MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 7

The angle of elevation of a ladder leaning against a wall is

$\displaystyle 60^{\circ}$ and the foot of the ladder is

$4.6\ m$ away from the wall The length of the ladder is

0%
$2.3\ m$
0%
$4.6\ m$
0%
$7.8\ m$
0%
$9.2\ m$

Explanation

Consider the given figure. Let

$AC$ denote the length of the ladder and

$AB$ denote the wall. Now

$cos60^{\circ}=\dfrac{BC}{AC}$

$AC=\dfrac{BC}{cos\ 60^{\circ}}$

$AC=BC.sec\ 60^{\circ}$

$=4.6\times 2$

$=9.2$
Hence the length of the ladder is

$9.2\ m$ .

Two objects are one side of a Tower. Their angles of depression from the top of the Tower are

$\displaystyle45^{\circ}$ and

$\displaystyle60^{\circ}$ respectively. If the distance between two objects will be-

0%
$6.34$ metre
0%
$63.4$ metre
0%
$63$ metre
0%
$64$ metre

Explanation

$\displaystyle \Delta ABC$

$\displaystyle \tan 45^{\circ}=\frac{AB}{BD}$

$\displaystyle 1=\frac{150}{BD}$

$\displaystyle BD=150m$
In

$\displaystyle \Delta ABC$ ,

$\displaystyle \tan 60^{\circ}=\frac{AB}{BC}$

$\displaystyle \sqrt{3}=\frac{150}{BC}$

$\displaystyle BC=50\sqrt{3}=86.60m$
Hence, the distance between two objects CD will be

$\displaystyle =150-86.60$
=63.4 metre

The angle of elevation of the top of a tower from two points distant a and b ( a > b )from its foot and in the same straight line from is are

$\displaystyle 30^{\circ}$ and

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

0%
$\displaystyle \frac{a}{b}$
0%
$\displaystyle \sqrt{}\frac{a}{b}$
0%
ab
0%
$\displaystyle \sqrt{ab}$

Explanation

From

$\displaystyle \bigtriangleup AOT$

$\displaystyle h=a\tan 30^{\circ}$
From

$\displaystyle \bigtriangleup BOT$

$\displaystyle h=a\tan 60^{\circ}$

$\displaystyle \therefore h^{2}=ab\tan 30^{\circ}\tan 60^{\circ}$

$\displaystyle =ab\frac{1}{\sqrt{3}}.\sqrt{3}$
or

$\displaystyle h=\sqrt{ab}$

$AB$ is a straight road leading to

$C$ , the foot of a tower.

$A$ is at a distance

$125\text{ m}$ from

$B$ and

$B$ at

$75\text{ m}$ meters from

$C$ . If the angle of elevation of the tower at

$B$ be double the angle of elevation at

$A$ , then the find the value of

$\cos2\alpha,$

0%
$\dfrac{4}{5}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{2}{5}$
0%
$\dfrac{3}{5}$

Explanation

$\triangle APB,$

$\angle CBP=\angle A+\angle APB$

$\Rightarrow2\alpha =\alpha +\angle APB$

$\Rightarrow\angle APB= \alpha$

This implies that

$AB=BP=125 m$ as sides opposite to equal angles in a triangle are equal.

$\triangle BCP,$

$\displaystyle \cos 2\alpha =\frac{75}{125}$

$\Rightarrow \cos 2\alpha =\dfrac{3}{5}$

1027570_374356_ans_7b78b6d1090440cab18224d975fa9124.png

The angle of elevation of the top of a tower ad observed from a point 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 if 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 height of the tower ,

$AB=h$

From

$\displaystyle \bigtriangleup ADB, \tan y =\frac{h}{BD}$

$\displaystyle BD= h \cot y$ ....

$(i)$

From

$\displaystyle \bigtriangleup ADB$ and

$\triangle ACB$

$\displaystyle \tan x=\frac{h}{d+DB}$ or

$d+DB =h \cot x$

$\displaystyle h(\cot x-\cot y)=d$

$\displaystyle \therefore h=\frac{d}{\cot x-\cot y}$

$\displaystyle =\dfrac{d}{\dfrac{1}{\tan x}- \dfrac{1}{\tan y}}\\=\dfrac{d\tan x.\tan y}{\tan y-\tan x}$

1292122_374196_ans_4f1f976790554054af03781125569795.png

The angles of elevation of a tower from a point on the ground is

$\displaystyle 30^{\circ}$ . At a point on the horizontal line passing through the foot of the tower and

$100$ meters closer to it than the previous point, if the angle of elevation is found to be

$\displaystyle 60^{\circ}$ , then height of the tower is

0%
$\displaystyle 50\sqrt{3}$ meters
0%
$\displaystyle \frac{50}{\sqrt{3}}$ meters
0%
$\displaystyle 100\sqrt{3}$
0%
$\displaystyle \frac{100}{\sqrt{3}}$

Explanation

Let

$AB=h$ be the height of the tower, and let

$AD=x$ m

From right angled

$\displaystyle \bigtriangleup ABD$

$\displaystyle \tan 60^{\circ}=\frac{AB}{AD}$

$\displaystyle \sqrt{3}=\frac{h}{x}$ or

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

From right angled

$\displaystyle \bigtriangleup BCA$

$\displaystyle \tan 30^{\circ}=\frac{AB}{CA}$

$\displaystyle \frac{1}{\sqrt{3}}=\frac{h}{100+x}$

$\displaystyle h=\frac{100+x}{\sqrt{3}}=\frac{100+\dfrac{h}{\sqrt{3}}}{\sqrt{3}}$

$\displaystyle \frac{2}{\sqrt{3}}h=100$

$\displaystyle \therefore h=50\sqrt{3}m$

1027558_374190_ans_96d81e77a31f40f68350a038ff6cbff0.png

From a point

$p$ on a level ground the angle of elevation of the top of a tower is

$\displaystyle 30^{0}$ If the tower is

$100\ m$ high the distance of point

$p$ from the foot of the tower is

0%
$149\ m$
0%
$156\ m$
0%
$173\ m$
0%
$200\ m$

Explanation

Let

$AB$ be the tower then

$\displaystyle \angle APB=30^{0}$ and

$\displaystyle AB=100m$

$\displaystyle \frac{AB}{AP}=\tan 30^{0}\\\ \ \ \ \ \ \ \ =\dfrac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow AP=\left ( AB\times \sqrt{3} \right )=100\sqrt{3}m$

$\displaystyle =\left ( 100\times 1.73 \right )m$

$\displaystyle = 173m$

An observer 1.6m tall is

$\displaystyle 20\sqrt{3}$ m away from a tower The angle of elevation from his eye to the top of the tower is

$\displaystyle 30^{0}$ The height of the tower is

0%
21.6m
0%
23.2m
0%
24.72m
0%
None of these

Explanation

Let

$AB$ be the observer and

$CD$ tower

Draw

$BE$ perpendicular to

$CD$

Then

$CE=AB=1.6 \ m$

And

$BE=AC=$

$20\sqrt{3}$ m

Then right angle triangle DEB

$\therefore tan 30^{o}=\frac{DE}{BE}$

$\Rightarrow \dfrac{1}{\sqrt{3}}=\dfrac{DE}{20\sqrt{3}}$

$\Rightarrow DE=20\sqrt{3}$ m

Then

$CD=CE+DE=1.6+20=21.6\ m$

A rope of length 5 meters is tightly tied with one end at the top of a vertical pole and other end to the horizontal ground. If the rope makes an angle

$\displaystyle 30^{\circ}$ to the horizontal , then the height of the pole is

0%
$\displaystyle \frac{5}{2}m$
0%
$\displaystyle \frac{5}{\sqrt{2}}m$
0%
$\displaystyle 5\sqrt{2}m$
0%
5m

The angle of elevation of the sun when the length of the shadow of a tree is

$\displaystyle \sqrt{3}$ times the height of the tree is:

0%
$\displaystyle 30^{0}$
0%
$\displaystyle 45^{0}$
0%
$\displaystyle 60^{0}$
0%
$\displaystyle 90^{0}$

Explanation

Let

$AB$ be the tree and

$AC$ be its shadow.

Also,

$AC=\sqrt3AB$ .
Let the angle of elevation of the sun

$=\displaystyle \angle ACB=\theta$
Then in

$\triangle ABC,$

$\displaystyle \tan \theta = \frac{AB}{AC}=\dfrac{1}{\sqrt{3}}$ {

$\because AC=\sqrt3AB$ }

$\tan 30^o =\dfrac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow \theta =30^{0}$

Two ships are sailing in the sea on the two sides of a lighthouse The angles of elevation of the top of the lighthouse as observed from the two
ships are

$\displaystyle 30^{0}$ and

$\displaystyle 45^{0}$ respectively If the lighthouse is

$100$ m high the distance between the two ships is

0%
$173$ m
0%
$200$ m
0%
$273$ m
0%
$300$ m

Explanation

Let

$AB$ be the height of lighthouse and

$C$ and

$D$ be the position of the ships.

Given

$AB=100$ m and

$\angle ACB=30^{0}$

$\angle ADB=45^{0}$

$\Delta BCA$

$\dfrac{AB}{AC}=tan 30^{0}\\\ \ \ \ \ \ \ =\dfrac{1}{\sqrt{3}}$

$\Rightarrow AC=\sqrt{3}AB\\\ \ \ \ \ \ \ \ \ \ =100\sqrt{3}$

$\Delta BAD$

$\dfrac{AB}{AD}=tan 45^{0}\\\ \ \ \ \ \ \ \ =1$

$\Rightarrow AD=AB=100 m$

$\therefore CD=AC+AD\\=100\sqrt{3}+100\\=100(\sqrt{3}+1)\\=100(1.73+1)\\=273 m$

The top of a

$15$ metre high tower makes an angle of elevation of

$\displaystyle 60^{0}$ with the bottom of an electric pole and angle of elevation of

$\displaystyle 30^{0}$ with the top of the pole. What is the height of the electric pole?

0%
$5$ metres
0%
$8$ metres
0%
$10$ metres
0%
$12$ metres

Explanation

Let

$AB$ be the tower and

$CD$ be the electric pole
Then

$\displaystyle \angle ACB=60^{0},\angle EDB=30^{0}$ and

$\displaystyle AB=15$ m
Let

$\displaystyle CD=h.$ Then

$\displaystyle BE=\left ( AB-AE \right )$

$\displaystyle =\left ( AB-CD \right )=\left ( 15-h \right )$
We have

$\displaystyle \frac{AB}{AC}=\tan 60^{0}=\sqrt{3}$

$\displaystyle \Rightarrow AC=\frac{AB}{\sqrt{3}}=\frac{15}{\sqrt{3}}$
And

$\displaystyle \frac{BE}{DE}=\tan 30^{0}=\frac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow DE=\left ( BE\times \sqrt{3} \right )$

$\displaystyle =\sqrt{3}\left ( 15-h \right )$

$\displaystyle \Rightarrow 3h=\left ( 45-15 \right )$

$\Rightarrow h=10$ m

If the height of a pole is

$\displaystyle 2\sqrt{3}$ metres and the length of its shadow is 2 metres then the angle of elevation of the sun is equal to

0%
$\displaystyle 30^{0}$
0%
$\displaystyle 45^{0}$
0%
$\displaystyle 60^{0}$
0%
$\displaystyle 90^{0}$

Explanation

Height of a pole is

$2\sqrt { 3 }$

length of its shadow is

$2$ metre

Let ABC be a triangle where AB be

$2\sqrt { 3 }$ height of a pole

AC be

$2$ metre length of shadow

Angle of elevation of the sun will be

$\tan { \theta = } \dfrac { AB }{ AC } =\dfrac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 }$

$\tan { { 60^ o } } =\sqrt { 3 }$

Angle of elevation of the sun is equal to

${ 60^o }$

The upper part of a tree broken by the wind makes an angle of

$\displaystyle 60^{0}$ with the ground and the distance from the roots to the point where the top of the tree meets the ground is

$20$ m. The length of the broken part of the tree is

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

Explanation

As shown in the fig. let length of the tree be

$AB$ and

$DC$ is the length of broken part.

$\displaystyle BC=DC=\dfrac{20}{\cos 60^{0}}=20\times 2m=40m$

If a flag-staff 6 metres high placed on the top pf a tower throws a shadow of

$\displaystyle 2\sqrt{3}$ metres along the ground then the angle that the sun makes with the ground is

0%
$\displaystyle 60^{0}$
0%
$\displaystyle 30^{0}$
0%
$\displaystyle 45^{0}$
0%
None of these

Explanation

OA and AB be the shadows of the tower OP and the flag-staff PQ respectively on the grounds Let the sum makes an angle

$\displaystyle \theta$ with the ground

$\displaystyle AB=2\sqrt{3};OA=x;OP=h;PQ=6$
In triangles OAP and OBQ We have

$\displaystyle \tan \theta =\frac{h}{x}$ and

$\displaystyle \tan \theta =\frac{h+6}{x+2\sqrt{3}}$

$\displaystyle \Rightarrow \frac{h}{x}=\frac{h+6}{x+2\sqrt{3}}$

$\displaystyle \Rightarrow x=\frac{h}{\sqrt{3}}$
or

$\displaystyle \frac{h}{x}=\tan \theta =\sqrt{3}$ so

$\displaystyle \theta=60^{0}$

A man standing on the bank of the river observes that the angle subtended by a tree on the opposite bank is

$\displaystyle 60^{0}$ . When he retires

$36$ metres from the bank he finds the angle to be

$\displaystyle 30^{0}$ . The breadth of the river is

0%
$\displaystyle 12\sqrt{3}$ m
0%
$18\ m$
0%
$12\ m$
0%
$27\ m$

Explanation

Let the breadth of the river be AC = 'x' metres
Clearly

$\displaystyle \tan 30^{0}=\frac{AB}{AD}=\frac{h}{x+36}$ and

$\displaystyle \tan 60^{0}=\frac{AB}{AD}=\frac{h}{x}$

$\displaystyle \Rightarrow \frac{\tan 60^{0}}{\tan 30^{0}}=\frac{h}{x}\div \frac{h}{x+36}$

$\displaystyle \Rightarrow 3=\frac{x+36}{x}$

$\displaystyle \Rightarrow x=18m$

In case of angle of depression the observer has to look _____ to view the object.

0%
Straight
0%
Anywhere
0%
Down
0%
Up

A man on the top of a rock lying on a seashore observes a boat coming towards it. If it takes

$10$ minutes for the angles of depression to change from

$\displaystyle 30^{\circ}$ to

$\displaystyle 60^{\circ}$ how soon will the boat reach the shore?

0%
$20$ minutes
0%
$15$ minutes
0%
$10$ minutes
0%
$5$ minutes

Explanation

AB is the rock C and D be the two positions of the boat

Given:

$\angle EAC=30^{\circ}$ and

$\angle EAD=60^{\circ}$

Now,

$\angle ADB=EAD$

$[\because AE \parallel BC, AD~ \text{is transversal}, ~\angle ADB$ and

$\angle EAD$ are alternate interior angles

$]$

$\implies \displaystyle \angle ADB=60^{\circ}$

$\angle EAC=ACB$

$[\because AE \parallel BC, AC~ \text{is transversal}, ~\angle EAC$ and

$\angle ACB$ are alternate interior angles

$]$

$\implies \displaystyle \angle ACB=30^{\circ}$

Let

$CD=x, ~DB=y$

From

$\triangle ABC$

$\displaystyle \tan 30^{0}=\frac{AB}{x+y}=\frac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow AB=\frac{x+y}{\sqrt{3}}$
From

$\triangle ABD$

$\displaystyle \tan 60^{0}=\frac{AB}{y}\Rightarrow AB=\sqrt{3}y$

$\displaystyle \therefore \frac{x+y}{\sqrt{3}}=\sqrt{3}y$ or

$\displaystyle \frac{x}{2}=y$
The boat takes

$10$ minutes to cover distance

$x,$

so it will take

$5$ minutes to cover

$\displaystyle \frac{x}{2}$

$\displaystyle \therefore$ Time to reach

$\displaystyle B=\left ( 10+5 \right )$ min

$= 15$ Minutes from point

$C$

A flagstaff

$5$ m high stands on a building

$25$ m high. As observed from a point at a height of

$30$ m, the flagstaff and the building subtend equal angles. The distance of the observer from the top of the flagstaff is

0%
$\displaystyle \frac{5\sqrt{3}}{2}$
0%
$\displaystyle 5\sqrt{\frac{3}{2}}$
0%
$\displaystyle 5\sqrt{\frac{2}{3}}$
0%
None of thses

Explanation

OB is the bisector of DOAC of DAOC

$\displaystyle \therefore \frac{OC}{OA}=\frac{BC}{AB}=\frac{x}{\sqrt{x^{2}+30^{2}}}=\frac{5}{25}$

$\displaystyle \Rightarrow x=\frac{30}{\sqrt{24}}=\frac{30}{2\sqrt{6}}$

$\displaystyle \therefore x=5\sqrt{\frac{3}{2}}$

The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of

$100\ m$ from its base is

$\displaystyle 45^{o}$ If the angle of elevation of the top of the complete pillar at the same point is to be

$\displaystyle 60^{o}$ then the height of the incomplete pillar is to be increased by

0%
$\displaystyle 50\sqrt{2}\ m$
0%
$100\ m$
0%
$\displaystyle 100\left ( \sqrt{3-1} \right )\ m$
0%
$\displaystyle 100\left ( \sqrt{3+1} \right )\ m$

Explanation

Let BC be the incomplete and BD be the complete pillar In Triangles ABC and ABD we have

Given

$BC=100$ m,

And

$\angle CAB=45^{0}$

$\angle DAB=60^{0}$

Let

$CD= x$ m

$\Delta ABC$

$tan 45^{o}=\frac{BC}{AB}$

$\Rightarrow 1=\frac{BC}{100}$

$\Rightarrow BC=100 m$

$ABD$

$tan60^{0}=\frac{AD}{AB}$

$\Rightarrow \sqrt{3}=\frac{BD}{100}$

$\Rightarrow BD=100\sqrt{3}$ m

Then

$BD=BC+CD=100+x$

$\therefore 100+x=100\sqrt{3}$

$\Rightarrow x=100(\sqrt{3}-1)$ m

The length of the shadow of a person is x when the angle of elevation of the sun is

$\displaystyle 45^{0}$ . If the length of the shadow increased by

$\displaystyle \left ( \sqrt{3}-1 \right )x$ then the angle of elevation becomes

0%
$\displaystyle 15^{0}$
0%
$\displaystyle 18^{0}$
0%
$\displaystyle 25^{0}$
0%
$\displaystyle 30^{0}$

Explanation

Let

$AB$ be the height of the person,

$AC$ and

$AD$ be his shadows at two instants.

$\displaystyle AC=x$

$\displaystyle AD= AC+CD =x+\left ( \sqrt{3}-1 \right )x$

$\displaystyle =\sqrt{3}x$

and

$\displaystyle \angle ACB=45^{0}$

$\displaystyle \frac{AB}{AC}=\tan 45^{0}=1\\ \Rightarrow AB=AC=x$

$\displaystyle \tan \theta =\frac{AB}{AD}\\ =\dfrac{x}{\sqrt{3x}}\\ =\dfrac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow \theta =30^{0}$

A tower subtends an angle

$\displaystyle \alpha$ at point 'A' in the plane of its base and the angle of depression of the foot of the tower at height 'b' just above A is

$\displaystyle \beta$ . The height of the tower is

0%
$\displaystyle b\tan \alpha \cot \beta$
0%
$\displaystyle b\cot \alpha \cot \beta$
0%
$\displaystyle b\tan \alpha \tan \beta$
0%
$\displaystyle b\cot \alpha \tan \beta$

Explanation

Let

$\displaystyle AP=BR=x$ then from

$\displaystyle \Delta APQ$ we find that
the height of the tower

$\displaystyle =PQ=x\tan \alpha =\left ( b\cot \beta \right )\tan \alpha$
(

$\displaystyle \because$ In triangle APB,

$\displaystyle \frac{x}{b}=\cot \beta$ )

From the top of the cliff

$150\ m$ high the angles of depression of the top and bottom of a tower are observed to be

$\displaystyle 30^{\circ}$ and

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

0%
$100\ m$
0%
$\displaystyle 50\sqrt3\ {m}$
0%
$\displaystyle 133\frac{1}{3}$
0%
$\displaystyle 100\left ( \sqrt{3}-1 \right )$

Explanation

$\displaystyle BF=\left ( 150-x \right )m.$ Let

$\displaystyle AC=y$

$\displaystyle \tan 30^{0}=\frac{150-x}{y}$

$\displaystyle \tan 60^{0}=\frac{150}{y}$

$\displaystyle \Rightarrow y=\left ( 150-x \right )\sqrt{3}=\frac{150}{\sqrt{3}}$

$\displaystyle \therefore 150-x=50$ or

$\displaystyle x=100\ m$

If the length of the shadow of a pole on ground is twice the length of the pole, then the angle of elevation of the sun is _____

0%
30 $^o$
0%
45 $^o$
0%
60 $^o$
0%
None of these

Explanation

Referring figure, if

$AB$ is pole and

$AC$ is length of the shadow of pole,

then

$AB = x$ ,

$AC = 2x$

and let

$\angle\; ACB=\theta$

Then,

$\tan$

$\theta$ =

$\displaystyle{\frac{AB}{AC} \Rightarrow \tan \theta = \frac{x}{2x} = \frac{1}{2}}$

$\therefore$ Value of

$\theta$ is neither 30

$^o$ , 60

$^o$ nor 45

$^o$ .

So, option D is correct.

A tree is broken by the wind and now its upper part touches the ground at a point

$10$ m from the foot of the tree and makes an angle of

$\displaystyle 60^{0}$ with the ground. The entire length of the tree is:

0%
$15$ m
0%
$20$ m
0%
$\displaystyle \left ( 10+\frac{20}{\sqrt3} \right )\sqrt{3}m$
0%
$\displaystyle \left ( 10+\frac{\sqrt{3}}{2} \right )m$

Explanation

Let say the Length of the tree

$= BT = BC + CT$ and

$CT = CA$ when the tree is broken.

Given:

$AB = 10m$

$\triangle ABC$ ,

$\tan60^o = \dfrac{BC}{AB}$ ............(1)

Also,

$sec\ 60^o = \dfrac{AC}{AB}$ ...................(2)

$BT=$

$BC + AC$

$=$

$AB \displaystyle \tan 60^{0}+AB\ \text{sec} 60^{0}$ {Using (1) and(2)}

$= AB\sqrt3 + AB\sec60^o$

$\displaystyle =AB\left ( \sqrt{3}+ 2 \right ) =10\left ( 1+\frac{2}{\sqrt3} \right )\sqrt{3}m$

$\displaystyle =\left ( 10+\frac{20}{\sqrt3} \right )\sqrt{3}m$

A flagstaff

$10\ m$ high stands at the center of a equilateral triangle which is horizontal at the top of the flagstaff. Each side subtends at an angle of

$\displaystyle 60^{o}$ . The length of each side of the triangle is:

0%
$\displaystyle \dfrac{6\sqrt{3}}{3}$
0%
$\displaystyle \dfrac{4\sqrt{6}}{3}$
0%
$\displaystyle \dfrac{20}{\sqrt{3}}$
0%
$\displaystyle \dfrac{6\sqrt{5}}{3}$

Explanation

$\Rightarrow$ Considering Right angle

$\triangle ADC$ we have,

$\Rightarrow$

$\tan 60^o=\dfrac{AD}{DC}$

$\Rightarrow$

$\sqrt{3}=\dfrac{10}{DC}$

$\Rightarrow$

$DC=\dfrac{10}{\sqrt{3}}$

$\Rightarrow$

$DC=\dfrac{10}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}$

$\therefore$

$DC=\dfrac{10\sqrt{3}}{3}$

$\Rightarrow$

$BC=2\times DC$

$\therefore$

$BC=2\times \dfrac{10\sqrt{3}}{3}$

$\therefore$

$BC=\dfrac{20\sqrt{3}}{3}$

$\therefore$ The length of each side of the triangle is

$\dfrac{20}{\sqrt{3}}\,m$

If the angle of elevation of an object from a point

$100 \text{ m}$ above a 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%
$100\left ( 2-\sqrt{3} \right )\text{ m}$
0%
$100\left ( 2+\sqrt{3} \right )\text{ m}$
0%
$100\left ( \sqrt{3}-1 \right )\text{ m}$
0%
$1000\left ( \sqrt{3}+1 \right )\text{ m}$

Explanation

Let

$AB$ the surface of the lake

$O$ be the point of observation

$P$ be the object and

$P'$ be its image.

Let the distance of object

$P$ above lake be

$x.$ So, it will be below the lake by same distance as they are mirror images.

$PE=P'E=x$

$PD=x-100$

$P'D=x+100$

$OC=DE=100 \text{ m}$

$\triangle OPD,$

$\tan 30^{\circ}=\dfrac{PD}{OD}$

$\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{x-100}{OD}$

$\Rightarrow OD=\sqrt3 (x-100)\text{ m}$

$\triangle OP'D,$

$\begin{aligned}{}\tan {45^\circ } &= \frac{{P'D}}{{OD}}\\OD& = P'D\\\sqrt 3 (x - 100) &= x + 100\\x& = \frac{{100(\sqrt 3 + 1)}}{{\sqrt 3 - 1}}\\ &= 100(2 + \sqrt 3 ){\text{ m}}\end{aligned}$

Hence, the distance of the object above the lake is

$100(2+\sqrt3)\text{ m}.$

The angle of elevation

$\displaystyle \theta$ of a vertical tower from a point on the ground is such that its tangent is

$\displaystyle \frac{5}{12}$ . On walking

$192$ meters towards the tower in the same straight line the tangent of angle of elevation

$\displaystyle \phi$ is found to be

$\displaystyle \frac{3}{4}$ . Find the height of the tower

0%
$170$ m
0%
$175$ m
0%
$180$ m
0%
$185$ m

Explanation

$\displaystyle \Delta PQR,\tan \theta =\frac{h}{192+x}=\frac{5}{12}$

$\displaystyle \Rightarrow x=\frac{1}{5}\left ( 12h-960 \right )$ ...(i)
In

$\displaystyle \Delta PQS,\tan \phi =\frac{h}{x}=\frac{3}{4}$

$\displaystyle \Rightarrow x=\frac{4h}{3}$ ...(ii)
From (i) and (ii), we get

$\displaystyle \frac{4h}{3}=\frac{1}{5}\left ( 12h-960 \right )$

$\displaystyle \Rightarrow h=180$

$\displaystyle \therefore$ The height of the tower is 180 m

From the top of a lighthouse, the angles of depression of two stations on opposite sides and

$a$ distance apart are

$\displaystyle \alpha$ and

$\displaystyle \beta$ . The height of the lighthouse is:

0%
$\displaystyle \frac{a}{\cot \alpha \cot \beta }$
0%
$\displaystyle \frac{a}{\cot \alpha +\cot \beta }$
0%
$\displaystyle \frac{a\cot \alpha \cot\beta }{\cot \alpha +\cot \beta }$
0%
$\displaystyle \frac{a\tan \alpha \tan \beta }{\cot \alpha +\cot \beta }$

Explanation

Let Height of Light House be

$h=AD$ , as shown in the figure above.

and

$\angle ABD=\alpha\ (Given)$

$\angle ACD =\beta\ (Given)$

Now, in

$\triangle ABD,\ BD=h\cot \alpha\quad \quad \left[\cot \theta=\dfrac {\text{Adjacent Side}}{\text{Opposite Side}} \right]$

Also, in

$\triangle ADC, \ CD=h\cot\beta\quad \quad \left[\cot \theta=\dfrac {\text{Adjacent Side}}{\text{Opposite Side}} \right]$

Given that,

$BC=a$

$\therefore \ a=BC=BD+CD\\$

$\Rightarrow a=h\cot\alpha+h\cot\beta\\$

$\Rightarrow a=h(\cot\alpha+\cot\beta)\\$

$\therefore \ h=\dfrac{a}{\cot\alpha+\cot\beta}\\$

Hence, option B is correct.

An aeroplane flying at a height of

$300\ metre$ above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on ground are

$\displaystyle 60^{\circ}$ and

$\displaystyle 45^{\circ}$ respectively. The height of the lower plane above the above the ground in meters is

0%
$\displaystyle 100\sqrt{3}$
0%
$\displaystyle \frac{100}{\sqrt{3}}$
0%
$50$
0%
$\displaystyle 150\left ( \sqrt{3+1} \right ).$

Explanation

$P$ and

$Q$ are positions of planes.

From

$\displaystyle \Delta OMP;MP=300\ m$

$\displaystyle OM=MP\cot 60^{0}=300\times \dfrac{1}{\sqrt{3}}\ m$

$\displaystyle =100\sqrt{3}$
In

$\displaystyle \Delta OMQ,MQ=OM$

Height of the lower plane,

$MQ = OM = 100\sqrt{3}\ m$

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 7 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 7 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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