When the angle of elevation of the sun is 45 the length of the shadow of a tower of height 10m is?

Question 11 Some Applications Of Trigonometry Exercise 12.1

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Answer:

Solution:

A common notion in trigonometry, specifically, is the angle of elevation, which has to do with height and distance. It is described as an angle formed by the horizontal plane and an oblique line between the observer's eye and a target above it.

Let the height of the tower(AB) = h m

Let the length of the shorter shadow be x m

Then, the longer shadow is (10 + x)m

So, from fig. In ΔABC

tan 60o = AB/BC

√3 = h/x

x = h/√3…. (i)

Next, in ΔABD

tan 45o = AB/BD

1 = h/(10 + x)

10 + x = h

10 + (h/√3) = h {using (i)}

10√3 + h = √3h

h(√3 -1) =10√3

h = 10√3/ (√3 -1)

After rationalising the denominator, we have

h = [10√3 x (√3 + 1)]/ (3 – 1)

h = 5√3(√3 + 1)

h = 5(3 + √3)

= 5( 3+ 1.732) {as √3 value is 1.732}

=5 (4.732)

= 23.66

Therefore, the height of the tower is 23.66 m.

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24.

The angle of elevation of aeroplane from a point on the ground is 45°. After flying 15 sec, the elevation changes to 30°. If the aeroplane is flying at a height of 2500 m, then the speed of the aeroplane in km/hr is

• 600

• 600( √3 + 1 )

• 600√3

• 600( √3 - 1 )

D.

600( √3 - 1 )

∴  Speed of the aeroplane

Let h be the height of tower AB and angle of elevation are 45° and 60° are given.

In a triangle OAC, given that AB = 10+x and BC = x

Now we have to find the height of the tower.

So we use trigonometrical ratios.

In a triangle OAB,

`=> tan A = (OB)/(AB)`

`=> tan 45^@ = (OB)/(AB)`

`=> 1= h/(10 + x)`

=> h = 10 + x

Therefore  x = h - 10

Again in a triangle OCB

`=> tan C = (OB)/(BC)`

`=> tan 60^@ = (OB)/(BC)`

`=> sqrt3 = h/x`

`=> h = sqrt3x`

Put x = h - 10

`=> h = sqrt3 (h - 10)`

`=> h = sqrt3h - 10sqrt3`

`=> 10sqrt3 = h(sqrt3 - 1)`

`=> h = (10sqrt3)/(sqrt3 - 1)`

`=>  h = (10 xx 1.732)/(1.732 - 1)`

`=> h = 17.32/0.327`

=> h = 23.66

Hence height of tower is 23.66 m