Question

the angle of elevation of a cliff from a fixed point is theta after going up a distance of k meter towards the top of the Cliff at an angle of Beta , it is found that the angle of elevation is alpha show that the height of Cliff is $k( \cos \beta - \sin \beta \cot \alpha ) \: upon \: \cot \: theta - \cot \alpha \: \: meter$​

Answer 1

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

Let assume that AB = h be height of the cliff and let assume that C be the fixed point such that angle of elevation ∠AOB = θ

As it going up a distance of k metres ( OC ) towards the top of the cliff at angle β.

Let from point C the angle of elevation of the top of the Cliff is α.

From C, draw perpendicular CD and CE on AB and OA respectively.

Now, CE = AD and AE = CD

Consider, triangle AOB

$\rm \: cot\theta = \dfrac{AO}{AB}$

$\rm \: cot\theta = \dfrac{AO}{h}$

$\rm\implies \:\boxed{\tt{ \: AO = h \: cot \theta \: }}$

Now, In triangle OCE

$\rm \: sin \beta \: = \: \dfrac{CE}{OC}$

$\rm \: sin \beta \: = \: \dfrac{CE}{k}$

$\rm\implies \:\boxed{\tt{ CE \: = \: k \: sin \beta \: }}$

Also,

$\rm \: cos \beta \: = \: \dfrac{OE}{OC}$

$\rm \: cos \beta \: = \: \dfrac{OE}{k}$

$\rm\implies \:\boxed{\tt{ OE \: = \: k \: cos \beta \: }}$

Now,

$\rm \: CD = AE = OA - OE = h \: cot \theta - k \: cos \beta$

and

$\rm \: BD = AB - AD = AB - CE = h - k \: sin \beta$

Now, In triangle BCD

$\rm \: cot \alpha = \dfrac{CD}{BD}$

$\rm \: cot \alpha = \dfrac{h \: cot \theta - k \: cos \beta }{h - k \: sin \beta }$

$\rm \: cot \alpha (h - k \: sin \beta ) = h \: cot \theta - k \: cos \beta$

$\rm \: h \: cot \alpha - k \: sin \beta \: cot \alpha = h \: cot \theta - k \: cos \beta$

$\rm \: h \: cot \theta - h \: cot \alpha = k \: cos \beta - k \: sin \beta \: cot \alpha$

$\rm \: h \:( cot \theta - \: cot \alpha) = k (\: cos \beta - \: sin \beta \: cot \alpha )$

$\rm\implies \:h = \dfrac{k (\: cos \beta - \: sin \beta \: cot \alpha )}{cot \theta - \: cot \alpha} \: metres$

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ADDITIONAL INFORMATION

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

Answer 2

Given :-

The angle of elevation of a cliff from a fixed point is theta after going up a distance of k meter towards the top of the Cliff at an angle of beta,

To Prove :-

$\sf \dfrac{k(\cos\beta-\sin\beta \cot\alpha)}{\cot\theta-\cot\alpha}\;m$

Solution :-

Given

Angle of elevation = α

Let us assume

Height = h = CE

Distance covered = AF = k meter

In ΔABF

We know that

cosθ = Perp/Hyp

cos β = AB/AF

cos β = AB/k

cosβ.k = AB (1)

Also,

sin β = opp/hyp

sin β = BF/AF

sin β = BF/k

sin β.k = BF (2)

In Δ ACE

We know that

tanθ = Opp/Adj

tanθ = CE/AC

tanθ = h/AC

1/cot θ = h/AC (tanθ = 1/cotθ)

AC = cot θ. h

Now

A/Q

FD = BC

Finding DF

DF + AB = AC

DF = AC - AB

From 1

DF = AC - cosβ. k

From 3

DF = cotθ. h - cosβ. k (iv)

Now

Finding DE

DE + DC = CE

DE = CE - DC

DE = h - BF (DC = BF)

DE = h - sin β. k (v)

In ΔDEF

tanα = DE/DF

As we know that

tan α = 1/cot α

1/cot α = DE/DF

1/cot α = h - sinβ.k/cotθ. h - cosβ. k

cotθ. h - cosβ. k = cot α (h - sinβ.k)

cotθ. h - cosβ. k = h cotα - sinβ. k cot α

h = k(cosβ - sinβ cotα)/cotθ - cotα