Question

Example 6. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height5m.From a point on the ground the angles of elevation of the top and bottom of the flagstaff are 60 and 30respectively.find the height of the tower and the distance of the point from the base of the tower​

Answer 1

Let the height of the tower be x m and distance between tower and point be y m

Solution-

Given that

Height of the flagstaff=5m

First In ∆TRP

Perpendicular (P)=x

Base (B)=y

$\rm we \: know \\ \rm tan \theta = \frac{p}{b} \\ \rm here \\ \rm \: tan30 = \frac{x}{y} \\ \implies \rm \frac{1}{ \sqrt{3} } = \frac{x}{y} \\ \implies \boxed{ \rm \: y = \sqrt{3}x }..1$

Now In ∆QRP,

Perpendicular(P)=QR=5+x

Base(B)=RP=y

$\rm\: we\:know \\ \rm tan \theta = \frac{p}{b} \\ \rm here \\ \rm tan60 = \frac{5 + x}{y} \\ \rm \implies \sqrt{3} = \frac{5 + x}{y} \\ \rm \implies \sqrt{3} y = 5 + x \\ \implies \rm \sqrt{3} \times \sqrt{3} x = 5 + x \: \: \: ......using \: 1 \\ \implies \rm 3x = 5 + x \implies \: 2x = 5 \\ \implies \boxed{ \rm \: x = \frac{5}{2}and \: y = \sqrt{3}x = \frac{5 \sqrt{3} }{2} }$

Therefore height of the tower is 5/2m and distance between point and base of tower is 5√3/2