Question

At the foot of mountain the elevation of its summit is 45°. After ascending 1000m
Towards the mountain up a slope of 30° inclination the elevation is found to be 60°.Find the height of the mountain.

Answer 1

Answer:

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

Then ang.OFS=45° and therefore

ang.OSF=45°.

Consequently, OF=OS=h km....(say)

Let FP=1000m=1km be the slope so that ang.OFP=30°.

Draw PM perpendicular to OS &

PL perpendicular to OF.

Join PS, it is given that ang.MPS=60°.

In Δ FPL we have,

sin 30°= PL/PF

$\\ = > PL = PF\: = sin \: 30 \: deg. = (1 \times \frac{1}{2} km) = \frac{1}{2} km \\ \\ OM = PL = \frac{1}{2} km \\ \\ = > MS = OS - OM = (h - \frac{1}{2})km........1$

Also,

cos 30°=FL/PF

$= > FL = PF \: cos \: 30 \: deg. = (1 \times \frac{ \sqrt[]{3} }{2}) km = \frac{ \sqrt{3} }{2} km \\ \\ h = OS = OF = OL + LF \\ \\ = > h = OL + \frac{ \sqrt{3} }{2} \\ \\ OL = (h - \frac{ \sqrt{3} }{2} )km \\ \\ PM = (h - \frac{ \sqrt{3} }{2})km.........2$

In Δ SPM we have,

tan 60°=SM/PM.

$= > SM = PM .tan \: 60 \: deg. \\ \\ = > (h - \frac{1}{2} ) = (h - \frac{ \sqrt{3} }{2} ) \sqrt{3} .........using \: (1) \: and \: (2) \\ \\ = > h - \frac{1}{2} = h \sqrt{3} - \frac{3}{2} \\ \\ = > \sqrt{3} h - h = \frac{3}{2} - \frac{1}{2} \\ \\ = > h( \sqrt{3} - 1) = 1 \\ \\ = > h = \frac{1}{ \sqrt{3 - 1} } \\ \\ = > h = \frac{ \sqrt{3} + 1}{( \sqrt{3 - 1}) (\sqrt{3 + 1}) } \\ \\ h = \frac{ \sqrt{3} + 1}{2} \\ \\ h = \frac{2.732}{2} \\ \\ h = 1.366km$

Therefore,

height of the mountain is 1.366km.