Question

There are two poles, one each on either Bank of a river just opposite to each other. One pole is 60m

high. From the top of this pole, the angle of depression of the top and foot of the other pole are 30

degree and 60 degree respectively. Find the width of the river and height of the other pole.​

Answer 1

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kindly refer to the attachment for the solution....it will surely help you so, make sure that u mark the answer

Answer 2

• Let BD is one pole of height = 60m
• And AE is another pole.

According to the question,

Image is in the attached file.

Now,

From the top of this pole, the angle of depression of the top and foot of the other pole is 30 and 60 degrees respectively.

AC ║ED, so AE = CD and AE║CD also.

• In triangle BCA,

$\boxed{\frac{BC}{AC}=tan \theta}$

$= \frac{(60-a)}{AC}=tan\ 30$

$\boxed{tan30^{o} = \frac{1}{\sqrt{3}}}$

$= \frac{(60-a)}{AC}=\frac{1}{\sqrt{3}}$

[By cross multiplication, we get,]

= AC = √3(60 - a)

= AC = 60√3 - √3a

Now,

• In triangle BED

$\boxed{\frac{BD}{ED}=tan \theta}$

As, AC = ED, we ca write AC in place of ED.

$\boxed{\frac{BD}{AC}=tan \theta}$

Putting the values,

$= \frac{60}{60\sqrt{3}-\sqrt{3a}}=tan\ 60^{o}$

$\boxed{tan60^{o} = \sqrt{3}}$

$= \frac{60}{60\sqrt{3}-\sqrt{3a}}=\sqrt{3}$

By cross multiplication, we get,

⇒ √3 (60√3 - √3a) = 60

⇒ 60*3 - 3a = 60

⇒ 180 - 3a = 60

⇒ 180 - 60 = 3a

[By transporting (-3a) to RHS and 60 to LHS]

⇒ 120 = 3a

⇒ 120 ÷ 3 = a

[By transporting 3 to LHS]

⇒ 40 = a

So,

• The height of the pole AE = 40m

[AE = CD = a = 40]

Now,

The width of the river,

$\boxed{\frac{BC}{AC}=tan \theta}$

BC = 60 - a

     = 60 - 40

     = 20m

$\implies \frac{(20)}{AC}=tan\ 30$

$\implies \frac{(20)}{AC}=\frac{1}{\sqrt{3}}$

⇒ AC = 20√3 m

• Hence,
• Height of another pole is 20m
• Width of the river is 20√3 m