A person, standing on the bank of a river, observes that the angle subtended by a tree on the opposite bank is 60°. When he retreated 20m from the bank, he finds the angle to be 30°. Find the height of the tree and the breadth of the river.

Here

$tan 60 = \frac {h}{b}$

or $h= \sqrt {3} b$

Now

$tan 30 = \frac {h}{20 +b}$

20 +b =3b

or b=10 m

and H= $10 \sqrt 3$ m

A tree is broken by the wind. The top struck the ground at an angle of 30° and at a distance of 30 meters from the roots. Find the whole height of the tree.

Here

$tan 30 = \frac {h}{30}$

or $h= 10 \sqrt 3$

Now Hypotnuse will be

$hyp= \sqrt { 30^2 + (10 \sqrt 3)^2 } = 34.64$ m

Total height of tree = 17.32 +34.64=51.96 m

The angles of elevation of the top of a tower from two points at distances p and q meters from the base and in the same straight line with it are complementary. Prove that the height of the tower is $\sqrt {pq}$

Here

$tan \beta = \frac {h}{p}$ -(1)

Now

$tan (90 - \beta)= \frac {h}{q}$

$cot \beta= \frac {h}{q}$

or $ \frac {1}{tab \beta} =\frac {h}{q}$

$tab \beta = \frac {q}{h}$ -(2)

Comparing 1 and 2
$\frac {h}{p}=\frac {q}{h}$
$h= \sqrt {pq}$

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is 5/12 . On walking 192 meters towards the tower, the tangent of the angle of elevation is 3/4. Find the height of the tower.

Here

$\frac {5}{12} = \frac {h}{x}$

Now

$\frac {3}{4}= \frac {h}{192 + x}$

Solving these we get h=180 m

A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60

Same as Question 1

h = 34. 64m, width = 20m

A man on the top of a vertical tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this, will the car reach the tower? Give your answer to the nearest second.

16 minutes 23 seconds

The shadow of a flag – staff is three times as long as the shadow of the flag – staff when the sun rays meet the ground at an angle of 60

30°

An aeroplane at an altitude of 200m observes the angles of depression of opposite points on the two banks of a river to be 45° and 60

315.4 meters

Two pillars of equal height and on either side of a road, which is 100m wide. The angles of elevation of the top of the pillars are 60

Distance from 1^{st} pillar = 25m, 2^{nd} = 75m, Height = 43.3m

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30

864km/hr

If the angle of elevation of a cloud from a point h meters above a lake is $\alpha$ and the angle of depression of its reflection in the lake is $\beta$, prove that the height of the cloud is

Let C be the cloud and C' be the cloud reflection point. The figure is shown as below

Now

$tan \alpha = \frac {x}{base}$

or $base = \frac {x}{tan \alpha}$

Also

$tan \beta = \frac {x + 2h}{base}$

or $base = \frac {x +2h}{tan \beta}$

Equating the above two , we get the value of x as

$x= \frac {2h tan \alpha}{tab \beta - tab \alpha }$

Now the height of the cloud = $x + h$

or

=$ \frac {2h tan \alpha}{tab \beta - tab \alpha } + h= \frac {h(tan \alpha + tan \beta}{tab \beta - tab \alpha }$

A round balloon of radius r subtends an angle α at the eye of the observer while the angle of elevation of its centre is $ \beta$. Prove that the height of the centre of the balloon is

Let OA=d

Now in Triangle OAN

$sin \frac {\alpha}{2} = \frac {r}{d}$

or $d = r cosec \frac {\alpha}{2} $

Also in AOB

$sin \beta = \frac {h}{d}$

or $ h= d sin \beta=r cosec \frac {\alpha}{2} sin \beta$

The angle of elevation of a cliff from a fixed point is $\theta$. After going up a distance of K meters towards the top of the cliff at an angle of $\phi$, it is found that the angle of elevation is $\alpha$. Show that the height of the cliff is meter

The figure is shown as above.
Here AD=K
In triangle ADB

$BD=K sin \phi $ and $AB= K cos \phi $

Now in Triangle ODE

$tan \alpha = \frac {h - K sin \phi}{DE}$

or $DE = \frac {H - K sin \phi}{tan \alpha}= (H -k sin \phi) cot \alpha $

Also in AOC

$tan \theta = \frac {H}{(H -k sin \phi) cot \alpha +K cos \phi }$

Solving for H

or $ H= \frac {K(cos \phi - sin \phi cot \alpha}{cot \theta - cot \alpha}$

The angle of elevation of the top of a vertical tower PQ from a point X on the ground is 60°. At a point Y, 40m vertically above X, the angle of elevation of the top is 45° Calculate the height of the tower.

Now

$tan 45= \frac {PR}{YR}$

or YR=PR

Also

$tan 60 = \frac {PR + 40}{XQ}$

Now XQ=YR=PR

So,

$\sqrt 3 = \frac {PR+40}{PR}$

$PR= \frac {40}{\sqrt 3 -1}$

Height of tower= $40 + \frac {40}{\sqrt 3 -1}=94.64$ m

If the angle of elevation of a cloud from a point h metre above a lake is $\alpha$ and the angle of depression of its reflection in the lake be $\beta$, prove that the distance of the cloud from the point of observation is

Let C be the cloud and C' be the cloud reflection point. The figure is shown as below

Now

$tan \alpha = \frac {x}{base}$

or $base = \frac {x}{tan \alpha}$

Also

$tan \beta = \frac {x + 2h}{base}$

or $base = \frac {x +2h}{tan \beta}$

Equating the above two , we get the value of x as

$x= \frac {2h tan \alpha}{tab \beta - tab \alpha }$

Now the distance of the cloud from the point of observation = $\frac {x}{sin \alpha} =\frac {2h sec \alpha}{tab \beta - tab \alpha }$

From an aeroplane vertically above a straight horizontal road, the angles of depression of two consecutive mile stones on opposite sides of the aeroplane are observed to be $\alpha$ and $\beta$. Show that the height in miles of aeroplane above the road is given by

Now

$tan \alpha= \frac {h}{x}$

or

$x= \frac {h}{tan \alpha}$ m

Now

$tan \beta = \frac {h}{y}$

or

$y= \frac {h}{tan \beta}$ m

Now

$x + y =1$

$\frac {h}{tan \alpha} + \frac {h}{tan \beta}=1$

or

$h=\frac {tan \alpha tan \beta}{tan \alph + tan \beta}$

A ladder rests against a wall at an angle $\alpha$ to the horizontal. Its foot is pulled away from the wall through a distance , so that its slides a distance b down the wall making an angle $\beta$ with the horizontal. Show that

A boy is standing on the ground and flying a kite with 100m of string at an elevation of 30°. Another boy is standing on the roof of a 10 m high building and is flying his kite at an elevation of 45°. Both the boys are on opposite sides of both the kites. Find the length of the string that the second boy must have so that the two kites meet.

40√2 m

Two boats approach a light house in mid- sea from opposite directions. The angles of elevation of the top of the light house from two boats are 30° and 45° respectively. If the distance between two boats is 100m, find the height of the light house.

Here

$tan 45= \frac {H}{100-y}$

or y=100 -H

Now

$tan 30 =\frac {H}{y} = \frac {H}{100 -H}$

$100- H= H \sqrt 3$

$H = \frac {100}{\sqrt 3 + 1} = 50 (\sqrt 3 -1}$

From the top of a light house, the angles of depression of two ships on the opposite sides of it are observed to be $\alpha$ and $\beta$. If the height of the light house be h meters and the line joining the ships passes through the foot of the light house, show that the distance between the ship is metres

Now

$tan \alpha= \frac {h}{x}$

or

$x= \frac {h}{tan \alpha}$ m

Now

$tan \beta = \frac {h}{y}$

or

$y= \frac {h}{tan \beta}$ m

Distance is given by

$=x + y = \frac {h}{tan \alpha} + \frac {h}{tan \beta}=\frac {h(tan \alpha + tan \beta)}{tan \alpha tan \beta}$

As observed from the top of a 150m tall light house, the angles of depression of two ships approaching it are 30° and °. If one ship is directly behind the other, find the distance between the two ships.

Now

$tan 45= \frac {PQ}{BQ}$

or

$BQ= PQ=150$ m

Now

$tan 30 = \frac {PQ}{AQ} = \frac {150}{x+ 150}$

or $x= 150 \sqrt 3 - 150=109.5$ m

This Class 10 Maths Important Questions for Application of Trigonometry with answers is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.You can download in PDF form also using the below links

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Class 10 Maths Class 10 Science