From a point on hill-side of constant inclination. the angle of elevation of the top of a flagstaff on its summit is observed to be α and a meter nearer the top of the hill. it is β. If h is the height of the flagstaff, the inclination of the hill to the horizontal is

Engineering

Mathematics

problems on height and distance

Question

${cos}^{- 1} (\frac{a sin α sin β}{h sin (β - α)})$

${sin}^{- 1} (\frac{a sin α sin β}{h sin (β - α)})$

${tan}^{- 1} (\frac{a sin α sin β}{h sin (β - α)})$

none of these

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Let OP be the flagstaff oh height h and Q be the inclination of the hill.



Then,         ∠ORP = α, ∠OQP = β

and                RQ = a

∴              ∠RPQ = β - α

From ΔPQR, we have

$\frac{PQ}{sin α} = \frac{PR}{sin (180^{ο} - β)} = \frac{RQ}{sin (β - α)}$

∴ $∴ PR = \frac{a sin β}{sin (β - α)}$

Also,            $∠ PQR = \frac{π}{2} + θ$

So, in Δ POR

$\frac{PO}{sin α} = \frac{PR}{sin (\frac{π}{2} + θ)}$

∴                 $PR = \frac{h cos θ}{sin α}$

So, from Eqs. (i) and (ii), we have

                       $\frac{a sin β}{sin (β - α)} = \frac{h cos θ}{sin α}$

∴                          $cos θ = \frac{a sin α sin β}{h sin (β - α)}$

⇒                          $θ = {cos}^{- 1} [\frac{a sin α sin β}{h sin (β - α)}]$