Question

A radio transmitter antenna of height 100m stands at the top of tallbuilding .at a point on grounf angle of elevation of bottom of antenna is 45° and top of antenna is 60° what is the height of building

Answer 1

see diagram.

AB = antenna, OA = buildingtan 45 = 1 = OA/OC     => OA = OC
tan 60 = OB/OC = (OA+100)/OC=>       OA = 100 / (√3 - 1) m

Answer 2

Answer:

136.6025 m

Step-by-step explanation:

Refer the attached figure

Height of Antenna = CB = 100 m

Height of Building = AB

At a point on ground angle of elevation of bottom of antenna is 45° and top of antenna is 60°

In ΔABC

$tan \theta = \frac{Perpendicular}{Base}$

$tan 45^{\circ} = \frac{AB}{AD}$

$1 = \frac{AB}{AD}$

AB = AD ---1

In ΔACD

$tan \theta = \frac{Perpendicular}{Base}$

$tan 60^{\circ} = \frac{AC}{AD}$

$\sqrt{3} = \frac{100+AB}{AD}$

$\sqrt{3}AD = 100+AB$

Using 1

$\sqrt{3}AB - AB= 100$

$AB(\sqrt{3}-1)= 100$

$AB= \frac{100}{(\sqrt{3}-1)}$

$AB= 136.6025$

Hence the height of building is 136.6025 m