Question

an aeroplane at an altitude 200m observes the angle of depression of opposite points on two banks of river to be 45 degree and 60 degree.find the width of river.​

Answer 1

Hi ,

Draw a rough diagram of given data

|A

|

|

| 60

|__________|___________45_

B. C. D

Join A to C and D

Let

Height of the aeroplane from the

ground = AB = 200m

Width of the river = CD

i) In right triangle ABD,

tan 45° = AB / BD

1 = 200 / BD

BD = 200m ---( 1 )

ii ) In triangle ABC ,

Tan 60° = AB / BC

√3 = 200 / BC

BC = 200 /√3

BC =( 200 √3)/( √3 × √3 )

BC = 200√3 / 3----( 2 )

iii ) width of the river =

CD = BD - BC

= 200 - 200√3/3

= 200( 1 - √3 /3)

= 200( 3 - √3 )/3

=( 200/3) ( 3 - √3 ) m

I hope this helps you.

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Answer 2

✯ Given :-

An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 40° and 60° .

✯ To Find :-

What is the width of the river.

✯ Solution :-

» Let, AD be the height of the aeroplane

» And, BC = x be the width of the aeroplane.

⋆ Given that, AD = 200 m

➟ In ∆ABD,

$\tt \ \: {tan45° = \dfrac{AD}{BD}}$

$\implies \: 1 = \dfrac{AD}{BD}$

⇒ AD = BD

⇒ BD = 200 m

Again,

➟ In ∆ACD

$\tt \ \: { \implies \: tan60° = \dfrac{AC}{CD}}$

$\tt \ \: { \implies \: \: \sqrt{3} \: \: = \: \: \frac{ac}{cd} }$

⇒ BC = BD + CD

⇒ BC = 200 + 115.4

➥ BC = 315.4 m

therefore the width of the river is $\boxed{\bold{\small{315.4\: m}}}$

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