Question

The angle of elevation of the top of a tower from a point on the ground is 45° . On walking 30m towards,the tower, the angle of elevation became 60° . Find the height of a tower and the original distance from the foot of the tower. (Root 3 = 1.73)

Answer 1

Given

• ∠ADB = 60°
• ∠ACB = 45°
• CD = 30m
• ABD = 90°

Let the height of tower(AB) be xm & distance between B and D be ym.

To Find

• Height of tower(x) = ?
• Distance (y) = ?

Solution

✏ In ∆ABC,

• Perpendicular = AB
• Base = BC

$\bf \implies tan \: 45° = \frac{perpendicular}{base} \\ \\ \tt \implies tan \: 45 = \frac{AB}{BC} \\ \\ \tt \implies 1= \frac{x}{(y + 30)} m \\ \\ \tt \implies \: x = y + 30 \\ \\ \tt \implies y = x - 30......eq(1)$

✏ In ∆ABD,

• Perpendicular = AB
• Base = BD

$\bf \implies tan \: 60° = \dfrac{perpendicular}{base} \\ \\ \tt \implies \: tan \: 60 = \dfrac{AB}{BD} \\ \\ \tt \implies \: 3 = \dfrac{x}{y} \\ \\ \tt \implies3 = \dfrac{x}{x - 30}......(From \: eq(1) \\ \\ \tt \implies x = \sqrt{3}x - 30 \sqrt{3} \\ \\ \tt \implies \sqrt{3} x - x = 30 \sqrt{3} \\ \\ \tt \implies x = \dfrac{30 \sqrt{3} }{ \sqrt{3} - 1} \\ \\ \tt \implies x = \dfrac{30 \sqrt{3} }{ \sqrt{3 } - 1} \times \dfrac{ \sqrt{3 } + 1}{ \sqrt{3} + 1 } \\ \\ \tt \implies x = \dfrac{30 \sqrt{3}( \sqrt{3} + 1)}{2} \\ \\\tt \implies x = \dfrac{30(1.73)(1.73 + 1)}{2} \\ \\ \tt \implies x = \dfrac{51.9 \times 273}{2}$

$\large \implies \boxed {\boxed {\tt \blue {x = 70.95m }}}$

➣ Height of tower = 70.95m

➣ Origin distance = BC = y + CD =(x-30)+30 = x = 70.95m

Answer 2

Correct question= The angle of elevation of the top of a tower from a point on the ground is 45° . On walking 30m towards,the tower, the angle of elevation became 60° . Find the height of a tower and the original distance from the foot of the tower. (Root 3 = 1.73)

Solution⬇️

Let tower ne AB

Let point be C

Distance of point C from foot of tower= 30m

Hence, BC = 30m

Angle of elevation= 30°

So AB= 30°

Since tower is vertical,

ABC= 90°