Question

At the foot of a mountain, the elevation of its summit is ${45}^{\circ }.$ After ascending $1000\text{m}$ 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

In $△ADE$

By $\sin {30}^{\circ }=\frac{DE}{AD}$

$\frac{1}{2}=\frac{DE}{1000}$

$DE=500m$

By $\cos {30}^{\circ }=\frac{AE}{AD}$

$\frac{\sqrt{3}}{2}=\frac{AE}{1000}$

$AE=500\sqrt{3}m$

Let $EB=xm$

then $EB=DF=xm$

In $△DLF,\tan {60}^{\circ }=\frac{CF}{DF}$

$\Rightarrow \sqrt{3}=\frac{CF}{DF}$

$\Rightarrow CF=x\sqrt{3}$

In $△ABC$

$\tan {45}^{\circ }=\frac{BC}{AB}$

$1=\frac{BC}{AB}$

$\Rightarrow AB=BC$

$x+500\sqrt{3}=500+\sqrt{3}x$

$x(\sqrt{3}-1)=500(\sqrt{3}-1)$

$x=500m$

$Height=x\sqrt{3}+500$

$=500\sqrt{3}+500$

$=500(\sqrt{3}+1)m$

$\therefore$ Height of the mountain is $500(\sqrt{3}+1)\text{m}$