# At the foot of a mountain the elevation of its summit is 45º; after ascending 1000 m towards the mountain up a slope of 30º inclination is found to be 60º. Find the height of the mountain. - Mathematics

Sum

At the foot of a mountain the elevation of its summit is 45º; after ascending 1000 m towards the mountain up a slope of 30º inclination is found to be 60º. Find the height of the mountain.

#### Solution

Let F be the foot and S be the summit of the mountain FOS.

Then, ∠OFS = 45º and therefore ∠OSF = 45º. Consequently,

OF = OS = h km(say).

Let FP = 1000 m = 1 km be the slope so that

∠OFP = 30º. Draw PM ⊥OS and PL ⊥OF.

Join PS. It is given that ∠MPS = 60º.

In ∆FPL, We have

\text{sin }30^\text{o}=\frac{PL}{PF}

\Rightarrow PL=PF\text{ sin }30^\text{o}=( 1\times \frac{1}{2})=\frac{1}{2}km.

\therefore OM=PL=\frac{1}{2}km

\Rightarrow MS=OS-OM=( h-\frac{1}{2})km

\text{Also, cos }30^\text{o}=\frac{FL}{PF}

\Rightarrow FL=PF\text{ cos }30^{o}=( 1\times\frac{\sqrt{3}}{2})=\frac{\sqrt{3}}{2}km

Now, h = OS = OF = OL + LF

\Rightarrow h=OL+\frac{\sqrt{3}}{2}

\Rightarrow OL=( h-\frac{\sqrt{3}}{2})km

\Rightarrow PM=( h-\frac{\sqrt{3}}{2})km

In ∆PSM, we have

\text{tan }60^\text{o}=\frac{SM}{PM}

⇒ SM = PM. tan 60º …..(ii)

\Rightarrow ( h-\frac{1}{2})=( h-\frac{\sqrt{3}}{2})\sqrt{3}

[Using equations (i) and (ii)]

\Rightarrow h-\frac{1}{2}=h\sqrt{3}-\frac{3}{2}

⇒ h(√3 – 1) = 1

\Rightarrow h=\frac{1}{\sqrt{3}-1}

\Rightarrowh=\frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{\sqrt{3}+1}{2}

=\frac{2.732}{2}=1.336km

Hence, the height of the mountain is 1.366 km.

Concept: Heights and Distances
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