Question 49

In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If $$\angle BAC = 45^{\circ}$$ and $$\angle ABC=\theta\ $$, then $$\frac{AD}{BE}$$ equals


It is given, Angle BAE = 45 degrees

This implies AE = BE

Let AE = BE = x 

In right-angled triangle ABD, it is given $$\angle ABC=\theta\ $$

$$\sin\theta=\frac{AD}{AB}\ $$

$$\sin\theta=\frac{AD}{x\sqrt{\ 2}}\ $$

$$\sqrt{\ 2}\sin\theta=\frac{AD}{BE}\ $$

The answer is option D.

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