Question

The angle of depression from the top of a tree to the two points on the same side of the tree at the distances of 9 m and 16 m from its base are complementary. Find the height of the tree.

Answer 1

★ Figure refer to attachment

GIVEN

The angle of depression from the top of a tree to the two points on the same side of the tree at the distances of 9 m and 16 m from its base are complementary.

TO FIND

Find the height of the tree.

SOLUTION

• CD = 9m
• BD = 16m
• AB = Height of tree = ?

★ Let ∠ADC = θ

★ ∠ACB = (90° - θ) {∠ADC and ∠ACB are complementary}

In ∆ ADC

→ tan θ = AB/BD

→ tan θ = AB/9 ---(i)

In ∆ ACB

→ tan (90° - θ) = AB/BC

→ cot θ = AB/16 ----(ii)

Multiply (i) and (ii)

→ AB/9 × AB/16 = tan θ × cot θ

→ AB²/144 = 1 {tan θ × cot θ = 1}

→ AB² = 144

→ AB = √144 = 12

Hence, the height of the tree is 12m

Answer 2

Given:-

• The angle of depression from the top of a tree to the two points on the same side of the tree at the distances of 9 m and 16 m from its base are complementary.

To find:-

• Height of tree.

Solution:-

• BC :- 9m
• BD :- 16m
• CD :- 16 - 9 = 7m

★ ∠ADC and ∠ACB are complimentary

∠ADC + ∠ACB = 90°

Let's ∠ADC = $\sf{ \theta}$

So, ∠ACB = (90 - $\sf{ \theta}$ )

✧ In ∆ ABC :-

$\implies \sf{ \tan{ \theta} = \dfrac{AB}{BC}}$

$\implies \sf{ \tan{ \theta} = \dfrac{AB}{9}}$ --(1)

✧ In ∆ ABD :-

$\implies \sf{ \tan{90 - \theta} = \dfrac{AB}{BD}}$

$\implies \sf{ \tan{90 - \theta} = \dfrac{AB}{16}}$

$\implies \sf{ \cot{ \theta} = \dfrac{AB}{16}}$ --(2)

★ Multiplying eq.(1) and (2) →

$\sf{\tan{ \theta} \times \cot{ \theta} = \dfrac{AB}{9} \times \dfrac{AB}{16}}$

$\sf{\tan{ \theta} \times \dfrac{1}{\tan{ \theta}} = \dfrac{AB^2}{144}}$

$\sf{ 144 = AB^2}$

★ Taking sqrt. both side →

$\sf{ \sqrt{144} = \sqrt{AB^2}}$

$\sf{ AB = 12}$

✧ Hence, Height of tree is 12m.

________________________