Question

The elevation of the summit of a mountain from its foot is 45°. After ascending 2 km towards the mountain upon an incline of 30°,the elevation changes to 60°. What is the approximate height of the mountain?

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that, A be the foot and C be the top of mountain and let height of the mountain BC be h metres.

Now, ∠BAC = 45°

So, ∠ACB = 45°

This, implies AB = BC = h metres.

[ Side opposite to equal angles are equal ]

Now, After ascending 2 km towards the mountain upon an incline of 30°, the elevation changes to 60°.

Let AD = 2 km = 2000 m

Also, ∠DAB = 30° and ∠EDC = 60°.

Now, From D, drop DF and DE perpendiculars on side AB and BC.

Now, In triangle ADF

$\rm \: sin30 \degree \: = \: \dfrac{DF}{AD}$

$\rm \: \dfrac{1}{2} \: = \: \dfrac{DF}{2000}$

$\rm\implies \:DF = 1000 \: m$

$\rm\implies \:DF = BE = 1000 \: m$

Now, In triangle ADF

$\rm \: cos30 \degree = \dfrac{AF}{AD}$

$\rm \: \dfrac{ \sqrt{3} }{2} \degree = \dfrac{AF}{2000}$

$\rm\implies \:AF = 1000 \sqrt{3} \: m$

Now,

$\rm \: DE = BF = AB - AF = h - 1000 \sqrt{3} \: m$

Also,

$\rm \: CE = BC - BE = BC - DF = h - 1000 \: m$

Now, In triangle DCE

$\rm \: tan60 \degree = \dfrac{CE}{DE}$

$\rm \: \sqrt{3} = \dfrac{h - 1000}{h - 1000 \sqrt{3} }$

$\rm \: \sqrt{3}h - 3000 = h - 1000$

$\rm \: \sqrt{3}h - h = 3000 - 1000$

$\rm \: (\sqrt{3} - 1)h = 2000$

$\rm \: h = \dfrac{2000}{ \sqrt{3} - 1}$

So, on rationalizing the denominator, we have

$\rm \: h = \dfrac{2000}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1 }{ \sqrt{3} + 1}$

We know,

$\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}$

So, using this, we get

$\rm \: h = \dfrac{2000( \sqrt{3} + 1)}{ (\sqrt{3})^{2} - 1}$

$\rm \: h = \dfrac{2000( \sqrt{3} + 1)}{ 3 - 1}$

$\rm \: h = \dfrac{2000( \sqrt{3} + 1)}{2}$

$\rm\implies \:h = 1000( \sqrt{3} + 1) \: m \: or \: \sqrt{3} + 1 \: km$

$\rm\implies \:h = 1.732 + 1 = 2.732 \: km \: (approx)$

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Solution:-

Let A be the foot and B be the summit of the mountain OB at height h.

In △AOB,

tan 45°=$\sf \dfrac{OB}{OA}$

1 =$\sf \dfrac{OB}{OA}$

OA=h

In △ACP,

sin 30°=$\sf \dfrac{PC}{AP}$

⇢$\sf \dfrac{1}{2}$ $\sf \dfrac{PC}{2}$

⇢PC=1

Also,

cos 30°=$\sf \dfrac{AC}{AP}$

⇢$\sf \dfrac{√3}{2}$ =$\sf \dfrac{AC}{2}$

⇢AC=√3

Now,

h=OA

⇢h=AC+OC

⇢h=OC+√3

⇢OC=h - √3

⇢PD=h−√3

Now, in △PDB,

tan 60°=$\sf \dfrac{BD}{PD}$

⇢√3=$\sf \dfrac{h-1}{h-√3}$

⇢√3h - 3=h-1

⇢h=$\sf \dfrac{2}{√3 - 1}$

⇒h= √3 +1

⇒h=2.732 km

Hence, the height of the mountain is 2.732 km...