Question

the angle of elevation of cliff from a fixed point is theta .After going up a distance of k metres towards the top of the cliff at an angle of a,it is found that the angle of elevation is alpha.show that the height iof cliff is

⚓⚓⚓⚓dont spamming ⚓⚓

⭐⭐⭐class 10⭐⭐⭐​

Answer 1

Given that,

Angle of elevation of cliff = θ

Distance = k meters

Angle = a

Found that the angle of elevation = α

Let CE be the cliff of height h m.

Angle of elevation of cliff from a fixed point A be θ

We need to prove the height of cliff is $\dfrac{k(\cos\phi-\sin\phi\cot\alpha)}{\cot\theta-\cot\alpha}$

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

$\tan\theta=\dfrac{h}{AC}$

$AC=\dfrac{h}{\tan\theta}$

$AC=h\cot\theta$....(II)

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

Open the above image to see the detailed answer