Question

A man standing on a deck of a ship which is 10m above the water level observes the angle of elevation of the top of the hill as 60 and the angle of depression as 30 find the distance of the hill from the ship and the height of the hill

Answer 1

$\textbf{Answer is in Attachment !}$

Answer 2

Answer:

the height of the hill is 10+30=40

the distance between the ship and the hill is  17.32

Step-by-step explanation:

let AC be the height of the hill

DE=10m

∠ADB=60°,∠BDC=30°

inΔBDC

tan60°=$\frac{AB}{DB}$

    $\sqrt{3} \frac}$ =$\frac{h}{x}$

    x=$\frac{h}{\sqrt{3} }$

in ΔBDC

tan30°=$\frac{BC}{DB}$

$\frac{1}{\sqrt{3} }$=$\frac{10}{x}$

$\frac{h}{\sqrt{3} }/=10\sqrt{3}$

h=10×3

h=30

x=$\frac{30}{\sqrt{3} }$×$\frac{\sqrt{3} }{\sqrt{3} }$

=$\frac{30\sqrt{3} }3}$

17.32

the height of the hill is 10+30=40

the distance between the ship and the hill is  17.32