From the top of a hill h metres high the angles of depression of the top and the bottom of a pillar are `alpha` and `beta` respectively. The height (in metres ) of the pillar is

Options

A. `(h(tan beta-tan alpha ))/(tan beta)`
B. `(h(tan alpha-tan beta))/(tan alpha )`
C. `(h (tan beta+tan alpha ))/(tan beta)`
D. `(h(tan beta+tan alpha ))/(tan alpha ) `

Answer

A

Explanation

Let AB be the pole of height H metres and Q be the top of the pillar. In `triangle's` APQ and BCQ, we have

tan `beta=(PQ)/(AP)` and `tan alpha =(QC)/(BC)`
`rArr tan beta=h/x` and `tan alpha =(h-H)/x`
`rArr (tan alpha )/(tan beta)=(h-H)/h`
`rArr H/h=(tan beta-tan alpha )/(tan beta) rArr H=(h(tan beta-tan alpha ))/(tan beta)`