Question

The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building.​

Answer 1

Step-by-step explanation:

$\huge\bf{\underline{\red{ANSWER}}}$

$tan \: 45\: = \frac{h - 50}{x} \\ x = h - 50 \\ \\ \tan \: 60 \: = \frac{h}{x} \\ x = \frac{h}{ \sqrt{3} } \\ h - 50 = \frac{h}{ \sqrt{3} } \\ \sqrt{3} h -50 \: \sqrt[]{3} = h \\ 3\sqrt{h } - h = 50\sqrt{3} \\ \\ h( \sqrt{3} - 1) = 50\sqrt{3} \\ h = \frac{ 50\sqrt{3} }{ \sqrt{3} - 1 } \\ h = \frac{ 50\sqrt{3}( \sqrt{3} + 1)}{3 - 1} \\ h = \frac{50(3 + \sqrt{3}) }{2} \\ \\ h = 75 + 25\sqrt{3} \\ h = 118.25m \\ \\ \\ \\ \tan45 = \frac{h - 50}{x} \\ x = h - 50 \\ \\ \tan60 = \frac{h}{x} \\ x = \frac{h}{ \sqrt{3} } \\ h - 50 = \frac{h}{ \sqrt{3} } \\ h = 75 + 25\sqrt{3} \\ 118.25m \\ \\ x = h - 50 \\ = 75 + 25 \sqrt{3} - 50 \\ = 25 + 25 \sqrt{3} = 25( {1} + \sqrt{3} ) \\ = 68.25m$

Height of the tower = 118.25 m

Distance between the tower and the building = 68.25 m

Answer 2

In ΔBTP,

tan 30° = TP/BP

1/√3 = TP/BP

BP = TP√3

In ΔGTR,

tan 60° = TR/GR

√3 = TR/GR

GR = TR/√3

As BP = GR

TP√3 = TR/√3

3 TP = TP + PR

2 TP = BG

TP = 50/2 m = 25 m

Now, TR = TP + PR

TR = (25 + 50) m

Height of tower = TR = 75 m

Distance between building and tower = GR = TR/√3

GR = 75/√3 m = 25√3 m