Question

$\huge{\underline{\underline{\bf Question}}}$

the angle of elevation of a cliff from a fixed point is [te\sf theta}[/tex] after going up a distance of ke metres towards the top of the Cliff at an angle of it is found that the angle of elevation is alpha show that the height of the Cliff :-

$\sf k \dfrac{( \cos(ϕ) - \sin(ϕ)) }{ \cot( \theta) - \cot( \alpha ) }$

Class - 10th

Chapter - Applications of Trigonometry

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Answer 1

$\sf\underline \pink{ Given\:that}$

• Angle of elevation of cliff = θ
• Distance = k meters
• Angle = a

$\sf\underline \red{ To\:Prove }$

height of cliff is

$\dfrac{k(\cos\phi-\sin\phi\cot\alpha)}{\cot\theta-\cot\alpha}$

$\sf\underline \purple{ step-by-step\:solution }$

• Found that the angle of elevation = α
• Let CE be the cliff of height .
• Angle of elevation of cliff from a fixed point A be θ

USING DIAGRAM-

In right angle ABF,

• → $\cos\phi=\dfrac{AB}{AF}$

Put the value of AF

• → $\cos\phi=\dfrac{AB}{k}$
• → $AB=k\cos\phi$------(I)

Now,

• → $\sin\phi=\dfrac{BF}{AF}$

Put the value of AF

• → $\sin\phi=\dfrac{BF}{k}$
• → $BF=k\sin\phi$

In right triangle ACE,

• → $\tan\theta=\dfrac{CE}{AC}$

Put the value of CE

• →$\tan\theta=\dfrac{h}{AC}$
• →$AC=\dfrac{h}{\tan\theta}$
• → $AC=h\cot\theta$--------(ll)

According to figure,

• FD=BC

We need to calculate the value of DF

Using equation (I) and (II)

• $DF=AC-AB$

Put the value of AB and AC

• $DF=h\cot\theta-k\cos\phi$....(III)

According to figure,

• DC=FB

We need to calculate the DE

Using diagram

• → $DE=EC-DC$
• → $DE=EC-FB$

Put the value into the formula

• → $DE=h-k\sin\phi$-----(IV)

In triangle DEF,

• → $\tan\alpha=\dfrac{ED}{FD}$

Put the value of ED and FD

• $\tan\alpha=\dfrac{h-k\sin\phi}{h\cot\theta-k\cos\phi}$  

• $\dfrac{1}{\cot\alpha}=\dfrac{h-k\sin\phi}{h\cot\theta-k\cos\phi}$  

• $h\cot\theta-k\cos\phi=h\cot\alpha-k\sin\phi\cot\alpha$

• $h\cot\theta-h\cot\alpha=k\cos\phi-k\sin\phi\cot\alpha$

• $h=\dfrac{k(\cos\phi-\sin\phi\cot\alpha)}{(\cot\theta-\cot\alpha)}$

Hence, Proved the height of cliff is $\dfrac{k(\cos\phi-\sin\phi\cot\alpha)}{\cot\theta-\cot\alpha}$

Answer 2

EXPLANATION.

The angle of elevation of a cliff from a fixed point = ∅.

Going up a distance of k meters towards the top of the cliff at an angle = Ф.

The angle of elevation is = α.

As we know that,

⇒ In ΔROT, we get.

⇒ sinФ = RT/OT.

⇒ sinθ = perpendicular/Hypotenuse. = p/h.

⇒ sinΦ = RT/k.

⇒ k sinФ = RT ----------------(1).

⇒ cosθ = Base/Hypotenuse = B/H.

⇒ cosΦ = OT/OR.

⇒ cosΦ = OT/k.

⇒ k cosΦ = OT. ---------------(2).

As we can see in figures, we get.

⇒ RT = PS.

⇒ k sinΦ = PS.

⇒ In ΔPOQ, we get.

⇒ tanθ = perpendicular/base = p/b.

⇒ tanθ = PQ/OP.

⇒ tanθ = h/OP.

⇒ OP = h/tanθ.

⇒ OP = h cotθ.  ----------------(3).

⇒ In ΔQRS, we get.

⇒ tanθ = perpendicular/base. = p/b.

⇒ tanα = QS/RS.

First we find QS & RS.

As we know that,

We write QS = PQ - PS.

⇒ RT = PS [ Equal ].

⇒ QS = h - k sinΦ. ---------------(4).

As we know that,

We write RS = TP [ Equal ].

⇒ RS = OP - OT.

⇒ RS = h cotθ - k cosΦ. -------------(5).

Put the value in equation, we get.

⇒ tanα = QS/RS.

⇒ tanα = h - k sinΦ/h cotθ - k cosΦ.

⇒ 1/cotα = h - k sinΦ/h cotθ - k cosΦ.

⇒ h cotθ - k cosΦ = cotα(h - k sinΦ).

⇒ h cotθ - k cosΦ = h cotα - k sinΦ.cotα.

⇒ h cotθ - h cotα = k cosΦ - k sinΦ.cotα.

⇒ h(cotθ - cotα) = k(cosΦ - sinΦ.cotα).

⇒ h = k(cosΦ - sinΦ.cotα)/(cotθ - cotα).

HENCE PROVED.