Question

at the foot of a 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

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

$\qquad\dashrightarrow~ \sf \angle OFS=45\degree$

$\qquad\dashrightarrow~\sf \angle OSF =45\degree$

$\qquad\sf \dashrightarrow~\angle MPS=60\degree$

$\:$

☀ Let's say OF = OS = h km. Let FP = 1km be the slope so that :

$\sf\qquad\dashrightarrow~ \angle OFP=30\degree$

$\:$

☀ Draw :

$\qquad\sf \dashrightarrow~ PM\perp OS$

$\qquad\sf \dashrightarrow~ PL\perp OF$

Join PS.

☀ In ∆FPL :

$\sf\qquad\dashrightarrow ~ sin~30\degree=\dfrac{PL}{PF}$

$\sf\qquad\dashrightarrow~ PL=PF~ sin~30\degree$

$\sf \qquad \dashrightarrow \: \bigg \lgroup1 + \dfrac{1}{2} \bigg \rgroup \: km$

$\pmb{\sf\qquad\dashrightarrow~ OM=PL=\dfrac{1}{2}~km}$

$\ \:$

☀ Finding MS :

$\sf\qquad\dashrightarrow ~MS=OS-OM$

$\pmb{\sf\qquad\dashrightarrow~ MS=\bigg\lgroup h-\dfrac{1}{2}\bigg\rgroup~km~~~. ~.~.~.~.~.~(I)}$

$\:$

☀ Also, we have :

$\sf\qquad\dashrightarrow~ cos~ 30\degree=\dfrac{FL}{PF}$

$\sf\qquad\dashrightarrow ~ FL=PF~cos~30\degree$

$\sf \qquad \dashrightarrow \: \bigg \lgroup1 \times \dfrac{ \sqrt{3} }{2} \bigg \rgroup \: km$

$\qquad\pmb{\sf\dashrightarrow~ FL=\dfrac{\sqrt{3}}{2}~km}$

$\:$

☀ Now, we have :

$\sf\qquad\dashrightarrow~ h=OS=OF=OL+LF$

$\sf\qquad\dashrightarrow~ h=OL+\dfrac{\sqrt{3}}{2}$

$\sf\qquad\dashrightarrow~ OL=\bigg\lgroup h-\dfrac{\sqrt{3}}{2}\bigg\rgroup~km$

$\sf\qquad\dashrightarrow~ PM=\bigg\lgroup h-\dfrac{\sqrt{3}}{2}\bigg\rgroup~km$

$\:$

☀ In ∆SPM :

$\sf\qquad\dashrightarrow~ tan~60\degree=\dfrac{SM}{PM}$

$\qquad\pmb{\sf \dashrightarrow~ SM=PM~tan~60\degree}$

$\:$

☀ Finding the value of ‘h’ :

$\sf \qquad \dashrightarrow \: \bigg \lgroup h - \dfrac{1}{2} \bigg \rgroup = \bigg \lgroup h - \dfrac{ \sqrt{3} }{2} \bigg \rgroup \sqrt{3}$

$\sf \qquad \dashrightarrow \: h - \dfrac{1}{2 } = h\sqrt{3} - \dfrac{3}{2}$

$\sf \qquad \dashrightarrow \: \sqrt{3} h - h = \dfrac{3}{2} - \dfrac{1}{2}$

$\qquad \sf \dashrightarrow \: h( \sqrt{3} - 1) = 1$

$\sf \qquad \dashrightarrow \: h = \dfrac{1}{ \sqrt{3} - 1 }$

$\qquad \sf \dashrightarrow \: h = \dfrac{ \sqrt{3} + 1 }{( \sqrt{3} - 1)( \sqrt{3} + 1) }$

$\qquad \sf \dashrightarrow \: h = \dfrac{ \sqrt{3} + 1}{2}$

$\pmb{ \qquad \sf \dashrightarrow \: h = 1.36 \: km}$

$\:$

☀ Therefore, the height of the mountain is 1.36km.