MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 2

At a point 15 metres away from the base of a 15 metres high house, the angle of elevation of the top is

0%
$45^0$
0%
$30^0$
0%
$60^0$
0%
$90^0$

Explanation

$\tan\theta = \dfrac{15}{15} = 1$

$\theta$ =

$45^0$

The angle of elevation of a cloud from a point h m above the level of water in a lake is

$\alpha$ and the angle of depression of its reflection in the lake is

$\beta$ .Then the height of the cloud above the water level is

$\dfrac{h\,sin\, (\beta - \alpha)}{sin\,(\beta + \alpha)}$

0%
True
0%
False

Explanation

From Figure

Point

$A$ is the cloud

Point

$D$ is view point

$FD$ is parallel to

$BE$ and

$CD$ transversal

$\angle FDC=\angle DCE$

From Reflection property

$\angle BCA =\angle DCE =\beta$

From

$\Delta ABC$

$\sin \beta =\dfrac {AB}{AC} \cdots(1)$

From

$\Delta CDE$

$\sin \beta = \dfrac {DE}{DC} \cdots(2)$

From

$(1 ) \& (2)$

$\implies \dfrac {AB}{AC} =\dfrac {DE}{DC}$

$\implies AB =\dfrac {DE}{DC}\times AC \cdots(3)$

From

$\Delta ACD$

By Exterior angle is sum of two interior opposite angles

$2\beta =\alpha +\beta +\angle DAC$

$\angle DAC = \beta -\alpha$

From

$\sin$ rule

$\implies \dfrac {AC}{\sin (\alpha +\beta)}=\dfrac {DC}{\sin (\beta -\alpha )}$

$\implies \dfrac {AC}{DC}=\dfrac {\sin (\beta -\alpha )}{\sin (\alpha +\beta)} \cdots(4)$

Put it in

$(3)$

$\implies AC =\dfrac{DE \sin (\beta -\alpha )}{\sin (\alpha +\beta)}$

From Question

$DE=h$

$\implies AC =\dfrac{h \sin (\beta -\alpha )}{\sin (\alpha +\beta)}$

1539082_1006600_ans_0f6a16fdbd274af8bfc0ddf95057784d.jpg

The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the same straight line are complementary. The height of the tower is

$\sqrt{ab}$ metres.

0%
True
0%
False

Explanation

Let AB be the tower and let C and D be the two points of the observer. Then,

AC = a metres and AD = b metres

Let

$\angle ACB =\theta$ . Then,

$\angle ADB =(90^{o}-\theta)$ .

Let AB = h metres

In right

$\triangle CAB$ , we have

$\tan \theta =\dfrac{AB}{AD}$

$\tan \theta =\dfrac{h}{a}$

$h=a\tan \theta$ ......(1)

In right

$\triangle DAB$ , we have,

$\tan(90^{o}-\theta)=\dfrac{AB}{AD}$

$\cot \theta =\dfrac{h}{b}$

$h=b\cot \theta$ .......(2)

From 1 and 2, we get,

$h^2=ab$

$h=\sqrt{ab} \ metres$

1325267_1053913_ans_d1e7ea9daaa646898d04ee3d749fc0f6.png

$AB$ is vertical pole with

$B$ at the ground level and

$A$ at the top. A man find that the angle of elevation of the point

$A$ from a certain point

$C$ on the ground is

$60^{o}$ . He moves away from the pole along the line

$BC$ to a point

$D$ such that

$CD=7\ m$ . From

$D$ the angle of elevation of the point

$A$ is

$45^{o}$ . Then the height of the pole is:

0%
$\dfrac { 7\sqrt { 3 } }{ 2 } \dfrac { 1 }{ \sqrt { 3 } -1 } m$
0%
$\dfrac { 7\sqrt { 3 } }{ 2 } \left( \sqrt { 3 } +1 \right) m$
0%
$\dfrac { 7\sqrt { 3 } }{ 2 } \left( \sqrt { 3 } -1 \right) m$
0%
$\dfrac { 7\sqrt { 3 } }{ 2 } \dfrac { 1 }{ \sqrt { 3 } +1 } m$

Explanation

Then in triangle

$ABC$

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

$x=\frac{h}{\sqrt3}$ ----- (i)

$\Delta ABD$

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

$1=\dfrac{h}{x+7}$

$h=x+7$

$h=\dfrac{h}{\sqrt3}+7$ from (i)

$h=\dfrac{7\sqrt3}{\sqrt3-1}$

$h=\dfrac{7\sqrt3}{2}(\sqrt3+1)$

Therefore the height of the pole is

$\dfrac{{7\sqrt 3 }}{2}(\sqrt 3 + 1)\;m$

1048171_1046470_ans_d37a437007ec484f8a81247ce6d6f55a.png

A bridge above the river makes an angle of

${45}^{o}$ with the bank of river. If length of bridge above the river is

$150\ m$ then breadth of river will be

0%
$75\ m$
0%
$50\sqrt {2}\ m$
0%
$150\ m$
0%
$75\sqrt {2}\ m$

Explanation

Let

$AC$ be the bridge and

$AB$ be the width of river.

$\Delta ABC$

$\dfrac{AB}{AC}=\sin 45^{\circ}$

$\dfrac{AB}{150}=\dfrac{1}{\sqrt2}$

$AB=\dfrac{150\sqrt2}{2}=75\sqrt2$

therefore width of the river is

$75\sqrt2\;m$ .

1078441_1052393_ans_44da1a74bf1c49508a32d4a49422957b.jpg

A flagstaff stands on the middle of a square tower. A man on the ground, opposite to the middle of one face and distant from it

$100$ m, just see the flag ; on his receding another

$100$ m, the tangents of the elevation of the top of the tower and the top of the flagstaff are found to be

$\dfrac {1}{2}$ and

$\dfrac {5}{9}$ . Find the height of the flagstaff, the ground being horizontal

0%
$20$
0%
$25$
0%
$30$
0%
$35$

Explanation

$\dfrac{h_2}{b}=\dfrac{h_1}{100}$

$\dfrac{h_1}{200}=\dfrac{1}{2}$

$\Rightarrow h_1=100$

$\dfrac{h_1+h_2}{200+b}=\dfrac{5}{9}$

$\dfrac{100+b}{200+b}=\dfrac{5}{9}$

$b=\dfrac{100}{4}$

$\Rightarrow b=h_2=25$

As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. An equation to model the motion is y=20cos(

$\frac {\pi}{4} (t-3))+23$ . Predict your height above the ground at a time of 1 seconds.

0%
$20.86 ft$
0%
$23 ft$
0%
$8.14 ft$
0%
$18.96$

Explanation

$y=20\ cos (\cfrac{\pi}{4}(t-3))+23$

$t=1$ second

Height above the ground,

$y=20\ cos (\cfrac{\pi}{4}(1-3))+23$

$y=20\ cos (\cfrac{-2\pi}{4})+23$

$\implies y=23 ft$ . [ Since ,

$\ cos (\cfrac{-\pi}{4})=0$ ]

The shadow of a tower, when the angle of elevation of the sun is

$45^{o}$ , is found to be 10 metres longer than when the angle of elevation is

$60^{o}$ . Find the height of the tower.

0%
$15+5\sqrt{3} \ m$
0%
$12+5\sqrt{3} \ m$
0%
$15+\sqrt{3} \ m$
0%
$12+\sqrt{3} \ m$

Explanation

Let AB be the tower and let AC and AD be its shadows when the angles of elevation of the sum are

$60^{o}$ and

$45^{o}$ , respectively. Then,

$\angle ACB=60^{o}, \angle ADB = 45^{o}, CD=10 \ m$

Let

$AB=h \ metres$ and

$AC=x \ metres$

In right

$\triangle$ CAB, we have,

$\tan 60^{o}=\dfrac{AB}{AC}$

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

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

In right

$\triangle$ DAB, we have,

$\tan 45^{o}=\dfrac{AB}{AD}$

$1=\dfrac{h}{10+x}$

$x=h-10$ .....(2)

From 1 and 2, we get,

$\dfrac{h}{\sqrt{3}}=h-10$

$h=\sqrt{3}h-10\sqrt{3}$

$\sqrt{3}h-h=10\sqrt{3}$

$h(\sqrt{3}-1)=10\sqrt{3}$

$h=5\sqrt{3}(\sqrt{3}+1)$

$h=15+5\sqrt{3} \ m$

1325247_1054068_ans_2e979461e77943dc8dd0e994d2309899.png

The shadow of a

$6\ m$ high tower is

$15\ m$ and at the same point of time length of shadow of a tree is

$25\ m$ . What is the height of the tree?

0%
$21\ m$
0%
$10\ m$
0%
$35\ m$
0%
$none\ of\ these$

Explanation

$\dfrac{Hight \ of \ the \ tower}{Length \ of \ shadow \ of \ the \ tower} = \dfrac{Hight \ of \ the \ tree}{Length \ of \ shadow \ of \ the \ tree}$

$\implies \dfrac{6}{15} = \dfrac{H}{25}$

$\implies H = 25 \times \dfrac{6}{15}$

$\therefore H = 10 \ m$

OPtion B is correct.

If a ladder

$13 m$ is placed against a wait such that its roots at a distance from the wall, then the height of the top of the ladder from the ground :

0%
$10 m$
0%
$11 m$
0%
$12 m$
0%
None of these

Explanation

If a ladder is

$13$ m long is placed at a distance of

$12$ m from a wall, then the weight of the wall is:

$\begin{array}{l} A.T.Q, \\ h=\sqrt { { { \left( { 13 } \right) }^{ 2 } }-{ { \left( { 12 } \right) }^{ 2 } } } \\ =\sqrt { 169-144 } \\ =\sqrt { 25 } \\ =5\, \, m \end{array}$

A tower stands vertically on the ground. From a point on the ground which is

$30\ m$ away from the foot of a tower, the angle of elevation of the top of the tower is found to be

$45^{o}$ . Find the height of tower.

0%
$15$
0%
$40$
0%
$30$
0%
$20$

Explanation

REF.Image.

$30 m$ away /

$45^{\circ}$ inclination.

Let the height of tower be h m

$AB = h \, m$

Distance of the point from

foot of tower =

$30 m$

$CB = 30 m$

Angle of elevation

$= 45^{\circ}$

$\angle ACB = 45^{\circ}$

Since tower is vertical to ground

$\angle ABC = 90^{\circ}$

Now

$tanC = \dfrac{side\, opp\, to\, angle\, C}{side\, adjacent \,to\, angle \,C}$

$\tan C = \dfrac{AB}{CB}$

$\tan45 = \dfrac{AB}{CB}$

$1 = \dfrac{H}{30}$

$H = 30$

Hence height of tower =

$30 m$

Answer = option

$C = 30 m$

1375736_1226831_ans_b7ef016d0830465da9127368aa51b194.JPG

A man observes the angle of elevation of a balloon to be

$30^{0}$ at a point

$A$ . He then walks towards the balloon and at a certain place

$B$ , he finds the angle of elevation to be

$60^{0}$ . He further walks in the direction of the balloon and finds it to be directly over him at a height of

$\dfrac12\ km$ , then the distance

$AB$ is:

0%
$\displaystyle \frac{1}{\sqrt{2}}\ km$
0%
$\displaystyle \frac{1}{\sqrt{3}}\ km$
0%
$\displaystyle \frac{1}{\sqrt{4}}\ km$
0%
$\displaystyle \frac{1}{\sqrt{5}}\ km$

Explanation

$h = 500\ m$

$\tan60^0 = \dfrac{h}{b}$

$\Rightarrow b = \dfrac{500}{\sqrt {3}}$

$\tan30^0 = \dfrac{h}{d+b}$

$\Rightarrow d+ \dfrac{500}{\sqrt {3}} = 500 \sqrt {3}$

$\Rightarrow d = \dfrac{1000}{\sqrt {3}}\ m$

$\Rightarrow d = \dfrac{1}{\sqrt {3}}\ km$

The horizontal distance between two towers is

$60\ m$ and angular depression of the top of the first as seen from the second, which is

$150\ m$ in height, is

$30^{0}$ . The height of the first tower is

0%
$(150+20\sqrt{3})\ m$
0%
$(150+15\sqrt{3})\ m$
0%
$(150-20\sqrt{5})\ m$
0%
$(150-20\sqrt{3})\ m$

Explanation

Let AB and CD be the two towers

Height of the second tower, AB =150m and

Distance between two towers BD =60m

Let's assume the height of the first tower is

$h\ m$ .

We know, In right angled

$\triangle$ ,

$\tan(\Theta ) = \dfrac {Opposite} {Adjacent}$

$\tan 30^0 = \dfrac{AE}{EC}$

$\tan 30^0 = \dfrac{150-h}{60}$ .....[AE=AB-ED] and [EC=BD=60]

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{150-h}{60}$

$\Rightarrow h \sqrt{3} = 150 \sqrt{3} -60$

$\Rightarrow h = 150-20 \sqrt{3}$

The height of the first tower is

$150-20 \sqrt{3}$ m.

A kite is flying with the string inclined at

$30^{o}$ to the horizon. The height of the kite above the ground, when the string is

$15\ m$ long is

0%
$15\ m$
0%
$30\ m$
0%
$\dfrac{15}2\ m$
0%
$\dfrac{15}3\ m$

Explanation

$\sin{30^0} = \dfrac{AB}{AC} = \dfrac h{15}$

$\Rightarrow \dfrac12 = \dfrac h{15}$

$\Rightarrow h = \dfrac {15}2\ m$

From the top of the tree, a man observes the angle of depression of a point which is at a distance of

$40\ m$ from the foot is

$75^{0}$ . The height of the tree is:

0%
$40\sqrt{3}\ m$
0%
$40(2+\sqrt{3})\ m$
0%
$21\sqrt{3}\ m$
0%
$3\sqrt{21}\ m$

Explanation

Let say AC be the tree and the Observer is at A who is observing point B.

Given: BC = 40m

$\angle B = 75^o$ = angle of depression {alternate angles}

$\triangle ABC$ ,

$\tan 75^0 = \dfrac{h}{40}$

$\Rightarrow \tan (45^o+30^o) = \dfrac{h}{40}$

$\Rightarrow \dfrac{\tan45^o+ \tan 30^o}{1-\tan45^o\tan 30^o} = \dfrac{h}{40}$

$\because \dfrac{\tan A+ \tan B}{1-\tan A\tan B} = tan(A+B)$

$\Rightarrow \dfrac{1+\dfrac{1}{\sqrt{3}}}{1- 1\times \dfrac{1}{\sqrt{3}}}= \dfrac{h}{40}$

$\Rightarrow \dfrac{(\sqrt{3}+1)}{(\sqrt{3}- 1)} = \dfrac{h}{40}$

$\Rightarrow\left(\dfrac{3+1+2\sqrt{3}}{2}\right) = \dfrac{h}{40}$ {rationalizing by multiplying and dividing by

$\sqrt 3+1$ }

$\Rightarrow (2+\sqrt{3}) = \dfrac{h}{40}$

$\Rightarrow h = 40 (2+\sqrt{3})m$

From the top of a hill

$h$ meters high, the angle of depression of the top and the bottom of a pillar are

$\alpha,\ \beta$ respectively. Then the height(in meters) of the pillar is

0%
$\displaystyle \dfrac{h(\tan\beta-\tan\alpha)}{\tan\beta}$
0%
$\displaystyle \dfrac{h(\tan\alpha-\tan\beta)}{\tan\alpha}$
0%
$\displaystyle \dfrac{h(\tan\beta+\tan\alpha)}{\tan\beta}$
0%
$\displaystyle \dfrac{h(\tan\beta+\tan\alpha)}{\tan\alpha}$

Explanation

$\tan \alpha = \dfrac{h-d}{a}$

$\tan \beta = \dfrac{h}{a}$

$\dfrac{\tan \alpha}{\tan \beta} = \dfrac{h-d}{h}$

$\Rightarrow \dfrac{\tan \alpha}{\tan \beta} = 1- \dfrac{d}{h}$

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

$\Rightarrow d= h\left(\dfrac{\tan \beta-\tan \alpha}{\tan \beta}\right)$

From the top of a building

$h$ metres, the angle of depression of an object on the ground is

$\alpha$ , the distance of the object from the foot of the building is

0%
$h\cot\alpha$
0%
$h\tan\alpha$
0%
$h\cos\alpha$
0%
$h\sin\alpha$

Explanation

$\tan \alpha = \dfrac{h}{d}$

$\Rightarrow d = \dfrac{h}{\tan \alpha}$

$\Rightarrow d = h \cot \alpha$

The angle of elevation of an object from a point

$P$ on the level ground is

$\alpha$ . Moving

$d$ meters on the ground towards the object, the angle of elevation is found to be

$\beta$ , then the height (in meters) of the object is

0%
$d\tan\alpha$
0%
$d\cot\beta$
0%
$\dfrac d{\cot\alpha+\cot\beta}$
0%
$\dfrac d{\cot\alpha-\cot\beta}$

Explanation

Let the height of the object

$AB$ be

$h$ metres.

Given,

$\angle ACB = \alpha, \angle ADB = \beta,\ CD = d$ metres

Let

$DB = x$ metres.

Then, in right-angled

$\triangle ABD,$

$\cot β=\dfrac{x}{h}$

$⇒ x = h \cot \beta$

$...(i)$

In right-angled

$\triangle ACB,$

$\cot \alpha =\dfrac{x+d}{h}$

$⇒ x + d = h \cot \alpha$

$...(ii)$

Let us subtract the equations:

$(ii)\ – (i)$

$\Rightarrow d = h \cot \alpha \ – h \cot \beta$

$\Rightarrow d=h(\cot \alpha \ -\cot \beta)$

$\Rightarrow h = \dfrac{d}{ \cot \alpha - \cot \beta }$

The flag staff of height

$10$ metres is placed on the top of a tower of height

$30$ metres. At the top of a tower of height

$40$ metres, the flag staff and the tower subtend equal angles then the distance between the two towers (in metres) is

0%
$40\sqrt{2}$
0%
$10\sqrt{2}$
0%
$20\sqrt{2}$
0%
$30\sqrt{2}$

Explanation

$\tan 2 \theta = \dfrac{40}{d}$ ...(i)

$\Rightarrow \dfrac{2\tan \theta}{1-\tan^{2} \theta} = \dfrac{40}{d}$

$\tan \theta = \dfrac{10}{d}$ ...(ii)

Substituting (ii) in (i), we get

$\dfrac{2\dfrac{10}{d}}{1-\dfrac{(10)^{2}}{d^{2}}}= \dfrac{40}{d}$

$\Rightarrow \displaystyle \frac{d}{d^{2}-10^{2}}= \frac{2}{d}$

$\Rightarrow d^{2} = 2d^{2} - 2 \times 10^{2}$

$\Rightarrow d^{2} = 2 \times 10^{2}$

$\Rightarrow d = 10 \sqrt{2}\ m$

Straight pole(AB) subtends a right angle at a point

$D$ of another pole at a distance of

$30$ meters from

$A$ , the top of

$A$ being

$60^{0}$ above the horizontal line joining the point

$B$ to the pole

$A$ . The length of the pole

$A$ is, in meters

0%
$20 \sqrt{3}$
0%
$40 \sqrt{3}$
0%
$60 \sqrt{3}$
0%
$\displaystyle \frac{40}{\sqrt{3}}$

Explanation

$\tan60=\dfrac{h_{2}}{30}$

$h_{2}= 30\sqrt{3}$

$\tan30^{0}=\dfrac{h_{1}}{30}$

$h_{1}=10\sqrt{3}$

$h_{1}+h_{2}$ = length of the pole =

$40\sqrt{3}$

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

$60^{0}$ with the ground and touches the ground at a distance of 50 metres from the foot. The height of the tree in metres is

0%
$124.2$
0%
$186.6$
0%
$243.2$
0%
$164.2$

Explanation

We will consider that the tree is broken at A.

So the total heigh of the tree will be

$AC + AB$

Let

$AC=L_1$ and

$AB =L_2$

$\Rightarrow$ Height of the tree

$= L_{1} + L_{2}$

In the

$\triangle ABC, \angle C = 90^\circ$ ,

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

$L_{2} \sin 60^{0} = L_{1}$ .....(1)

$\tan 60^{0} = \displaystyle\frac{L_{1}}{50}$

$\Rightarrow L_{1} = 50 \sqrt{3} m$

Substituting the value of

$L_1$ in (1)

$L_2 \times \dfrac{\sqrt 3}{2} = 50\sqrt3$

$\Rightarrow L_{2} = 100 m$

$\Rightarrow h =L_{1}+L_{2}$

$\Rightarrow h = 100+50 \sqrt{3}$

$=186.6 m$

$\therefore$ Height of the tree =

$=186.6 m$

The angle of elevation of the top of a flagstaff when observed from a point at a distance

$60$ meters from its foot is

$30^{0}$ . The height of the flagstaff (in meters) is:

0%
$20\sqrt{3}$
0%
$10\sqrt{3}$
0%
$60\sqrt{3}$
0%
$30\sqrt{3}$

Explanation

Let

$AB=h$ be the height of the flagstaff and

$C$ be the point at a distance of

$60m$ from the foot.

$\triangle ABC$ ,

$\tan{30^0} = \dfrac{AB}{BC} = \dfrac{h}{60}$

$\Rightarrow \dfrac1{\sqrt3} = \dfrac h{60}$

$\Rightarrow h = \dfrac{60}{\sqrt3}$

$\Rightarrow h = 20\sqrt3m$

A tower subtends an angle

$\alpha$ at a point

$A$ on the same level as the foot of the tower

$B$ is a point vertically above

$A$ and

$AB=h$ metres. The angle of depression of the foot of the tower from

$B$ is

$\beta$ . The height of the tower is

0%
$h\tan\alpha\cot\beta$
0%
$h\tan\alpha\tan\beta$
0%
$h\cot\alpha\cot\beta$
0%
$h\cot\alpha\tan\beta$

Explanation

$\tan\alpha =\dfrac{h_{1}}{d}$

$\tan\beta =\dfrac{h}{d}$

$\dfrac{\tan\alpha }{\tan\beta }=\dfrac{h_{1}}{h}$

$h_1=h \tan \alpha \cot \beta$

In a prison wall there is a window of

$1$ metre height,

$24$ metres from the ground. An observer at a height of

$10 m$ from ground, standing at a distance from the wall finds the angle of elevation of the top of the window and the top of the wall to be

$45^{ 0}$ and

$60^{0}$ respectively. The height of the wall above the window is

0%
$15\sqrt{3}$
0%
$15\left(1-\displaystyle \frac{1}{\sqrt{3}}\right)$
0%
$15(\sqrt{3}-1)$
0%
$14(\sqrt{3}-1)$

Explanation

We can see that ,

$CE=DF=10$

Therefore,

$BC=BE-CE=24-10=14$

$\triangle BCD$

$\tan45^o=\dfrac{14}{b}\Rightarrow b=14$

Now, In

$\triangle ACD$

$\tan60^o=\dfrac{a+14}{14}\Rightarrow \sqrt3=\dfrac{a+14}{14}\Rightarrow a=14(\sqrt3-1)$

Therefore, Answer is

$14(\sqrt3-1)$

The horizontal distance between two towers is

$30$ meters. From the foot of the first tower the angle of elevation of the top of the second tower is

$60^{o}$ . From the top of the second tower the angle of depression of the top of the first is

$30^{o}$ . The height of the small tower is:

0%
$20(\sqrt{3}+1)$ mts
0%
$20(\sqrt{3}-1)$ mts
0%
$20\sqrt{3}$ mts
0%
$20$ mts

Explanation

$\tan 60^o= \dfrac{h_{2}}{30}$ ...{1}

$\tan 30^o=\dfrac{h_{2}-h_{1}}{30}$ ...{2}

$\tan 30^{o}=\dfrac{h_{2}}{30}-\dfrac{h_{1}}{30}$

From {1} and {2}

$\Rightarrow \dfrac{h_{1}}{30}=\tan 60^o - \tan 30^o$

$\Rightarrow \dfrac{h_{1}}{30}=\dfrac{3-1}{\sqrt{3}}$

$\Rightarrow h_1=\dfrac{30\times2}{\sqrt{3}}$

$\Rightarrow h_{1}=20\sqrt{3} \,m$

If from the top of a tower of

$60$ metre high, the angles of depression of the top and floor of a house are

$\alpha$ and

$\beta$ respetivley and if the height of the house is

$\displaystyle \frac{60\sin(\beta-\alpha)}{x}$ , then

$x=$

0%
$\sin\alpha\sin\beta$
0%
$\cos\alpha\cos\beta$
0%
$\sin\alpha\cos\beta$
0%
$\cos\alpha\sin\beta$

Explanation

$\tan \alpha=\dfrac{60-h}{d}$

$\tan \beta=\dfrac{60}{d}$

$\dfrac{\tan \alpha}{\tan \beta}=\dfrac{60-h}{60}$

$\dfrac{\tan \alpha}{\tan \beta}=1-\dfrac{h}{60}$

$\dfrac{h}{60}=\dfrac{\tan \beta-\tan \alpha}{\tan \beta}$

$\dfrac{h}{60}=\dfrac{\sin \beta \ \cos\alpha-\sin\alpha \ \cos\beta}{\cos\alpha \ \sin\beta}$

$\dfrac{h}{60}=\dfrac{\sin(\beta-\alpha)}{\sin\beta \ \cos\alpha}$

$h=\dfrac{60\ \sin(\beta-\alpha)}{\cos\alpha \sin\beta }$

$\therefore x=\cos\alpha \sin \beta$

On the level ground the angle of elevation of the top of a tower is

$30^{0 }$ On moving 20 metres nearer tower, the angle of elevation is found to be

$60^{0}$ The height of the towerin metres is

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

Explanation

$\tan\ 60=\dfrac{h}{d}$

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

$\tan\ 30=\dfrac{h}{20+d}$

$\dfrac{1}{\sqrt{3}}=\dfrac{h}{20+\dfrac{h}{\sqrt{3}}}$

$20\sqrt{3}+h=3h$

$h=10\sqrt{3}$

lf the shadow of a tower is

$\sqrt{3}$ times of its height, the altitude of the sun is

0%
$15^{0}$
0%
$30^{0}$
0%
$45^{0}$
0%
$60^{0}$

Explanation

Let the height of tower be

$h$ .

$\tan \theta = \dfrac{h}{\sqrt{3}h}$

$\Rightarrow \tan \theta = \dfrac{1}{\sqrt{3}}$

$\Rightarrow \theta = 30^{o}$

Two poles of heights

$6$ m and

$11$ m stand vertically on a plane ground. If the distance between their feet is

$12$ m, find the distance between their tips.

0%
$13$ m
0%
$15$ m
0%
$14$ m
0%
none of the above

Explanation

From the given figure, the distance between the tips of the poles forms a right angled triangle with

$5$ m and

$12 m$ as its hypotenuse.
Pythagoras theorem for right triangle,

${Hyp}^{2} = ({Side1})^{2} + ({Side2})^{2}$
So, required distance

$= \sqrt {{5}^{2} + {12}^{2} } = \sqrt {25 + 144} = \sqrt {169} = 13 m$

The angle of elevation of the top of a hill when observed from a certain point on the horizontal plane through its base is

$30^{0}$ . After walking 120 meters towards it on level ground the elevation is found to be

$60^{0}$ . Find the height of the hill(in meters).

0%
120
0%
$60\sqrt{3}$
0%
$120\sqrt{3}$
0%
60

Explanation

From the triangle

$ABC$ , let

Height of the hill

$(BC) = h$

$AD = 120\,\mathrm{m}$ and

$DB = d$

In the

$\triangle DBC, \angle B = 90^\circ$

$\tan 60^\circ = \dfrac{BC}{DB}$

$\,\Rightarrow\sqrt{3} = \dfrac{h}{d}$ ........(1)

In the

$\triangle ABC, \angle B = 90^\circ$

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

$\,\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{h}{d+120}$ ......(2)

Dividing (1) by (2)

$\dfrac{\dfrac{h}{d}}{\dfrac{h}{d+120}} = \dfrac{\sqrt 3}{\dfrac{1}{\sqrt3}}$

$\Rightarrow \dfrac{d+120}{d} =3$

$\Rightarrow 3d = d + 120$

$\Rightarrow d = 60$

Substituting the value of

$d$ in (1), we get

$\sqrt{3} = \dfrac{h}{60}$

$\Rightarrow h = 60\sqrt{3}$

$\therefore$ Height of the hill

$=60\sqrt3 \mathrm{m}$

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 2 - 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 2 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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