question_answer

The length of the shadow of a pole inclined at 10o to the vertical towards the sun is 2.05 metres, when the elevation of the sun is 38o. The length of the pole is [Roorkee 1976]

A) \[\frac{2.05\,\,\sin \,\,{{38}^{o}}}{\sin \,{{42}^{o}}}\]

B) \[\frac{2.05\,\,\sin \,\,{{42}^{o}}}{\sin \,{{38}^{o}}}\]

C) \[\frac{2.05\,\,\cos \,\,{{38}^{o}}}{\cos \,\,{{42}^{o}}}\]

D) None of these

Correct Answer: A

Solution :

\[\frac{\sin {{38}^{o}}}{l}=\frac{\sin (SPO)}{2.05}\] = \[\frac{\sin ({{180}^{o}}-{{38}^{o}}-{{90}^{o}}-{{10}^{o}})}{2.05}\Rightarrow l=\frac{2.05\sin {{38}^{o}}}{\sin {{42}^{o}}}\].