
✔ Answer: ∠NPM = 60°
*(This is an equilateral triangle!)*
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Problem 6
Given: Triangle HFW with ∠H = 39°, ∠W = 26°
Find: ∠HFX (exterior angle at F)
> First, find interior ∠F:
> ∠F = 180° - 39° - 26° = 115°
> ∠HFX is exterior at F → linear pair with ∠F:
> ∠HFX = 180° - 115° =
65°
> *Or by Exterior Angle Theorem*:
> ∠HFX = ∠H + ∠W = 39° + 26° =
65°
✔ Answer: ∠HFX = 65°
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Problem 7
Given: Triangle PQR with ∠Q = 51°, ∠R = 46°
Find: ∠RPQ (interior angle at P)
> ∠RPQ = 180° - 51° - 46° =
83°
✔ Answer: ∠RPQ = 83°
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Problem 8
Given: Triangle ABC with ∠B = 47°, ∠C = 60°
Find: ∠XAC (exterior angle at A)
> First, find interior ∠A:
> ∠A = 180° - 47° - 60° = 73°
> ∠XAC is exterior at A → linear pair with ∠A:
> ∠XAC = 180° - 73° =
107°
> *Or by Exterior Angle Theorem*:
> ∠XAC = ∠B + ∠C = 47° + 60° =
107°
✔ Answer: XAC = 107°
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✔ Final Answers:
1.
∠ECD = 128°
2.
∠TQR = 86°
3.
∠MNP = 154°
4.
∠MOQ = 150°
5.
∠NPM = 60°
6.
∠HFX = 65°
7.
∠RPQ = 83°
8.
XAC = 107°
Let me know if you’d like a diagram explanation for any problem!
Parent Tip: Review the logic above to help your child master the concept of interior exterior angles worksheet.
