A) \[\frac{h\sin \alpha +a\cos \alpha }{9\sin \alpha }\]
B) \[\frac{h\cos \alpha -a\sin \alpha }{9\cos \alpha }\]
C) \[\frac{h\cos \alpha -a\sin \alpha }{9\sin \alpha }\]
D) \[\frac{h\sin \alpha +a\cos \alpha }{9\cos \alpha }\]
Correct Answer: B
Solution :
\[\frac{h}{a+9x}=\frac{y}{a}\]\[y=a\tan \alpha \]\[\Rightarrow \]\[\frac{h}{a+9x}=\frac{a\tan \alpha }{a}\] \[\Rightarrow \]\[a+9x=\frac{h}{\tan \alpha }\]\[\Rightarrow \]\[x=\frac{h-a\tan \alpha }{9}\] \[=\frac{(h\cos \alpha -a\sin \alpha )}{9\cos \alpha }\]
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