Question 60
In the trapezium ABCD the sides AB and CD are parallel. The value of $$\frac{sin \angle{BAC}}{sin \angle{BAD}}$$ is
Solution
Construct a perpendicular in the trapezium. Let its height be 'h'
Now, $$sin\angle BAC = \dfrac{h}{AC}$$
and $$sin\angle BAD = \dfrac{h}{AD}$$
Thus, $$\dfrac{sin\angle BAC}{sin\angle BAD} = \dfrac{\frac{h}{AC}}{\frac{h}{AD}} = \dfrac{AD}{AC}$$
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