Question

A man on the deck of a ship is 10 m above the water level he observed that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.​

Answer 1

Answer:

5° is the answer

mark me as brainliest

Answer 2

$answer$

Let AB be the deck of the ship. AB=10m

Let AB be the deck of the ship. AB=10mSuppose CE be the cliff. C and E are the top and bottom of the cliff.

CD=AB=10m

∴ DE=CE−CD=(h−10)m

$⇒  In △ADE,tan45 \: degree = \frac{de}{ad}$

$⇒  1= \: \frac{h - 10}{ ad}$

∴ AD=(h−10)m ------- ( 1 )

$⇒  In △ADC,tan30degree = \frac{cd}{ad}$

$⇒ \frac{1}{ \sqrt{3 \: } } = \frac{10}{ad}$

$therefore$

$ad = 10 \sqrt{3m \: } - - (2)$

$( From 1 and 2)</u></p><p></p><p></p><p><u>[tex]( From 1 and 2)$

$⇒  h−10=10 \: \sqrt{3}$

$⇒  h=10( \sqrt{3} + 1) = 10(1.732+1)=27.32m</u></p><p></p><p></p><p></p><p><u>[tex]⇒  h=10( \sqrt{3} + 1) = 10(1.732+1)=27.32m$

⇒ Height of the cliff is 27.32m and distance of cliff from the ship BC=AD=17.3m.

:)