Question 91
In triangle ABC, ∠ABC = $$15^o$$ .D is a point on BC such that AD = BD. What is the measure of ∠ADC (in degrees)?
Solution
Given : ∠ABC = $$15^o$$ and AD = BD
To find : ∠ADC = $$\theta$$ = ?
Solution : AD = BD, => $$\angle$$ ABD = $$\angle$$ DAB = $$15^o$$
=> $$\angle$$ ABD + $$\angle$$ DAB = $$15+15=30^\circ$$
Using, exterior angle property of a triangle,
=> $$\angle$$ ADC = $$\angle$$ ABD + $$\angle$$ DAB
=> $$\theta=30^\circ$$
=> Ans - (B)
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