Question

the angle of depression from the top of a vertical tower to a point on the ground is found to be 60°and from a point 50m above the foot of the tower angle of depression to the same point to be 30° find the height of the tower​

Answer 1

tan60° = \frac{(h+10)}{AB} ( in ∆

ABC)

AB

(h+10)

AB = \frac{(h+10)}{\sqrt{3}} ...(i)AB=

3

(h+10)

...(i)

In △DCO,

tan45° = \frac{h}{ab}

ab

h

∴ AB = h⋯(ii)

comparing eq.(i) and (ii), we get :

h = \frac{(h+10)}{\sqrt{3} }

3

(h+10)

∴ \sqrt{3}h

3

h = h+10

⇒ \sqrt{3}h

3

h - h = 10

h \sqrt{3-1} = 10

3−1

=10

∴ h = \frac{10}{\sqrt{3-1} }

3−1

10

Height of tower = h + 10

= \frac{10}{\sqrt{3-1} }

3−1

10

+10

= \frac{10+10\sqrt{3}-10}{\sqrt{3}-1 }

3

−1

10+10

3

−10

H = \frac{10 \sqrt{3}}{\sqrt{3} -1}

3

−1

10

3

m