Question

From the top of a light house 60m height with its base at the sea level, the angle of depression of a boat is 15. The distance of the boat from the foot of the light house is​

Answer 1

Consider any right angle triangle ABC. where AC is the perpendicular height of the light house and AB is the base and the distance between the foot of the light house and the boat. From C, draw a straight line || to AB and Mark a point D on it (horizontal) and the angle between CB AND CD is the angle of depression.

Given :-

Height of the tower = AC = 60m

Angle of depression = angle BCD = 15°

Therefore angle ABC = 15° (corresponding angles)

Now,

AC/AB = tan15°

➡ 60/AB = 2 - √3

➡ 60 = AB(2 - √3)

➡ AB = 60/(2 - √3)

By rationalizing, we get

= 60/(2 - √3) × (2 + √3)/(2 + √3)

= [60(2 + √3)]/[(2)² - (√3)²]

= 120 + 60√3

= 60(2 + 1√3) m

Therefore AB = 60(2 + √3)m and Hence, the distance between the boat and foot of the light house is 60(2 + √3)m

Answer 2

$\huge\text{Solution}$

Given:

From the top of a light house 60m height with its base at the sea level, the angle of depression of a boat is 15.

Find:

Finf the distance of the boat from the foot of the light house.

Answer:

$\bf \tan(15) ° = \frac{AC}{AB}$

$\bf = \frac{60}{AB} = 2 - \sqrt{3} \\ \bf = ab = \frac{60}{2} - \sqrt{3}$

By Rationalizing:

$\bf = \frac{60}{(2 - \sqrt{3) } } \times \frac{(2 + \sqrt{3)} }{(2 + \sqrt{3)} } \\ \bf = \frac{60(2 + \sqrt{3)} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} } \\ \bf = 120 + 60 \sqrt{3}$

$\sf = 60(2 + 1 \sqrt{3} )m$

60(2 + √3) m the distance between the boat and foot of the light house.