In a triangle ABC , if angle A +angle B = 108 degree ,angle B +angle C =130 degree find angle A , angle B , and angle C .
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Answered by
36
Answer :
- ∠A = 50°
- ∠B = 58°
- ∠C = 72°
Given :
- ∠A + ∠B = 108°
- ∠B + ∠C = 130°
To Find :
- The value of ∠A, ∠B and ∠C.
Step-by-step explanation :
We have ∠A + ∠B = 108° and ∠B + ∠C = 130°
On adding we get,
⟹ (∠A + ∠B) + (∠B + ∠C) = 108 + 130
⟹ ∠A + ∠B + ∠B + ∠C = 238
⟹ (∠A + ∠B + ∠C) + ∠B = 238
⟹ 180 + ∠B = 238
⟹ ∠B = 238 - 180
⟹ ∠B = 58
•°• ∠B = 58°
Now, ∠A + ∠B = 108
⟹ ∠A = 108 - ∠B
= 108 - 58
= 50°
Also, ∠B + ∠C = 130
⟹ ∠C = 130 - ∠B
= 130 - 58
= 72
So, ∠A = 50°
∠B = 58°
∠C = 72°
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