Question

At the foot of a mountain the elevation of its summit is ${45}^{\circ };$ after ascending 1000 m towards the mountain up a slope of ${30}^{\circ }$ inclination, the elevation is found to be ${60}^{\circ }$. Find the height of the mountain.

Answer 1

$DE=1000sin30=1000\times \frac{1}{2}=500m=FBEC=1000cos30=1000\times \frac{\sqrt{3}}{2}=500\sqrt{3}mLetAF=xmDF=\frac{x}{\sqrt{3}}m=BE$

$\text{We know,}From\Delta ABC,\Rightarrow 1=\frac{AF+FB}{BE+FC}\Rightarrow 1=\frac{x+500}{\frac{x}{\sqrt{3}}+500\sqrt{3}}\Rightarrow \frac{x}{\sqrt{3}}+500\sqrt{3}=x+500\Rightarrow x+1500=x\sqrt{3}+500\sqrt{3}\Rightarrow 1500-500\sqrt{3}=x\sqrt{3}-x\Rightarrow 500\sqrt{3}(\sqrt{3}-1)=x(\sqrt{3}-1)\therefore x=500\sqrt{3}m\text{The height of the triangle is}AB=AF+FB=500(\sqrt{3}+1)m$