the angles of depression of two ships from the top of a light house are 45 and 30 towards east - Maths - Some Applications of Trigonometry

Question

the angles of depression of two ships from the top of a light
house are 45 and 30 towards east if the ship some 100 m apart find
the height of light house?

Ajanta Trivedi · Accepted Answer

let the height of the light house be AB= h m.

let the ships be at point C and D such that CD= 100 m

and the angle of depression from the light house to the ships C and D are 45 deg and 30 deg respectively.

$∠ P A D = ∠ A D B = 30 ∠ P A C = ∠ A C B = 45$

in the triangle ABC,

$\tan 45 = \frac{A B}{B C} 1 = \frac{h}{B C} B C = h m .......... (1)$

in the triangle ABD,

$\tan 30 = \frac{A B}{B D} \frac{1}{\sqrt{3}} = \frac{h}{B D} B D = \sqrt{3} h m ........... (2)$

since B, C and D lie on the same horizontal level.

BD=BC+CD

$\sqrt{3} h = h + 100 (\sqrt{3} - 1) h = 100 h = \frac{100}{\sqrt{3} - 1} m .$

rationalising the denominator we have,

$h = \frac{100}{\sqrt{3} - 1} * \frac{\sqrt{3} + 1}{\sqrt{3} + 1} h = \frac{100 (\sqrt{3} + 1)}{3 - 1} = \frac{100 (\sqrt{3} + 1)}{2} h = 50 (\sqrt{3} + 1) m .$

thus the height of the light house is $50 (\sqrt{3} + 1) m .$

hope this helps you.

the angles of depression of two ships from the top of a light house are 45 and 30 towards east.if the ship some 100 m apart find the height of light house?