Question

4. A pole 5m height is fixed on the top of a tower. The angle of elevation of the top
of the pole observed from a point A on the ground is 60° and the angle of
depression of the point A from the top of the tower is 45º. Find the height of the
tower.
Ans: 6.83m)​

Answer 1

Solution

From Pole to A

→ tan 60° = (5 + h)/x

→ √3 = (5 + h)/x

→ √3x = 5 + h

From Tower to A

→ tan 45° = h/x

→ 1 = h/x

→ x = h

By substituting values we get

→ √3x = 5 + x

→ x(√3 - 1) = 5

→ x = 5(√3 + 1)/2

→ x = (8.66 + 5)/2

→ x = 13.66/2

→ x = 6.83

Hence height of tower is 6.83 m.

Answer 2

*Refer the attachment for figure.

Assume : BD = h = height of the tower and BC = x.

GIVEN : AD = 5 m = height of pole.

FIND : Height of the tower.

Solution :

In ∆DBC

→ tan45° = $\dfrac{BD}{CB}$

→ 1 = $\dfrac{h}{x}$

→ x = h _________ (eq 1)

Similarly;

In ∆ABC

→ tan60° = $\dfrac{AB}{CB}$

→ √3 = $\dfrac{AD\:+\:BD}{CB}$

→ √3 = $\dfrac{5\:+\:h}{x}$

Cross-multiply them

→ x√3 = 5 + h

→ x√3 = 5 + x [From (eq 1)]

→ x√3 - x = 5

→ x(√3 - 1) = 5

→ x = $\dfrac{5}{\sqrt{3}\:-\:1}$

Rationalize it..

→ x = $\dfrac{5}{\sqrt{3}\:-\:1}$ × $\dfrac{\sqrt{3}\:+\:1}{\sqrt{3}\:+\:1}$

Now..

(a + b) (a - b) = a² - b²

→ x = $\dfrac{5(\sqrt{3}\:+\:1)}{(\sqrt{3})^{2}\:-\:(1)^{2}}$

→ x = $\dfrac{5\sqrt{3}\:+\:5}{3\:-\:1}$

→ x = $\dfrac{5\sqrt{3}\:+\:5}{2}$

→ x = $\dfrac{8.67\:+\:5}{2}$

→ x = $\dfrac{13.67}{2}$

→ x = 6.835 m

Put value of x in (eq 1)

→ h = 6.835 m

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Height of tower = 6.83 m

________ [ ANSWER ]

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