Question

(i)the point A,B,C on the exterior of /\PQR.
(ii) the point X,Y,Z in the interior of/\PQR
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Answer 1

Step-by-step explanation:

Given,

∠DCE = 53°

∠DCE = ∠ACB = 53° (vertically opposite angles)

In triangle ABC,

∠ABC + ∠ACB = ∠CAF (by exterior angle property)

∠ABC + 53° = 112°

∠ABC = 112° – 53° = 59°

Again, in triangle ABC,

By angle sum property of a triangle,

∠BAC + ∠ABC + ∠BCA = 180°

∠BAC + 59° + 53° = 180°

∠BAC = 180° – 59° – 53° = 68°

Therefore, ∠BAC = 68°, ∠ABC = 59°, ∠ACB = 53°