Advertisement Remove all ads

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.

Advertisement Remove all ads

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.

  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×