Question

From the top of a building, 15 m bigh, the angle of elevation of the top of a tower is found to
be 30°. From the bottom of the same building the angle of elevation of the top of the tower is
found to be 60°. Find the height of the tower and the distance between the tower and building.​

Answer 1

Step-by-step explanation:

In triangle ABC teta=30

tanteta =opp/hyp

tan30=AB/BC

$1 \div \sqrt{3 } = 15 \div bc$

$bc = 15 \sqrt{3}$

in triangle DBC teta =60

tan teta =opp/hyp

tan60=Dc/Bc

$\sqrt{3 } = dc \div 15 \sqrt{3}$

dc=15*3

dc=45m