Question

From the top of a rock which rises vertically 100m out of the water, the angle of depression of the

boat is 30°

. Find the distance of the boat from the base of the rock. [Take√3 = 1.732]
​

Answer 1

Given:

From the top of a rock which rises vertically 100m out of the water, the angle of depression of the  the boat is 30°

To find:

The distance of the boat from the base of the rock

Solution:

To solve the above-given problem we will use the following trigonometric ratio of a triangle:

$\boxed{\bold{tan\: \theta = \frac{Perpendicular}{Base} }}$

Referring to the figure attached below we get,

"AB" = the height of the vertical rock = 100 m

"BC" = the distance of the boat from the base of the rock

"θ" = "∠DAC" = "∠ACB" = the angle of depression = 30°

Now, considering ΔABC, we have

AB = perpendicular

BC = base

θ = 30°

∴ $tan\: 30\° = \frac{Perpendicular\: Height \:of\:the\:rock}{Distance \:between\:boat\:\& \:base \:of \:rock} }}$

$\implies tan\: 30\° = \frac{AB}{BC} }}$

substituting AB = 100 m

$\implies \frac{1}{\sqrt{3} } = \frac{100}{BC} }}$

$\implies BC = \sqrt{3} \times 100$

taking √3 = 1.732

$\implies BC = 1.732 \times 100$

$\implies \bold{BC = 173.2\:m}$

Thus, the distance of the boat from the base of the rock is → 173.2 m.

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Answer 2

Answer:

173.2m is the answer of the question