MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 6

The length of a string between a kite and a point on the ground is 90m. The string makes an angle of

$\displaystyle 60^{\circ}$ with the level ground if there is no slack in the string, the height of the kite is

0%
$\displaystyle 90\sqrt{3}m$
0%
$\displaystyle 45\sqrt{3}m$
0%
$180 m$
0%
$45 m$

Explanation

Let

$A$ be the kite and

$AC$ is

$90$ m

$ΔABC$

$\sin 60^{0}=\dfrac{AB}{AC}$

$=\dfrac{h}{90}$

$\therefore \dfrac{\sqrt 3}{2}=\dfrac{h}{90}$

$\therefore h = 45\sqrt 3$ m

The angle of elevation of the top of a tower at a distance of

$\displaystyle \frac{50\sqrt{3}}{3}$ metres from the foot is

$\displaystyle 60^{\circ}$ . Find the height of the tower

0%
$\displaystyle 50\sqrt{3}$ metres
0%
$\displaystyle \frac{20}{\sqrt{3}}$ metres
0%
$-50$ metres
0%
$50$ metres

Explanation

Let the height of the tower be

$h$ .

$ΔABC$

$\tan 60^{0}=\dfrac{AB}{BC}$

$=\dfrac { h }{ \dfrac { 50\sqrt { 3 } }{ 3 } }$

$=>\sqrt 3=\dfrac { 3\times h }{ { 50\sqrt { 3 } } }$

$=\dfrac { 3\times h }{ { 50\sqrt { 3 } \times \sqrt { 3 } } }$

$=\dfrac { 3\times h }{ { 50\times 3 } }$

∴

$h=50$ m

If from the top of a tower 50 m high, the angles of depression of two objects due north of the tower are respectively

$\displaystyle 60^{\circ}$ and

$\displaystyle 45^{\circ}$ , then the approximate distance between the objects is :

0%
$\displaystyle 50\left ( \sqrt{2}-2 \right )m$
0%
$\displaystyle 50\left ( \sqrt{3}-3 \right )m$
0%
$31 \ m$
0%
None of these

Explanation

Let the two objects be C, D and the height of the tower is

$50$ m

$ΔABC$

$tan60^{0}=\dfrac{AB}{BC}$

$=\dfrac{50}{x}$

$\therefore x =\dfrac{ 50}{\sqrt 3}$ ... {i}

Now, In

$ΔABD$

$tan45^{0}=\dfrac{AB}{BD}$

$\Rightarrow 1 =\dfrac{50}{x+y}$

$\Rightarrow x + y = 50$

$\Rightarrow y=x-50$

$\Rightarrow y=50 - \dfrac{50}{\sqrt 3}$ ...[putting the value from {i}]

$\Rightarrow y = \dfrac{50}{\sqrt 3}(\sqrt 3−1)$

∴ Distance between two objects is

$=$

$\dfrac{50}{\sqrt 3}(\sqrt 3−1)$

Rationalising the denominator we get the Distance

$=\dfrac{50 \sqrt{3}(\sqrt{3}-1)}{3}$ m

Option D is correct.

A man on a cliff observes a boat at an angle of depression of

$\displaystyle 30^{\circ}$ which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later the angle of depression of the boat is found to be

$\displaystyle 60^{\circ}.$ Find the total time taken by the boat from the initial point to reach the shore.

0%
$9$ min
0%
$7$ min
0%
$10$ min
0%
$6$ min

The angle of elevation of the top of tower as observed from a pint on the horizontal ground is 'x' if we move a distance 'd' towards the foot of the tower the angle of elevation increases to 'y' then the height of the tower is

0%
$\displaystyle \frac{d\tan x\tan y}{\tan y-\tan x}$
0%
$\displaystyle d(\tan y+\tan x)$
0%
$\displaystyle d(\tan y-\tan x)$
0%
$\displaystyle \frac{d\tan x\tan y}{\tan y+\tan x}$

Explanation

Let the height of the tower be h.

$\displaystyle \therefore$ In

$\displaystyle \Delta ABC$

$\displaystyle \tan y=\dfrac{h}{b}$

b =

$\displaystyle \dfrac{h}{\tan y}$

$\displaystyle \Delta ABD$

$\displaystyle \tan x=\dfrac{h}{b+d}$

$\displaystyle \tan x=\dfrac{h}{\dfrac{h}{\tan y}+d}$

$\displaystyle \dfrac{h}{\tan x}=\dfrac{h}{\tan y}+d$

$d =$

$\displaystyle h\left ( \dfrac{1}{\tan x}-\dfrac{1}{\tan y} \right )$

$d =$

$\displaystyle h\left ( \dfrac{\tan y-\tan x}{\tan x\tan y} \right )$

$h =$

$\displaystyle \dfrac{d\tan x\tan y}{\tan y-\tan x}$

The angels of elevation of the top of a vertical tower from two points

$30$ meter apart and on the same straight line passing through the base of tower are

$\displaystyle 30^{\circ}$ and

$\displaystyle 60^{\circ}$ respectively The height of the tower is

0%
$10 m$
0%
$15 m$
0%
$\displaystyle 15\sqrt{3}m$
0%
$30 m$

Explanation

$ΔAPQ$ ,

$\angle PAQ={ 30 }^{ ∘ }$

Therefore,

$\tan { { 30 }^{ ∘ } } =\dfrac { PQ }{ AQ }$

$\Rightarrow$

$\dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ x }$

$\Rightarrow$

$x=\dfrac { h }{ \sqrt { 3 } }$ ............(eqn 1)

Similarly,

$ΔPQB$ ,

$\angle PBQ={ 60 }^{ ∘ }$

Therefore,

$\tan { { 60 }^{ ∘ } } =\dfrac { h }{ 30-x }$

$\Rightarrow$

$\sqrt { 3 } =\dfrac { h }{ 30-x }$ ............(eqn 2)

From equations 1 and 2,

$\sqrt { 3 } =\dfrac { h }{ 30-\dfrac { h }{ \sqrt { 3 } } }$

$\Rightarrow$

$\sqrt { 3 } \left( 30-\dfrac { h }{ \sqrt { 3 } } \right) =h$

$\Rightarrow$

$2h=30\sqrt { 3 }$

$\Rightarrow$

$h=15\sqrt { 3 } m$

Thus height of the tower is

$15\sqrt { 3 } m$ .

Hence option C is correct.

The angle of elevation of the top of a tower from a point on the ground

$\displaystyle 30^{\circ}$ after walking

$200$ m towards the tower the angle of elevation becomes

$\displaystyle 60^{\circ}$ . The height of the tower is

0%
$\displaystyle 100\sqrt{3}$ m
0%
$\displaystyle 200\sqrt{3}$ m
0%
$100$ m
0%
$200$ m

Explanation

Let

$h$ be the height of the tower

$ΔABC$

$tan 60^{0}=\dfrac{AB}{BC}$

$=>\sqrt 3=\dfrac{h}{x}$

$=>x = \dfrac{h}{\sqrt 3}$ .......(i)

$ΔABD$

$tan 30^{0}=\dfrac{AB}{BD}$

$=>\dfrac{1}{\sqrt 3}=\dfrac{h}{x+200}$

$=>x + 200 = \sqrt 3h$

$=>x = \sqrt 3h - 200$ .......(ii)

From (i) and (ii) we get,

$\dfrac{h}{\sqrt 3}=\sqrt 3h−200$

$=>h = 3h - 200 \sqrt 3$

$=>2h = 200\sqrt 3$

$h=> 100\sqrt 3$ m

$=>h = 100 x 1.732$

$=>h = 173.2$ metre

∴ Height of tower =

$173.2$ m.

The angles of elevation of the top of a vertical tower from two points 30 metres apart and on the same straight line passing through the base of tower are

$\displaystyle 30^{\circ}$ and

$\displaystyle 60^{\circ}$ respectively. The height of the tower is

0%
$10 m$
0%
$15 m$
0%
$\displaystyle 15\sqrt{3}m$
0%
$30 m$

Explanation

Let the height of the tower is h

$ΔABC$

$\tan 60^{0}=\dfrac{h}{x}$

$=>\sqrt 3=\dfrac{h}{x}$

$=>x = \dfrac{h}{\sqrt 3}$ 4 ......(i)

$ΔABD$

$\tan 30^{0}=\dfrac{h}{x+30}$

$=>\dfrac{1}{\sqrt 3} =\dfrac{h}{x+30}$

$=>x+30 = h\sqrt 3$ ........(ii)

$=>x=h\sqrt 3-30$

From (i) and (ii)

$\dfrac{h}{\sqrt 3}=h\sqrt 3−30$

$=>h = 3h - 30\sqrt 3$

$=>2h=30\sqrt 3$

$=>h = 15\sqrt 3$

So, the height of tower =

$15\sqrt 3$ m

If the altitude of the sun is at

$\displaystyle 60^{\circ}$ then the height of the vertical tower that will cast a shadow of length 20 m is

0%
$\displaystyle 20\sqrt{3}m$
0%
$\displaystyle \frac{20}{\sqrt{3}}m$
0%
$\displaystyle \frac{15}{\sqrt{3}}m$
0%
$\displaystyle 40\sqrt{3}m$

Explanation

Let the height of the tower be

$h$

$\tan 60^{0}=\dfrac{h}{20}$

$\Rightarrow \sqrt 3=\dfrac{h}{20}$

$\Rightarrow h = 20\sqrt 3$ m

$\Rightarrow h=20\times 1.732$ m

$\therefore h= 34.64$ m

The length of a ladder is exactly equal to the height of the wall it is leaning against. If the lower end of the ladder is kept on a bench of height 3 m. and the bench is kept 9 m. away from the wall, the upper end of the ladder coincides with the top of the wall. The height of the wall is

0%
$11 m$
0%
$12 m$
0%
$15 m$
0%
$18 m$

Explanation

$AB^2+BC^2=AC^2$

$(h-2)^2+9^2=h^2$

$h=15 m$

The shadow of a tower is 30 meters when the sun's altitude is

$\displaystyle 30^{\circ}$ . When the sun's altitude is

$\displaystyle 60^{\circ}$ then the length of shadow will be

0%
$60 m$
0%
$15 m$
0%
$10 m$
0%
$5 m$

Explanation

In fig. AB is the tower , BC is the length of the shadow of tower when altitude is

$30^{\circ}$ i.e the angle of elevation of the top of the tower from the tip of the shadow is

$30^{\circ}$ and DB is the length of the shadow, when the angle of elevation is

$60^{\circ}$
Given, BC = 30 m
In

$\triangle ABC$

$\tan 30^{\circ}= \dfrac {AB}{BC}$

$\Rightarrow \dfrac 1{\sqrt 3}= \dfrac {AB}{30}$

$\Rightarrow AB= \dfrac {30}{\sqrt 3} = 10\sqrt 3 m$
Now,
In

$\triangle ABD$

$\tan 60^{\circ}= \dfrac {AB}{BD}$

$\Rightarrow \sqrt 3= \dfrac {10\sqrt 3}{BD}$

$\Rightarrow BD = 10 m$
Option C is correct.

The angle of depression of a car moving with uniform speed towards the building as observed from the top of the building is found to be

$30^{\circ}$ . The same angle of depression changes to

$60^{\circ}$ after

$12$ seconds. How much more time would the car take to reach the base?

0%
$6$ sec
0%
$8$ sec
0%
$4$ sec
0%
$12$ sec

Explanation

Let

$'h'$ be the height of the building.

$=x$ and CD

$=y$

$\triangle ABC,$

$\Rightarrow \tan { 60° } = \dfrac { h }{ x }$

$\Rightarrow \sqrt { 3 } = \dfrac { h }{ x }$

$\Rightarrow h=\sqrt { 3 }x \longrightarrow \left( 1 \right)$

$\triangle ADB,$

$\Rightarrow \tan { 30° } =\dfrac { h }{ x+y }$

$\Rightarrow \dfrac { 1 }{ \sqrt { 3 } } =\dfrac { h }{ x+y }$

$\Rightarrow x+y=\sqrt { 3 } h$

From

$(1),$

$\Rightarrow x+y=\sqrt { 3 } \times \sqrt { 3 } x$

$\Rightarrow x+y=3x$

$\Rightarrow y=2x$

$\Rightarrow x=\dfrac{y}{2}$

Since

$y$ is the distance traveled in

$12$ sec from point D to point C.

so, the distance

$x$ is travelled in

$\dfrac{12}{2}=6$ sec.

$[\because$ distance

$x$ is half the distance

$y]$

Hence, the answer is

$6$ seconds

The angle of elevation of the top of a building from the foot of the tower is

$30$ and the angle of the elevation of the top of the tower from the foot of the building is

$60$ . If the tower is

$50$ m high, then the height of the building is :

0%
$\displaystyle \frac { 50 }{ 3 }$ m
0%
$\displaystyle \frac { 35 }{ 3 }$ m
0%
$\displaystyle \frac { 47 }{ 3 }$ m
0%
$\displaystyle \frac { 52 }{ 3 }$ m

Explanation

Let

$AB$ be the tower and

$CD$ be the building.

Since the angle of elevation of the top of the building from the foot of the tower is

$30^{0}$ .

$∴∠CBD=306{0}$

Again, the angle of elevation of the top of the tower from the foot of the building is

$60^{0}$ .

$∴∠ADB=60^{0}$

In right-angled

$ΔABD$ , we have

$\tan60^{0}=\dfrac{AB}{BD}$

$⇒\sqrt 3=\dfrac{50}{BD}$

$⇒BD=\dfrac{50}{\sqrt 3}$ ........(i)

In right-angled

$ΔBDC$ , we have

$\tan 30^{0}=\dfrac{CD}{BD}$

$\Rightarrow CD=\dfrac{1}{\sqrt 3}\times \dfrac{50}{\sqrt 3}$ ....{from {i}}

$\Rightarrow CD=\dfrac{50}{3}$ m

The angles of elevation of the top of a tower from two points a and b from the base and in the same straight line with it are complementary. The height of the tower is :

0%
$\displaystyle ab$
0%
$\displaystyle \sqrt { ab }$
0%
$\displaystyle { a }^{ 2 }{ b }^{ 2 }$
0%
None of these

Explanation

Let AB be the tower and C and D be the two points such that BD = a and BC = b.
If

$\displaystyle \angle BDA=\theta ,$ then

$\displaystyle \angle BCA={ 90 }^{ o }-\theta$
Now in right-angled

$\displaystyle \Delta ABD$ ,

$\displaystyle \frac { AB }{ BD } =\tan { \theta } \Rightarrow \frac { h }{ a } =\tan { \theta }$

$\displaystyle \therefore \quad h=a\tan { \theta } ......(i)$
In right-angled

$\displaystyle \Delta ABC$ ,

$\displaystyle \frac { AB }{ BC } =\tan { \left( { 90 }^{ o }-\theta \right) } =\cot { \theta }$

$\displaystyle \Rightarrow \quad \frac { h }{ b } =\cot { \theta } \Rightarrow h=b\cot { \theta } ......(ii)$
multiplying (i) and (ii), we get

$\displaystyle { h }^{ 2 }=ab\tan { \theta } \cot { \theta } =ab\tan { \theta } \times \frac { 1 }{ \tan { \theta } } =ab$

$\displaystyle \therefore \quad h=\sqrt { ab }$

Two poles of equal heights are standing opposite to each other on either side of a road, which is

$100$ metres wide. From a point between them on the road, the angles of elevation of their tops are

$30$ and

$60$ . The height of each pole is :

0%
$44$ m
0%
$43.25$ m
0%
$50$ m
0%
$40.5$ m

Explanation

Let

$AB$ and

$CD$ be the two poles and

$M$ be the point where the angle of elevation is made of

$30^{0}$ and

$60^{0}$ respectively.

Let the road

$CM$

$=x$ m and

$AM$

$=100-x$ m

In right angled

$\triangle MCD$ , we have

$\tan60^{0}=\dfrac{DC}{CM}$

$\Rightarrow \sqrt 3=\dfrac{h}{x}$

$\Rightarrow h=x\sqrt 3$ (i)

In right angled

$\triangle MAB$ , we have

$\Rightarrow \tan30^{0}=\dfrac{BA}{AM}$

$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{h}{100-x}$

$\Rightarrow h \sqrt 3=100-x$

$\Rightarrow (x\sqrt 3)\sqrt 3=100-x$ ....(from {i})

$\Rightarrow 3x=100-x$

$\Rightarrow 4x=100$

$\Rightarrow x=25$ m

Therefore, height of pole

$CD$ ,

$\Rightarrow CD=25\times 1.732$

$\Rightarrow CD=43.25$ m

Thus, height of each pole is

$43.25$ m.

The angle of elevation of the top of a tower from a point on the ground, which is

$30$ m away from the foot of the tower is

$30$ . The height of the tower is :

0%
$\displaystyle 8\sqrt { 3 }$ m
0%
$\displaystyle 9\sqrt { 3 }$ m
0%
$\displaystyle 10\sqrt { 3 }$ m
0%
$\displaystyle 12\sqrt { 3 }$ m

The shadow of a vertical tower on level ground increases by

$10$ metres when the altitude of the sun changes from the angle of elevation

$45^0$ to

$30^0$ . Find the height of the tower correct to one place of decimal.
(take

$\displaystyle \sqrt { 3 } =1.732$ )

0%
$13.67\ m$
0%
$15\ m$
0%
$18.67\ m$
0%
$20\ m$

Explanation

Let height of tower

$CD$ be

$h$ metres
Now in right-angled

$\displaystyle \Delta BCD$ , we have

$\displaystyle \frac { h }{ x } =\tan { { 45 }^{ o } }$

$\displaystyle \Rightarrow \frac { h }{ x } =1$

$\Rightarrow h=x$ ......(i)
Again in right-angled

$\displaystyle \Delta ACD$ , we have

$\displaystyle \frac { h }{ x+10 } =\tan { { 30 }^{ o } }$

$\displaystyle \Rightarrow \frac { h }{ x+10 } =\frac { 1 }{ \sqrt { 3 } }$

$\displaystyle \Rightarrow \sqrt { 3 } h=x+10$

$\displaystyle \Rightarrow \sqrt { 3 } h=h+10$ ......using (i)

$\Rightarrow \displaystyle h\left( \sqrt { 3 } -1 \right) =10$

$\displaystyle \Rightarrow h=\frac { 10 }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 }$

$\displaystyle =\frac { 10\left( \sqrt { 3 } +1 \right) }{ { \left( \sqrt { 3 } \right) }^{ 2 }-{ \left( 1 \right) }^{ 2 } } =\frac { 10\left( 1.732+1 \right) }{ 3-1 }$

$\displaystyle =5\times 2.732=13.67$ m

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle

$30$ with it. The distance between the foot of the tree to the point where the top touches the ground is

$8$ m. The height of the tree is:

0%
$\displaystyle 5\sqrt { 3 }$ m
0%
$\displaystyle 8\sqrt { 3 }$ m
0%
$\displaystyle 10\sqrt { 3 }$ m
0%
$\displaystyle 6\sqrt { 3 }$ m

Explanation

Let

$AB$ be the tree.
Let it be broken by the wind at the point

$C$ and

$CB$ taking the position

$CD$ strikes the ground at

$D$ .
Now,

$\displaystyle \angle ADC={ 30 }^{ o }$
Let

$AC = x$ and

$CD = y$
In right

$\displaystyle \Delta ADC$ , we have

$\displaystyle \frac { AC }{ AD } =\tan { { 30 }^{ o } }$

$\displaystyle \Rightarrow \frac { x }{ 8 } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow x=\frac { 8 }{ \sqrt { 3 } }$
Again

$\displaystyle \frac { AD }{ CD } =\cos { { 30 }^{ o } } \Rightarrow \frac { 8 }{ y } =\frac { \sqrt { 3 } }{ 2 } \Rightarrow y=\frac { 16 }{ \sqrt { 3 } }$

$\displaystyle \therefore$ Height of the tree

$= (x + y)$ metres.

$\displaystyle =\frac { 8 }{ \sqrt { 3 } } +\frac { 16 }{ \sqrt { 3 } } =\frac { 24 }{ \sqrt { 3 } } =8\sqrt { 3 }$ metres.

From the top of a building h metres high, the angle of elevation of a monument is

$45^o$ and angle of depression of its foot is

$30^o$ . The height of the monument is

0%
$\sqrt 3h$
0%
$\dfrac {(\sqrt 3+1)}{2}h$
0%
$\dfrac {\sqrt 3-1}{2}h$
0%
$(\sqrt 3+1)h$

Explanation

$\Delta ADB$

$AB=\dfrac {h}{tan 30^\circ}$

$\Delta DCF, FC=CD tan 45^\circ$

But

$AB=CD$

$\rightarrow\dfrac {h}{tan 30^\circ}=\sqrt 3h$

Thus,

$FC=\sqrt3h$

Total height of monument

$=BC+FC$

$=h+\sqrt 3h$

$=(\sqrt 3+1)h$

1098265_340173_ans_db56eb9d4b62478d97c1cd2ef3b6f231.png

From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a

$20$ m high building are 45 and 60 respectively. The height of the tower is:

0%
$\displaystyle 25\left( \sqrt { 3 } -1 \right)\ m$
0%
$\displaystyle 20\left( \sqrt { 3 } -1 \right)\ m$
0%
$\displaystyle 20\ m$
0%
$\displaystyle 10\ m$

A kite is flying at a height of

$60$ m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is

$60$ . Find the length of the string, assuming that there is no slack in the string.

0%
$\displaystyle 40\sqrt { 3 }$ m
0%
$\displaystyle 30\sqrt { 3 }$ m
0%
$\displaystyle 20\sqrt { 3 }$ m
0%
$\displaystyle 10\sqrt { 3 }$ m

Explanation

Given, the height of the kite

$AB$ is

$60$ m and it makes an angle of elevation at point

$C$ of

$60^{0}$ .

In right angled

$\triangle ABC$ , we have

$cosec 60^{0}=\cfrac{AC}{AB}$

$\Rightarrow \cfrac{2}{\sqrt 3}=\cfrac{AC}{60}$

$\Rightarrow AC=60\times \cfrac{2}{\sqrt 3}$

$\Rightarrow AC=60\times \cfrac { 2\times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } }$

$\Rightarrow AC=40\sqrt 3$ m

The angle of elevation of the top of a hill at the foot of a tower is

$60$ and the angle of elevation of top of the tower from the foot of the hill is

$30$ . If the tower is

$50$ m high, then the height of the hill is :

0%
$148$ m
0%
$150$ m
0%
$152$ m
0%
$160$ m

Explanation

Let

$AB$ be the hill and

$CD$ be the tower.

Since the angle of elevation of the top of the tower from the foot of the hill is

$30^{0}$ .

$∴∠CBD=30^{0}$

Again, the angle of elevation of the top of the hill from the foot of the tower is

$60^{0}$ .

$∴∠ADB=60^{0}$

In right-angled

$ΔBDC$

$\tan30^{0}=\dfrac{CD}{BD}$

$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{50}{BD}$

$⇒BD=50\sqrt3$ m .....(i)

In right-angled

$ΔABD$ , we have

$\tan60^{0}=\dfrac{AB}{BD}$

$\Rightarrow \sqrt 3=\dfrac{AB}{50\sqrt 3}$ ....{from (i)}

$\Rightarrow AB={\sqrt 3}\times 50\sqrt 3$ ........(ii)

$\Rightarrow AB=150$ m

The shadow of a pole standing on a horizontal plane is

$a$ meters longer when the sun's elevation is

$\theta$ than when it is

$\phi$ . The height of the pole will be: {Use

$\sin(A-B) = \sin A \cos B - \sin B \cos A$ }

0%
$a\displaystyle\frac{\cos\theta\cdot\cos\phi}{\cos(\theta-\phi)}$ meter
0%
$a\displaystyle\frac{\sin\theta\sin\phi}{\sin(\phi-\theta)}$ meter
0%
$a\displaystyle\frac{\sin\theta\cos\phi}{\sin(\theta-\phi)}$ meter
0%
$a\displaystyle\frac{\sin\phi\cos\theta}{\cos(\theta-\phi)}$ meter

Explanation

Let the height of the pole standing on a horizontal plane be

$h$

Let the length of the shadow be

$x$ and

$(x+a)$ , when the sun's elevation is

$\phi$ and

$\theta$ respectively.

$\triangle APQ$ ,

$\dfrac{x}{h}=\cot \phi$

$\Rightarrow x=h \cot \phi$ ....(1)

$\triangle BPQ$ ,

$\dfrac{x+a}{h}=\cot \theta$

$\Rightarrow (x+a)=h \cot \theta$ ....(2)

$\Rightarrow (x+a)-x=h \cot \theta-h \cot \phi$ {using (1)}

$\Rightarrow a=h(cot \theta-cot \phi)$

$\Rightarrow h=\cfrac { a }{ \left( \dfrac { \cos \theta }{ \sin \theta } -\dfrac { \cos \phi }{ \sin \phi } \right) }$

$\Rightarrow h=a\cfrac { \sin \theta \sin \phi }{ \left( \sin \phi \cos \theta -\cos \phi \sin \theta \right) }$

$\Rightarrow h=a\cfrac { \sin \theta \sin \phi }{ \sin\left( \phi - \theta \right) }$

From the top of house

$32$ meter high, if the angle of elevation of the top of a tower is

$\displaystyle 45^{\circ}$ and the angle of depression of the foot of the tower is

$\displaystyle 30^{\circ}$ , then the height of the tower is

0%
$\displaystyle \frac{32}{\sqrt{3}}(\sqrt{3+1})$ meter
0%
$\displaystyle 32(\sqrt{3+1})$ meter
0%
$\displaystyle 32\sqrt3$ meter
0%
$\displaystyle \frac{32}{3}(\sqrt{3+1})$ meter

Explanation

From

$\displaystyle \bigtriangleup OBH$

$\cot 30^{\circ}=\dfrac{BH}{OB}$

$\Rightarrow \sqrt 3 =\dfrac{BH}{OB}$

$\Rightarrow \sqrt 3 =\dfrac{BH}{32}$
or

$\displaystyle BH =32\sqrt{3}$
Height of Tower

$\displaystyle =OT=OB+BT=32(\sqrt{3}+1)\ meters$

An aeroplane flying horizontally at a height of

$1.5$ km above the ground is observed at a certain point on earth to subtend an angle of

$60$ . After

$15$ seconds, its angle of elevation is observed to be

$30$ . Calculate the speed of the, aeroplane in km/h.

0%
$\displaystyle 240\sqrt { 3 }$ km/h
0%
$\displaystyle 230\sqrt { 3 }$ km/h
0%
$\displaystyle 210$ km/h
0%
$\displaystyle 220$ km/h

Explanation

Let

$P$ and

$Q$ be the two position of the plane.

Given angles of elevation of the plane in two position

$P$ and

$Q$ as

$\angle PAB=60^{0}$ and

$\angle QAC=30^{0}$

$PB=QC=$

$15$ km

Now in right angled

$\triangle ABP$

$\tan 60^{0}=\dfrac{BP}{AB}$

$\Rightarrow \sqrt 3=\frac{1.5}{AB}$

$\Rightarrow AB=\dfrac{1.5}{\sqrt 3}$

$\Rightarrow (0.5)\sqrt 3$ km

Again in right angled triangle

$\triangle ACQ$ ,

$\tan30^{0}=\dfrac{QC}{AC}$

$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{1.5}{AB}$

$\Rightarrow AC=(1.5)\sqrt 3$ km

$PQ=BC=AC-AB$

$=(1.5)\sqrt 3-(0.5)\sqrt 3$

$=(1.5-0.5)\sqrt 3$

$=\sqrt 3$ km

The plane travels

$PQ=$

$\sqrt 3$ km in

$15$ secs

$\therefore$ Speed of the aeroplane

$=$

$\dfrac{distance}{time}$

$\Rightarrow \dfrac { \sqrt { 3 } }{ \frac { 15 }{ 3600 } }$

$\Rightarrow 240 \sqrt 3$ km/h

If the angle of elevation of an object from a point 200 meter above the lake is found to be

$\displaystyle 30 ^{\circ}$ and the angle of depression of its image in the lake is

$\displaystyle 45^{\circ}$ , then the height of the object above the lake is

0%
$\displaystyle \frac{200\left ( \sqrt{3-1} \right )}{\left ( \sqrt{3+1} \right )}$ meter
0%
$\displaystyle \frac{200\left ( \sqrt{3-1} \right )}{\sqrt{3}}$
0%
$\displaystyle \frac{200\left ( \sqrt{3+1} \right )}{\sqrt{3}}$
0%
$\displaystyle \frac{200\left ( \sqrt{3+1} \right )}{\left ( \sqrt{3+1} \right )}$ meter

Explanation

Let OL=h, therefore LO'=H
From

$\displaystyle \bigtriangleup PQO',PQ=QO'=QL+LO'$
=200+h [OL=LO'=h]
From

$\displaystyle \bigtriangleup PQO,\frac{OQ}{PQ}=\tan 30^{\circ}$
or

$\displaystyle OQ=PQ \tan 30^{\circ}$
or

$\displaystyle h-200=(200+h)\frac{1}{\sqrt{3}} \therefore h=\frac{200(\sqrt{3}+1)}{\sqrt{3}-1}$

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height

$h$ . At a point on the plane, the angle of elevation of the bottom of the flagstaff is

$\displaystyle \alpha$ and that of the top of the flagstaff is

$\displaystyle \beta$ Then the
height of the tower is :

0%
$\displaystyle \frac { h }{ \tan { \alpha } }$
0%
$\displaystyle \frac { h\tan { \alpha } }{ \tan { \beta } -\tan { \alpha } }$
0%
$\displaystyle \frac { \tan { \beta } }{ h }$
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$\displaystyle \frac { \tan { \alpha } }{ \tan { \beta } -\tan { \alpha } }$

Explanation

Let

$BC$ be the tower and

$CD$ be the flagstaff.

The angle of elevation of the bottom of the flagstaff is

$\alpha$ and that of the top of flagstaff is

$\beta$ .

Let

$h$ be the height of the tower

$\triangle ABC$ , we have

$\Rightarrow \tan \alpha=\dfrac{BC}{AB}$ ....{i}

$\triangle ABD$

$\Rightarrow \tan \beta=\dfrac{BD}{AB}$

$\Rightarrow \dfrac{BC+h}{AB}=\dfrac{BD}{AB}$ ....{ii}

Now dividing {ii} by {i}, we get

$\dfrac{BC+h}{BC}=\dfrac{\tan \beta}{\tan \alpha}$

$\Rightarrow (BC+h)\tan \alpha=BC \tan \beta$

$\Rightarrow BC(\tan \beta-\tan \alpha)=h \tan \alpha$

$\Rightarrow BC=\dfrac{h \tan \alpha}{\tan \beta- \tan \alpha}$

The angles of elevations of the top of the tower from two points in the same straight line and at a distance of

$9 \text{ m}$ and

$16\text{ m}$ from the base of the tower are complementary. The height of the tower is :

0%
$18\text{ m}$
0%
$16\text{ m}$
0%
$10\text{ m}$
0%
$12\text{ m}$

Explanation

From the given figure,

$\displaystyle \tan { \theta } =\frac { h }{ 9 }$ -------

$(1)$

$\displaystyle \tan { \left( 90-\theta \right) } =\frac { h }{ 16 }$

$\implies\cot \theta=\dfrac{h}{16}$ .......

$(2)$

Multiplying

$(1)$ and

$(2)$

$\implies \displaystyle \tan { \theta } \times \cot { \theta } =\frac { { h }^{ 2 } }{ 9\times 16 }$

$[\because \tan \theta \times \cot \theta=1]$

$\implies \displaystyle h^2=9\times 16 \\\implies h=\displaystyle \sqrt{9\times 16}\\\implies h=12\text{ m}$

$\therefore h=12\text{ m}$

1170087_352232_ans_0a20c8fb5a4e411fb54923438ee1e959.png

The angle of elevation of the top of a tower at a distance of

$\displaystyle\frac{50\sqrt{3}}{3}$ metres from the foot is

$60^\circ$ . Find the height of the tower.

0%
$50\sqrt{3}\:m$
0%
$\displaystyle\frac{20}{\sqrt{3}}\:m$
0%
$-50 m$
0%
$50 m$

Explanation

Let the height of the tower be h metres

$\displaystyle\therefore\dfrac{h}{\displaystyle\left(\frac{50\sqrt{3}}{3}\right)}=\tan{60^\circ}=\sqrt{3}$

$\therefore h=50\:m$
1160813_371924_ans_62e6182a8806402bac7f99e2a6b5b1ba.png

The angles of elevation of the top of a tower from two points at the distances of

$4\ m$ and

$9\ m$ from the base of the tower and in the same straight line with it are complementary. The height of the tower is :

0%
$8\ m$
0%
$5\ m$
0%
$6\ m$
0%
$4\ m$

Explanation

Let AB be the tower and C and D be the two points such that BD = 4 m and BC = 9 m.
If

$\displaystyle \angle BDA=\theta ,then\quad \angle BCA={ 90 }^{ o }-\theta$
Now in right-angled

$\displaystyle \Delta ABD$

$\displaystyle \frac { AB }{ BD } =\tan { \theta } \Rightarrow \frac { h }{ 4 } =\tan { \theta }$

$\displaystyle \therefore \quad h=4\tan { \theta } .....(i)$
In right-angled

$\displaystyle \Delta ABC$

$\displaystyle \frac { AB }{ BC } =\tan { \theta } \left( { 90 }^{ o }-\theta \right) =\cot { \theta }$

$\displaystyle \Rightarrow \quad \frac { h }{ 9 } =\cot { \theta } \Rightarrow h=9\cot { \theta } .....(ii)$
multiplying (i) and (ii), we get

$\displaystyle { h }^{ 2 }=36\tan { \theta } \cot { \theta } =36\tan { \theta } \times \frac { 1 }{ \tan { \theta } } =36\\$

$\displaystyle \therefore \quad h=\pm \sqrt { 36 } =6m$

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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