Question

From a point on the horizontal plane, the elevation of the top of a hill is ${45}^{\circ }$. After walking $500\text{m}$ in towards its summit up a slope inclined at an angle of ${15}^{\circ }$ to the horizon the elevation is ${75}^{\circ }$, the height of the hill is:

• A. $500\sqrt{6}\text{m}$
• B. $500\sqrt{3}\text{m}$
• C. $250\sqrt{6}\text{m}$
• D. $250\sqrt{3}\text{m}$

Answer 1

The correct option is A $500\sqrt{6}\text{m}$

In $△ABD,$

$\tan {75}^{\circ }=\frac{AD}{BD}=\frac{AD}{OE}\dots \left(1\right)$

In $△ACO,$

$\tan {45}^{\circ }=\frac{AO}{OC}$

$\frac{AO}{OC}=1$

$\Rightarrow AO=OC$

In $△BCE$

$\tan {15}^{\circ }=\frac{BE}{CE}=\frac{OD}{CE}\dots \left(2\right)$

From $\left(1\right)$,

$AD=OE\tan {75}^{\circ }$

From $\left(2\right)$,

$OD=CE\tan {15}^{\circ }$

$OA=AD+OD=OE\tan {75}^{\circ }+CE\tan {15}^{\circ }$

In $△CBE,$

$\cos {15}^{\circ }=\frac{CE}{BC}$

$CE=BC\cos {15}^{\circ }$

$OA=OE\tan {75}^{\circ }+BC\tan {15}^{\circ }\times \cos {15}^{\circ }$

$OA=(OC-CE)\tan {75}^{\circ }+BC\sin {15}^{\circ }$

$OA=(OA-BC\cos {15}^{\circ })\tan {75}^{\circ }+BC\sin {15}^{\circ }$

$OA(1-\tan {75}^{\circ })=BC(\sin {15}^{\circ }-\cos {15}^{\circ }.\tan {75}^{\circ })$

$OA=\frac{BC(\sin {15}^{\circ }-\cos {15}^{\circ }.\frac{\sin {75}^{\circ }}{\cos {75}^{\circ }}}{1-\frac{\sin {75}^{\circ }}{\cos {75}^{\circ }}}$

$OA=\frac{BC(\sin {15}^{\circ }.\cos {75}^{\circ }-\cos {15}^{\circ }.\sin {75}^{\circ })}{\cos {75}^{\circ }-\sin {75}^{\circ }}$$=\frac{BC(-\sin {60}^{\circ })}{\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)-\left(\frac{\sqrt{3}+1}{2\sqrt{2}}\right)}=\frac{BC\left(\frac{-\sqrt{3}}{2}\right)}{\frac{-1}{\sqrt{2}}}=500\sqrt{6}$