MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 8

From the top of the house 18 m 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%
$\displaystyle 18\sqrt{3}\ \text{m}$
0%
$\displaystyle 18(\sqrt{3}+1)\ \text{m}$
0%
$\displaystyle 18(\sqrt{3}-1)\ \text{m}$
0%
$\displaystyle 6\sqrt{3}\ \text{m}$

Explanation

Let

$AB$ be the house and

$CD$ the tower.

Then in

$\triangle ABC$

$\displaystyle AC=AB\cot 30^{\circ}$

Given that:

$AB=18\ \text{m}$

$AC=18\sqrt{3}\ \text{m}$

$\displaystyle BE=AC=18\sqrt{3}\ \text{m}$

Also

$\displaystyle DE=BE\tan 45^{\circ}$

$=BE=18\sqrt{3}\ \text{m}$

$\displaystyle \therefore CD=CE+DE=AB+DE$

$=\left ( 18+18\sqrt{3} \right )\ \text{m}$

$\displaystyle = 18\left ( \sqrt{3}+1 \right )\ \text{m}$

A flagstaff of height

$\displaystyle \frac{1}{5}$ of the height of a tower id mounted on the top of the flagstaff as seen from the ground is

$\displaystyle 45^{0}$ and the angles of elevation of the top of a tower as seen from the same place is

$\displaystyle \theta$ , then the value of

$\displaystyle \tan \theta$ is

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

Explanation

Let AB be the tower and BC the flag -staff.Let
AB=h,

$\displaystyle BC=\frac{1}{5}h$

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

$\displaystyle \frac{OA}{AC}=\cot 45^{0}=1$

$\displaystyle \Rightarrow OA=AC=\left ( h+\frac{1}{5}h \right )$

$\displaystyle \tan \theta =\frac{AB}{AO}=\left ( h\div \frac{6}{5}h \right )=\frac{5}{6}$

From the top of a church spire

$96$ m high the angle of depression of two vehicles on the road at the same level as the base of the spire and on the same of it are

$\displaystyle x^{0}$ and

$\displaystyle y^{0}$ , where

$\displaystyle \tan x^{0}$ =

$\displaystyle \frac{1}{4}$ and

$\displaystyle y^{0}=\frac{1}{7}$ . The distance between the vehicles is

0%
$384$ m
0%
$672$ m
0%
$288$ m
0%
None of these

Explanation

$\displaystyle \frac{p}{96}=\cot x^{0}=4$

$\displaystyle \Rightarrow p=384m$

$\displaystyle \frac{p+q}{96}=\cot y^{0}=7$

$\displaystyle \Rightarrow p+q=96\times 7m=672m$

$\displaystyle \therefore CD=q=\left ( 672-384 \right )m=288m$

A ladder of a fire truck is elevated to an angle of

$60^{\circ}$ and extended to a length of

$70\ feet$ . If the base of the ladder is

$7\ feet$ above the ground, how many feet above the ground does the ladder reach?

0%
$35\ ft$
0%
$42\ ft$
0%
$35\sqrt { 3 }\ ft$
0%
$7+35\sqrt { 3 }\ ft$
0%
$7+42\sqrt { 3 }\ ft$

Explanation

$\triangle BAC,$

$\sin { 60° } =\dfrac { BC }{ AB } =\dfrac { BC }{ 70 }$

$\Rightarrow \dfrac { \sqrt { 3 } }{ 2 } =\dfrac { BC }{ 70 }$

$\Rightarrow BC=\dfrac { 70\sqrt { 3 } }{ 2 }$

$\Rightarrow BC=35\sqrt { 3 }$

$\therefore$ The ladder reaches

$\left( 7+35\sqrt { 3 } \right) m$ above the ground.

Hence, the answer is

$\left( 7+35\sqrt { 3 } \right).$

A kite flying at a height of 75 m from the ground is attached to a straight string Which is inclined at an angle of

$\displaystyle 60^{\circ}$ to the ground The length of the string is

0%
75 m
0%
$\displaystyle 50\sqrt{3}m$
0%
50 m
0%
$\displaystyle 75\sqrt{3}m$

A man in a boat rowed away from a cliff

$50$ m high It takes

$2$ minutes to change the elevation at the top of the cliff from

$\displaystyle 60^{\circ}$ to

$\displaystyle 45^{\circ}$ The speed of the boat is

0%
$\displaystyle \frac{9-3\sqrt{3}}{2}km/h$
0%
$\displaystyle \frac{9+3\sqrt{2}}{2}km/h$
0%
$\displaystyle \frac{9\sqrt{3}}{2}km/h$
0%
$\displaystyle 9\sqrt{3}km/h$

Explanation

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

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

$\displaystyle \therefore y=\frac{150}{\sqrt{3}}$

$\displaystyle \tan 45^{\circ}=\frac{150}{x+y}$ or

$x+y=150$

$\displaystyle \therefore x=150-\frac{150}{\sqrt{3}}$
Distance travelled in

$2$ min=

$\displaystyle 150-\frac{150}{\sqrt{3}}m$

$\displaystyle \therefore$ Speed=

$\displaystyle \frac{150\sqrt{3}-150}{\sqrt{3}}\times 30$
=

$\displaystyle \frac{150\left ( \sqrt{3}-1 \right )\times 30}{1000\sqrt{3}}$ km/h
=

$\displaystyle \frac{9-3\sqrt{3}}{2}$ km/h
1035094_380264_ans_e919ad90ef3b443ca03ec8dd44f33590.bmp

If the object to be viewed is below the level of the observer, then the angle by which the observer lowers his head down is called _____.

0%
Angle of elevation
0%
Angle of depression
0%
Line of sight
0%
Horizontal level

A kite is flying with the string inclined at

$\displaystyle 45^{\circ}$ to the horizontal If the string is straight and 50 m long the height at which the kite is flying is

0%
$\displaystyle 25\sqrt{2}m$
0%
$\displaystyle 50\sqrt{2}m$
0%
25 m
0%
50 m

Explanation

$\displaystyle \sin 45^{\circ}=\frac{h}{50}$ or

$\displaystyle \frac{1}{\sqrt{2}}=\frac{h}{50}$
or h=

$\displaystyle \frac{50}{\sqrt{2}}=\frac{2\times 25}{\sqrt{2}}=25\sqrt{2}$
1297519_380241_ans_e7b79fd79193422b995039036af8066a.jpg

If the given object is above the level of the observer, then the angle by which the observer raises his head is called _____.

0%
Angle of depression
0%
Angle of elevation
0%
Line of sight
0%
Horizontal level

The angle of elevation at the top of a tower from a point 10 m from the foot of the tower is

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

0%
$\displaystyle 10 \sqrt{3}$
0%
$\displaystyle 10/ \sqrt{3}$
0%
$\displaystyle 10 \sqrt{3}/3$
0%
10

Explanation

$\displaystyle \tan 30^{\circ}=\frac{h}{10}=\frac{1}{\sqrt{3}}=\frac{h}{10}$

$or$

$h=\displaystyle \frac{10}{\sqrt{3}}$

1297486_380171_ans_7e73bbd8334c4a5d85a53284e0766ed3.bmp

An observer

$1.5$ m tall is

$28.5$ m away from a tower and the angle of elevation from the eye of the observer is

$45^o$ . The height of the tower is:

0%
$27$ m
0%
$30$ m
0%
$28.5$ m
0%
None of these

Explanation

Let

$AB$ be the observer and

$CD$ be the tower also refer the given figure :

$AC = BE = 28.5 m$

$AB$ =

$CE$ =

$1.5$ m,

Let

$DE$ =

$h$ m

$\therefore$ From right angled

$\triangle BED$ ,

tan 45

$^o=\displaystyle{\frac{DE}{BE} \Rightarrow 1 = \frac{h}{28.5}}$

$h = 28.5$

$\therefore$ Height of tower $$CD

$= h + 1.5$

$= 28.5 + 1.5$

$= 30\ m$

At a certain instant the ratio of the length of a pilar and its shadow are in ratio

$1:$

$\sqrt3$ . At that instance, the angle of elevation of the sun is _____

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

Explanation

Here, AB is the pillar and AC is the shadow,

$\angle$ ACB =

$\angle \theta$
then,

$\displaystyle{\frac{AB}{AC} = \frac{1}{\sqrt3}}$ = tan 30

$^o$

$\therefore$

$\theta$ = 30

$^o$ .

$\therefore$ Angle of elevation of sun is 30

$^o$ .

SO, option A is correct.

A pole

$6\ \text{m}$ high casts a shadow of

$2$

$\sqrt 3\text{ m}$ on the ground. At that instance, the sun's elevation is :

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

Explanation

Let the sun's elevation be

$\theta$

$\triangle ABC, \angle A =90^{\circ}$ ,

$\therefore \tan \theta = \dfrac {AB}{AC} = \dfrac {6}{2\sqrt 3}$

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

$\therefore \theta = 60^{\circ}$

The height of a tower is

$100 \sqrt3$ . The angle of elevation of its top from a point

$100 m$ away from its foot(base) is ______.

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

Explanation

$AB =$ tower,

$O$ is the point of observation.

Given that height of the tower is $$100

\sqrt3$$

$\therefore$

$AB = 100$

$\sqrt3$ ,

$OA = 100\ m$

$\therefore$

$\tan$

$\theta$ =

$\displaystyle{\frac{AB}{OA} = \frac{100\sqrt3}{100}}$

$=$

$\sqrt3$

$\therefore \theta$

$= 60$

$^o$

$\therefore$ Angle of elevation is

$60^o$ .

So, option C is correct.

The string of a kite is

$100$ m long and it makes an angle of 60

$^o$ with the horizontal. If there is no slack in the string, the height of the kite above the ground is _____

0%
$100$ $\sqrt3$ m
0%
$50$ $\sqrt3$ m
0%
$50$ $\sqrt2$ m
0%
$100$ m

Explanation

Let

$AB$ be the string of the kite

$B$ and

$AC$ be the horizontal line.

$\therefore$ BC

$\perp$ AC, AB

$=$ 100 m,

$\angle$

$=$ 60

$^o$

$\sin 60^o= \displaystyle{\frac{BC}{AB} \Rightarrow \frac{\sqrt3}{2} = \frac{h}{100} \Rightarrow} h = 50\sqrt 3$ m

$\therefore$ height of the kite above ground is 50

$\sqrt3$ m

So, option B is correct.

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

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

$^o$ . The height of the tower is _____.

0%
$30$ m
0%
$10$ $\sqrt3$ m
0%
$10$ m
0%
$10$ $\sqrt2$ m

Explanation

AB = height of tower

$= h, \theta = 30^o, OA = 30$ m

$\therefore$

$\displaystyle{\frac{AB}{OA}}$ = tan

$\theta$

$\Rightarrow$

$\displaystyle{\frac{1}{\sqrt3} = \frac{h}{30}}$

$\Rightarrow$ h = 10

$\sqrt3$

$\therefore$ Height of the tower is 10

$\sqrt3$ m.

So, option B is correct.

The angle of elevation of the top of a tower at a distance 30 m from its foot on a horizontal plane is found to be 30

$^o$ . The height of the tower is _____

0%
$17.3$ m
0%
$57.96$ m
0%
$17.8$ m
0%
$173$ m

Explanation

Let AB be the tower and O be the point of observation,

then OA = 30 m and

$\angle$ AOB = 30

$^o$

$\therefore$ In

$\triangle$ OAB, tan 30

$^o$ =

$\displaystyle{\frac{AB}{OA} \Rightarrow \frac{1}{\sqrt3} = \frac{AB}{30} \Rightarrow AB = \frac{30}{\sqrt3}}$

$\therefore$ AB =

$\displaystyle{\frac{30 \times \sqrt3}{\sqrt3 \times \sqrt3}}$ = 10

$\sqrt3$ = 10 x 1.73 = 17.3

$\therefore$ Height of the tower is 17.3 m.

From the top of a tower, the angle of depression of a man standing 40 m away from the tower isThen the height of the tower is:

0%
40 $\sqrt3$ m
0%
$40$ m
0%
$20$ m
0%
$45$ m

Explanation

Referring figure, let AB be the tower and C be the position of the man.

$\therefore BC = 40$ m and

$\angle ACB = 45$

$^o$ ...(alternate angles)
In

$\triangle$ ABC

$\tan 45^o$ =

$\displaystyle{\frac{AB}{BC}}$

$\therefore$ 1 =

$\displaystyle{\frac{AB}{40}}$

$\therefore$ The length of the tower (height of tower) is 40 m

From a certain point

$100m$ away from the foot of the tower, the angle of elevation of the top of the tower is

$60$

$^o$ . The height of the tower is _____.

0%
$50$ $\sqrt3$ m
0%
$100$ $\sqrt3$ m
0%
$\dfrac{100}{\sqrt3}$ m
0%
$\dfrac{200}{\sqrt3}\ m$

Explanation

REF image given in the question.

Let say C be the point and AB =

$h$ be the height of the tower.

$\triangle ABC$ ,

$\angle C$

$= 60^o, AC = 100\ m$ .

$\therefore$

$\displaystyle{\dfrac{AB}{AC}}$

$= \tan 60^o = \sqrt3$

$\Rightarrow$

$\displaystyle{\dfrac{h}{100}}$

$=$

$\sqrt3$

$\therefore$

$h = 100$

$\sqrt3$

Height of the tower is

$100$

$\sqrt3$ m.

So, option B is correct.

$P$ is a point on the segment joining the feet of two vertical poles of heights

$a$ and

$b$ . The angles of elevation of the tops of the poles from

$P$ are

$45^0$ each. Then, the square of the distance between the tops of the poles is

0%
$\displaystyle \frac {a^2\, +\, b^2}{2}$
0%
$a^2\, +\, b^2$
0%
$2(a^2\, +\, b^2)$
0%
$4(a^2\, +\, b^2)$

Explanation

Let

$AD$ be the pole of height

$a$ and

$BC$ be the pole of height

$B$

$\Delta$

$APD$ , we have

$\tan\, 45^o\, =\, \displaystyle \frac {AD}{AP}$

$\Rightarrow 1\, =\, \displaystyle \frac {a}{AP}$

$\, \Rightarrow\, AP\, =\, a$

And in

$\Delta$

$BPC$ , we have,

$\tan 45^o$

$=\, \displaystyle \frac {BC}{PB}\,$

$\Rightarrow 1\, =\, \displaystyle \frac {b}{BP}$

$\, \Rightarrow\, BP\, =\, b$

$\therefore$

$DE=AB$

$= AP+BP$

$=a + b$

and

$CE =BC-BE= b - a$

$\Delta$

$DEC$ , applying pythagoras theorem,

$DC^2=DE^2\, +\, EC^2$

$=\, (a\, +\, b)^2\, +\, (b\, -\, a)^2$

$=a^2+2ab+b^2+a^2-2ab+b^2$

$=\, 2(a^2\, +\, b^2)$

A vertical pole subtends an angle

$\tan^{-1}\left (\dfrac {1}{2}\right )$ at a point P on the ground. If the angles subtended by the upper half and the lower half of the pole at P are respectively

$\alpha$ and

$\beta$ then

$(\tan \alpha, \tan \beta) =$

0%
$\left (\dfrac {1}{4}, \dfrac {1}{5}\right )$
0%
$\left (\dfrac {1}{5}, \dfrac {2}{9}\right )$
0%
$\left (\dfrac {2}{9}, \dfrac {1}{4}\right )$
0%
$\left (\dfrac {1}{4}, \dfrac {2}{9}\right )$

Explanation

From the diagram,

$\tan(\alpha +\beta) = \dfrac12= \dfrac{h}{d}$ ,

$\tan\beta = \dfrac{h}{2d}=\dfrac14$

$\therefore \dfrac{\tan{\alpha}+\frac14}{1-\frac{\tan{\alpha}}{4}} = \dfrac12$

$\Rightarrow \tan{\alpha}= \dfrac29$

The horizontal distance between two towers is

$60\text{ m}$ and the angle of depression of the top of the first tower as seen from the top of the second is

$30^o$ . If the height of the second tower be

$150\text{ m}$ , then the height of the first tower is

0%
$(150 + 60\sqrt{3}) \text{ m}$
0%
$(150 - 60\sqrt{3}) \text{ m}$
0%
$(150 + 20 \sqrt{3}) \text{ m}$
0%
None of these

Explanation

As the Angle of Depression is given from the top of Second Tower

Hence, Second Tower is Larger than the First,so we can clearly see that, Height of First Tower must be less than the Height of the Second Tower

$(=150 \ m)$

We can see that,

$BCDE$ is a Rectangle

so,

$BC=ED=60$

$BE=CD=h\ (Let)$

Now, in

$\triangle AED$

$\tan60^o=\dfrac{AE}{ED}\Rightarrow \sqrt3=\dfrac{AE}{60}\Rightarrow AE=60\sqrt3$

So, Height of First Tower is

$h=BE=AB-AE=150-60\sqrt3$

A person observes the top of a tower from a point

$A$ on the ground. The elevation of the tower from this point is

$60^{\circ}$ . He moves

$60\ m$ in the direction perpendicular to the line joining

$A$ and base of the tower. The angle of elevation of the tower from the point is

$45^{\circ}$ . Then the height of the tower (in meters) is

0%
$60\sqrt {\displaystyle \frac{3}{2} }$
0%
$60\sqrt {2 }$
0%
$60\sqrt {3 }$
0%
$60\sqrt {\displaystyle \frac{2}{3} }$

Explanation

Let the base of tower be

$B$ and the top be

$T$

Initial position is

$A$ . Let the new position be

$C$

$BT=\text{height of the tower}=h$

$\angle TAB=60^{\circ }$ and

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

$tan60^{\circ }=\dfrac{h}{AB}\\\Rightarrow AB=\dfrac{h}{\sqrt{3}}$

Now in

$\Delta TCB$

$tan45^{\circ }=\dfrac{h}{CB}\Rightarrow CB=h$

Applying Pythagoras theorem in

$\triangle CAB,$

$CB^{2}=CA^{2}+AB^{2}$

$\Rightarrow h^{2}=60^{2}+(\dfrac{h}{\sqrt{3}})^{2}\Rightarrow h=60\sqrt{\dfrac{3}{2}}m$

So, Option (A) is correct

An aeroplane leaving from Bismarck travels on a bearing of

$120^o$ , as shown in the figure. If the plane is

$295$ miles directly east of Pierre, how far apart are Bismarck and Pierre? select your answer to the nearest mile.

0%
$170$
0%
$160$
0%
$150$
0%
$140$

Explanation

The angle which the plane makes with the line joining Bismarck and Pierre is

$60^o$

So,

$\tan(60^o) = \cfrac{295}{x}$

where,

$x$ is the distance between Bismarck and Pierre.

$\therefore x = \cfrac{295}{\sqrt{3}} = 170.31$ miles, the nearest integer being

$170$ miles

In the figure, the length of

$\overline{QR}$ is

$4$ , and

$T$ is the midpoint of

$\overline{QS}$ . What is the length of

$\overline{RS}$ ?

0%
$4$
0%
$2\sqrt{3}$
0%
$2\sqrt{13}$
0%
$4\sqrt{13}$

Explanation

In right angled triangle

$\Delta RQT, QR = 4$ and

$\angle RTQ = 30^o$ ,

$\therefore RT = \dfrac{4}{\sin(30^o)} = 8$

Also,

$QT = \dfrac{4}{\tan(30^o)}$

$\Rightarrow QT = 4\sqrt{3}$

$\Rightarrow QS = 2 \times QT = 8\sqrt{3}$

Since

$\angle SQR = 90^o, (RS)^2 = (QR)^2 + (QS)^2$

$\therefore (RS)^2 = 16 + (8\sqrt{3})^2$

$(RS)^2 = 16 + 192 = 208$

$RS = 4\sqrt{13}$ units

The angle of elevation of a stationary cloud from a point

$2500$ m above a lake is

$15^0$ and from the same point the angle of depression of its reflection in the lake is

$45^o$ . The height (in meters) of the cloud above the lake, given that

$\cot\,15^o=2+\sqrt{3}$ , is

0%
$2500$
0%
$2500\sqrt{2}$
0%
$2500\sqrt{3}$
0%
$5000$

Explanation

Let

$EB=x$ m, then

$\tan 15^0=\dfrac{h}{x}$

$\Rightarrow x=h \cot 15^0$ ...(1)

and from triangle EBD, we can get

$\tan 45^o=\dfrac{h+5000}{x}$

$\rightarrow x\cdot 1=h+5000$ ...(2)

From (1) and (2)

$\Rightarrow h \cot 15^0=h+5000$

$\Rightarrow h (2+\sqrt 3)=h+5000$

$\Rightarrow h=\dfrac{5000}{\sqrt{3}+1} = \dfrac{5000(\sqrt{3}-1)}{3-1}=2500(\sqrt{3}-1)m$

$\therefore$ height of cloud is

$2500+2500(\sqrt{3}-1)$

$=2500\sqrt{3}$ m

An aeroplane leaving from Bismarck travels on a bearing of

$120^o$ , as shown in the figure. If the plane flew at an average rate of

$400$ miles per hour, how many minutes (approx) had it been in the air when it was

$295$ miles from Pierre?

0%
$51$
0%
$56$
0%
$29$
0%
$67$

Explanation

The angle that the plane makes with the line joining Bismarck and Pierre is

$60^o$

We thus have

$\sin (60^o) = \cfrac{295}{x}$ where

$x$ is the distance flown by the plane.

That distance becomes

$x = \cfrac{295 \times 2}{\sqrt{3}}$

$x = 340.63$ mi

Since the speed is

$400$ miles per hour, the time becomes

$340.63/400 \times 60 = 51.0945$ minutes

The angle of depression of a car parked on the road from the top of a

$150$ m high tower is

$\displaystyle { 30 }^{ \circ }$ . The distance of the car from the tower (in metres) is

0%
$\displaystyle 50\sqrt { 3 }$
0%
$\displaystyle 150\sqrt { 3 }$
0%
$\displaystyle 150\sqrt { 2 }$
0%
$\displaystyle 75$

Explanation

Given: Height $= 150$ m

Angle of depression $= 30^o$

Consider the diagram,

In $\Delta ABC$ ,

Let, distance of the car from tower $BC = x$ m,

In $\Delta ABC, \tan 30^o = \dfrac{AC}{AB}$

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

Therefore, $AC = 150\sqrt{3}$

So, distance between the tower and car is $150\sqrt{3}$ m

A pole is

$20$ feet high. A taut wire that is

$46$ feet extends from the top of the pole to the ground. What is the angle of depression, to the nearest degree, from the top of the pole to the bottom of the wire?

0%
$23$
0%
$26$
0%
$43$
0%
$64$
0%
$67$

Explanation

The given setup will form a right angled triangle with hypotenuse length

$46$ feet and height of pole is

$20$ fee

Let the angle of depression be

$\alpha$

$\because \sin \alpha=\dfrac{opposite}{hypotenuse}$

$\therefore$ Angle of depression

$=\alpha =\sin ^{ -1 }{ \left( \dfrac { 20 }{ 46 } \right) } =26°$

Hence, the angle of depression is

$26^{o}$

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

$36\ m$ and

$64\ m$ from its base and in the same straight line with it are complementary. What is the height of the tower?

0%
$50\ m$
0%
$48\ m$
0%
$25\ m$
0%
$24\ m$

Explanation

Given

$AB$ is the tower.

$P$ and

$Q$ are the points at distance of

$36m$ and

$64m$ respectively.

$PB = 36m, QB = 64m.$

Let angle of elevation from

$P$ be

$\alpha$ and angle of elevation from

$Q$ be

$\beta$ .

Given that

$\alpha$ and

$\beta$ are supplementary. Thus,

$\alpha +\beta = 90$

In triangle

$ABP$ ,

$\tan \alpha = \dfrac{AB}{BP}$ – (i)

In triangle

$ABQ$ ,

$\tan \beta = \dfrac{AB}{BQ}$

$\tan (90 – \alpha ) = \dfrac{AB}{BQ}$

$\cot \alpha =\dfrac{ AB}{BQ}$

$\dfrac{1}{\tan \alpha } = \dfrac{AB}{BQ}$

So,

$\tan \alpha =\dfrac{ BQ}{AB}$ – (ii)

From (i) and (ii)

$\dfrac{AB}{BP} = \dfrac{BQ}{AB}$

$AB^2 = BQ \times BP$

$AB^2 = 36 \times 64$

$AB^2 = 2304$

Therefore,

$AB = 48$ .

Hence, height of tower is

$48m$

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 8 - 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 8 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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