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Question 129
In a triangle ABC, $$\angle{A} = 90^o , \angle{C} = 55^o , \overline{AD}$$ is perpendicular to $$\overline{BC}$$. What is the value of $$\angle{BAD}$$ ?
Solution
$$\angle A = 90^o$$
$$\angle C = 55^o$$
$$\angle B$$ will be $$180 - (90+55) = 35^o$$
As AD is perpendicular to BC Hence $$\angle BAD=180-(90+35)=55^o$$
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