Question

From the top of a tower 60m high , the angle of depression of the top and the bottom of a vertical post are observed to br 30 and 60 respectively . Find the height of the post and horizontal distance between the tower and post .
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Answer 1

Understanding the question :-

Here this is a question related to the application of trigonometry. We have to find the horizontal distance between the post & tower and height of the post as mentioned in the question. As it is a question of application of trigonometry, we have to use trigonometric ratios to find the required distances.

As per the given information, diagram can be constructed as shown in the attached figure.

Step-by-step explanation:

Let's assume the horizontal distance between post & tower be y and height of post be x.

In △ ABD

$\sf\dashrightarrow \tan \theta = \dfrac{Perpendicular}{Base}$

$\sf\dashrightarrow \tan 60^{\circ}= \dfrac{AD}{AB}$

$\sf\dashrightarrow \sqrt 3 = \dfrac{60\: cm}{y}$

$\sf \dashrightarrow y = \dfrac{60\:cm}{\sqrt 3}$

$\sf \dashrightarrow y = \dfrac{60\:cm}{\sqrt 3} \times \dfrac{\sqrt3}{\sqrt3}$

$\sf \dashrightarrow y = \dfrac{60\sqrt3\:cm}{3}$

$\sf \dashrightarrow y = \cancel\dfrac{60\sqrt3\:cm}{3}$

$\sf \dashrightarrow y = \dfrac{20\sqrt3\:cm}{1}$

$\sf \dashrightarrow y = 20\sqrt 3\:cm$

y = AB = 20√3 cm = Horizontal distance between Tower and Post.

Now finding height of the post:

$\leadsto$ AB = EC = y cm

$\leadsto$ BC = AE = x cm

$\leadsto$ ED = AD - AE = (60 - x) cm

In △ ECD

$\sf\dashrightarrow \tan \theta = \dfrac{Perpendicular}{Base}$

$\sf\dashrightarrow \tan 30^{\circ}= \dfrac{ED}{EC}$

$\sf\dashrightarrow \dfrac{1}{\sqrt3} = \dfrac{(60-x)\: cm}{y}$

$\sf\dashrightarrow \dfrac{1}{\sqrt3} = \dfrac{(60-x)\: cm}{20\sqrt3\:cm}$

$\sf\dashrightarrow \dfrac{20\sqrt3\:cm}{\sqrt3} = \dfrac{(60-x)\: cm}{1}$

$\sf\dashrightarrow \cancel\dfrac{20\sqrt3\:cm}{\sqrt3} = \dfrac{(60-x)\: cm}{1}$

$\sf\dashrightarrow \dfrac{20\:cm}{1} = \dfrac{(60-x)\: cm}{1}$

$\sf\dashrightarrow 20\:cm = (60-x)\: cm$

$\sf\dashrightarrow 20\:cm = 60-x\: cm$

$\sf\dashrightarrow x = 60-20\: cm$

$\sf\dashrightarrow x= 40\: cm$

x = BC = height of the post.

So the required answer is :-

• Height of post = 40 cm

• Distance between post and tower =  20√3 cm