Question

the angle of elevation of the top of a vertical Tower from a point on the ground is 60 degree. From the another point 10 m vertically above the first it's angle of elevation is 30 degree find the height of the tower​

Answer 1

Answer:

15 m

Step-by-step explanation:

In △DBC:

$Tan 60^{o} =\frac{BC}{DC}\\\sqrt{3} =\frac{h+10}{b}\\b = \frac{h+10}{\sqrt{3}} --- (1)$

In △BAE:

$Tan 30^{o} =\frac{BE}{AE}\\\frac{1}{\sqrt{3}} =\frac{h}{b}\\b = \sqrt{3}h --- (2)$

After equating equations (1) & (2):

$\frac{h+10}{\sqrt{3}} = \sqrt{3}h\\h+10 = 3h\\10 = 2h\\h = 5$

Height of tower

= h + 10

= 5 +10

= 15 m