A) 10 m
B) 20 m
C) 30 m
D) 40 m
Correct Answer: B
Solution :
[b] Let PQ=h and PB=x=PA then tan \[30{}^\circ =\frac{h}{x}\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x}\] \[\Rightarrow x=\sqrt{3}h\] using sine rule in APAB \[\frac{PA}{\sin 30{}^\circ }=\frac{AB}{\sin 120{}^\circ }\] \[\Rightarrow x=\frac{60.\sin 30{}^\circ }{\sin 120{}^\circ }\] \[\Rightarrow \sqrt{3}h=60\times \frac{1}{2}\times \frac{2}{\sqrt{3}}\] \[\Rightarrow \] \[h=20\,m\]You need to login to perform this action.
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