Question

at the foot of the mountain the elevation of its summit is 45 degree after ascending 1000m towards the mountain up a slope of 30 degree inclination the elevation is 60 degree find the ht of the mountain
use the diagram provided√√√

Answer 1

Here your answer goes

Height of the mountain = AB = H

In ΔACB ,

tan( ABC ) = $\frac{AC}{CB}$

tan 45° = $\frac{AC}{CB}$

1 = $\frac{AC}{CB}$

CB = AC = H

In ΔFCD ,

sin 30° = $\frac{FD}{FC}$

$\frac{1}{2} =\frac{y}{1000}$

$y = 1000 *(\frac{1}{2})=500m$

⇒ $Thus, AE = AB - BE= (H-500)m$

Cos 30° = $\frac{CD}{FD}$

⇒ √3 / 2= x/1000

⇒ DC =  1000×(√3 / 2)

⇒ $DC =500$√3m

$Thus , AE = AB - BE= (H-500\sqrt{3})m\\Thus ,FE = (H-500\sqrt{3})m(since DE =FE )$

In ΔAFE

$tan 60 = \frac{AE}{FE}$

$\sqrt{3} = \frac{H-500}{H - 500\sqrt{3}}$

$\sqrt{3} (H-500\sqrt{3} ) = H -500 \\\\\sqrt{3}H - 1500 = H - 500 \\\\\sqrt{3} H - H = -500 +1500\\\\ H (\sqrt{3} -1 ) = 1000\\\\H = \frac{1000}{\sqrt{3} -1 }\\\\H = \frac{1000\sqrt{3}+1}{2}\\\\H = 500(\sqrt{3} + 1 )m$

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$Be$ $Brainly$

Together We go far →