From a tower 126 m high, the angles of depression of two rocks which are in a horizontal line through the base of the tower are 16° and 12° 20'. Find the distance between the rocks if they are on

(i) the same side of the tower

(ii) the opposite sides of the tower.

(i) Let the rocks be at point A and D.

From figure a,

∠CAB = ∠ECA = 12° 20' (Alternate angles are equal)
∠CDB = ∠ECD = 16° (Alternate angles are equal)

Considering right angled △ABC, we get

$\Rightarrow \text{tan 12° 20}' = \dfrac{BC}{AB} \\[1em] \Rightarrow 0.2186 = \dfrac{126}{AB} \\[1em] \Rightarrow AB = \dfrac{126}{0.2186} = 576.29$

Considering right angled △BCD, we get

$\Rightarrow \text{tan 16°} = \dfrac{BC}{BD} \\[1em] \Rightarrow 0.2867 = \dfrac{126}{BD} \\[1em] \Rightarrow BD = \dfrac{126}{0.2867} = 439.48$

Distance between two rocks (AD) = AB - BD = 576.29 - 439.48 = 136.81

Hence, the distance between two rocks when they are on same side of the tower is 136.81 meters.

(ii) Let the rocks be at point A and B.

From figure b,

∠CAD = ∠XCA = 12° 20' (Alternate angles are equal)
∠CBD = ∠YCB = 16° (Alternate angles are equal)

Considering right angled △ADC, we get

$\Rightarrow \text{tan 12° 20}' = \dfrac{CD}{AD} \\[1em] \Rightarrow 0.2186 = \dfrac{126}{AD} \\[1em] \Rightarrow AD = \dfrac{126}{0.2186} = 576.29$

Considering right angled △BCD, we get

$\Rightarrow \text{tan 16°} = \dfrac{CD}{DB} \\[1em] \Rightarrow 0.2867 = \dfrac{126}{DB} \\[1em] \Rightarrow DB = \dfrac{126}{0.2867} = 439.48$

Distance between two rocks (AB) = AD + DB = 576.29 + 439.48 = 1015.7

Hence, the distance between two rocks when they are on opposite sides of the tower is 1015.7 meters.