Question

From a fixed point the angle of evation is theta.At phi angle after walking k km it is found that angle is alpha.Prove height is k(cosphi-sinphicotalpha/cottheta-cotalpha​

Answer 1

Step-by-step explanation:

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}$

Let us take that AB is the height and P is the fixed point.

Let us take AB = h.

By question

PQ=k km

/_APB = $\theta$

/_APB = $\phi$

/_AQR=$\alpha$

Now, in ∆ PQS,

cos$\phi$ =$\dfrac{PS}{PQ}$

=> PS=PQcos$\phi$

and

sin$\phi$=$\dfrac{QS}{PQ}$

=>PQsin$\phi$ =QS

=>QS = ksin$\phi$

Again in ∆ABP,

=>cot$\theta$=$\dfrac{PB}{AB}$

=>PB=hcot$\theta$

Now,

QR = SB=PB-PS= hcot$\theta$- kcos$\phi$

(just proved)

AR=AB-RB=h -QS = h -ksin$\phi$

Now, in ∆ AQR,

=>cot$\alpha$=$\dfrac{QR}{AR}$

=> hcot$\theta$- kcos$\phi$= hcot$\alpha$-ksin$\phi$cot$\alpha$

=>hcot$\theta$-hcot$\alpha$=kcos$\phi$-ksin$\phi$cot$\alpha$

=>h(cot$\theta$-cot$\alpha$)=k(cos$\phi$-sin$\phi$cot$\alpha$)

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

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