Question

A man standing on the deck of a ship is 12 m above water levels he observed that the angle of elevation of the top of a cliff is 45 degree and the angle of depression of the base of a quick is a 30 degree then calculate the difference distance of the field from the ship and the height of the cliff

Answer 1

in triangle ABC ,

$\cot {30}^{0} = \frac{bc}{ab} \\ = > \sqrt{3} = \frac{x}{12} \\ = > x = 12 \sqrt{3} \\ \: here \: bc = ad$

in triangle ADE,

$\frac{ed}{ad} = \tan {45}^{0} \\ \frac{y}{12 \sqrt{3} } = 1 \\ y = 12 \sqrt{3}$

height of the cliff is

$= 12 + 12 \sqrt{3}$

distance between height of cliff and and distance of field is

$= 12 - 12 \sqrt{3 } - 12 \sqrt{3} \\ = > 12m$

$mark \: as \: brainlist$