Question

At the foot of a mountain, the angle of elevation of its summit is `45^(@)`. After ascending 1 km towards the mountain up an incline of `30^(@)`, the elevatioon changes to `60^(@)` (as shown in the given figure). Find the height of the mountain. [Given: `sqrt3=1.73.`]

Answer 1

Correct Answer - `1.365 km`
`(SD)/(FD)=tan45^(@)=1rArrSD=FD=hkm`
`rArr" "/_FSD=/_SFD=45^(@)`
`/_ASC=180^(@)-(60^(@)+90^(@))=30^(@)`
`rArr/_FSA=(45^(@)-30^(@))=15^(@)and/_SFA=15^(@).`
By the sine formula on `DeltaSAF`, we have
`(AF)/(sin/_FSA)=(FS)/(sin/_FAS)rArr1/(sin15^(@))=(sqrt2h)/(sin15^(@))=(sqrt2h)/(sin150^(@))rArr(2sqrt2)/((sqrt3-1))=2sqrt2h[becauseFS=sqrt(h^(2)+h^(2))=sqrt2h"and sin"15^(@)=((sqrt3-1))/(2sqrt2)]`
`rArrh=1/((sqrt3-1))xx((sqrt3+1))/((sqrt3+1))=((sqrt3+1))/2=((1.73+1))/2=(2.73)/2=1.365km.`