MCQ Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 13

From the top of a cliff

$24\mathrm { mt }$ height, a man observes the angle of depression of a boat is to be

$60 ^ { \circ } .$ The distance of the boat from the foot of the cliff is

0%
$8 \sqrt { 3 } \mathrm { mt }$
0%
$8 \sqrt { 2 } \mathrm { mt }$
0%
$8 \sqrt { 5 } \mathrm { mt }$
0%
$8 \mathrm { mt }$

Explanation

$\begin{array}{l} In\, ABC, \\ \tan { 60^{ 0 } } =\cfrac { { AB } }{ { BC } } \\ \sqrt { 3 } =\cfrac { { 24 } }{ { BC } } \\ BC=\cfrac { { 24 } }{ { \sqrt { 3 } } } \times \cfrac { { \sqrt { 3 } } }{ { \sqrt { 3 } } } \\ =8\sqrt { 3 } \end{array}$

Hence, this is the answer.

1228638_1322880_ans_4b3f85481c7a4f53837b452cca7ed4ad.JPG

If the angle of elevation of a cloud from a point P which is

$25$ m above a lake be

$30^0$ and the angle of depression of reflection of the cloud in the lake from P be

$60^0$ , then the height of the cloud (in meters) from the surface of the lake is:

0%
$42$
0%
$50$
0%
$45$
0%
$60$

Explanation

$tan 30^0 = \frac{x}{y} \Rightarrow y = \sqrt{3}x$ .......... (i)

$tan 60^0 = \frac{25 + x + 25}{y}$

$\Rightarrow \sqrt{3}y = 50 + x$

$\Rightarrow 3x = 50 + x$

$\Rightarrow x = 25 \ m$

$\therefore$ Height of cloud from surface = 25 + 25 = 50 m

1143900_1332503_ans_880c0ebdc56a4c69ad7824ca9c49a7f6.jpg

A tower of height $'h'$ standing vertically at the center of a square f side length $'a'$ subtends the same angle $'\theta '$ at all the corner points of the square. Then

0%
$2h^{2} = a^{2} tan^{2} \theta$
0%
$a^{2} = 2h^{2}tan^{2} \theta$
0%
$a^{2} = 2h^{2}cot^{2} \theta$
0%
$2a^{2} = h^{2}cot^{2} \theta$

Explanation

$\triangle EFC$

$\tan { \theta } =\cfrac { EF }{ EC } =\cfrac { h }{ \cfrac { a }{ \sqrt { 2 } } } \\ \Rightarrow \cfrac { a }{ \sqrt { 2 } } \tan { \theta } =h\\ \Rightarrow a\tan { \theta } =\sqrt { 2 } h$

Squaring both the sides

$\Rightarrow { 2h }^{ 2 }={ a }^{ 2 }\tan ^{ 2 }{ \theta }$

1222827_1381942_ans_9063b0a3395a4bfc9327049c77fcdfac.png

The angle of elevation of the top of a tower at a distance of 30 m from its foot is

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

0%
$20$ m
0%
$30\sqrt{3}$ m
0%
$15\sqrt{2}$ m
0%
$\cfrac{15}{\sqrt{2}}$ m

Explanation

Let the height of the tower be

$AB$ and the point be

$C$

Hence

$BC=30\ m$

and

$\angle ACB=60^\circ$

$\triangle ABC,$

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

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

$\therefore AB=30\sqrt 3\ m$

1477347_1418986_ans_46af0f4818ab4570836604e23c6b4953.jpg

If a pole

$12\ { m }$ high casts a shadow

$4\sqrt { 3 }\ { m }$ long on the ground then the angle of elevation is

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

Explanation

Here,

Height of pole =

$12$

$m$

Length of shadow =

$4\sqrt{3}$

$m$

Now, angle of elevation,

$\tan \theta = \dfrac{perpendicular}{base}$

$\tan \theta = \dfrac{12}{4\sqrt{3}} =\dfrac{3}{\sqrt{3}} =\sqrt{3}$

$\tan \theta = \tan 60^o$

Hence,

$\theta = 60^o$

Hence, option (c) is correct.

1673233_1453066_ans_83dbe7c21ace43babf67c5c6557a0ad0.PNG

The shadow of a tower of height

$(1+\sqrt { 3 } )$ m standing on the ground is found to be 2 m longer when the sun's elevation is

$30^0$ , then when the sun's elevation was

0%
$30^0$
0%
$45^0$
0%
$60^0$
0%
$75^0$

Explanation

From given, we have,

$\tan 30 =\dfrac{1+\sqrt 3}{x+2}$

$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{1+\sqrt 3}{x+2}$

$\Rightarrow x+2=\sqrt 3+3$

$\therefore x=1+\sqrt 3$

now,

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

$\Rightarrow \tan \theta =1$

$\therefore \theta =\tan ^{-1}1=45^{\circ}$

A man is standing on the deck of a ship, which is

$10\ m$ above water level. He observes the angle of elevation of the top of a light house as

$60^0$ and the angle of depression of the base of lighthouse as

$30^0$ . Find the height of the light house.

0%
$40\ m$
0%
$50\ m$
0%
$60\ m$
0%
$30\ m$

Explanation

Let

$CD$ be the lighthouse and suppose the man is standing on the deck of a ship at point

$A$

The angle of depression of the base

$C$ of the lighthouse

$CD$ observed from

$A$ is

${30}^{\circ}$ and the angle of elevation of the top

$D$ of the lighthouse

$CD$ observed from

$A$ is

${60}^{\circ}$

$\therefore\,\angle{EAD}={60}^{\circ}$ and

$\angle{BCA}={30}^{\circ}$

$\triangle{AED},$

$\tan{{60}^{\circ}}=\dfrac{DE}{EA}$

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

$\therefore\,h=\sqrt{3}x$ ..........

$(1)$

$\triangle{ABC},$

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

$\therefore\,\dfrac{1}{\sqrt{3}}=\dfrac{10}{x}$

$\therefore\,x=10\sqrt{3}$

Substituting

$x=10\sqrt{3}$ in equation

$(1)$ we get

$h=\sqrt{3}x=\sqrt{3}\times 10\sqrt{3}=30$

$\therefore\,DE=30\ m$

$\therefore\,CD=CE+ED=(10+30)\ m=40\ m$

Thus, the height of the lighthouse is

$40\ m$

1463805_1465535_ans_4438fca6b6894267beecc77b75a41c32.png

A ladder

$9m$ long reaches a point

$9m$ below the top of a vertical flagstaff. From the foot of the ladder, the elevation of the flagstaff is

$60^0$ . What is the height of the flagstaff ?

0%
$9$ m
0%
$10.5$ m
0%
$13.5$ m
0%
$15$ m

Explanation

$CD = AD = 9 \; m$

$\because \triangle{ACD}$ is isosceles.

$\angle{CAB} = 60°$

$\therefore \angle{ACB} = \angle{CAD} = 30°$

$\therefore \angle{BAD} = 30°$

$\triangle{ABD},$

$\sin{30°} = \cfrac{BD}{AD}$

$\Rightarrow BD = AD \sin{30°} = 4.5$

Therefore,

$BC = BC + DC = 4.5 + 9 = 13.5 \; m$

1441117_1688835_ans_bd99e22fffdc4274b506876e3ce38bbd.png

The tops of two poles of height

$18$ m and

$10$ m are connected by wire. If the wire makes an angle of measure

$45^o$ with the horizontal, then the length of wire is _________.

0%
$16$ m
0%
$18$ m
0%
$8\sqrt{2}$ m
0%
$8$ m

Explanation

Let

$AB$ be the pole having height

$18\ m$ and

$CD$ be the pole of height

$10\ m$

$AE=AB-EB$

$[DC=EB]$

$AE=18-10=8\ m$

In

$\triangle ADE,$

$\sin 30^\circ=\dfrac{AE}{AD}$

$\Rightarrow \dfrac{1}{2}=\dfrac{8}{AD}$

$\Rightarrow AD=2\times 8$

$\Rightarrow AD=16\ m$

Hence, the length of the wire is

$16\ m$

1496252_1599053_ans_7346c98556e44b9dbfcd0e79ce364bdb.jpg

From two points A and B on the same side of a building the angles of elevation of the top of the building is

$\displaystyle 30^{\circ}$ and

$\displaystyle 60^{\circ}$ respectively. If height of the building is 10 m, find the distance between A and B correct to two decimal places

0%
10.66 m
0%
13.43 m
0%
11.55 m
0%
12.26 m

The top of a broken tree touches the ground at a distance of

$12 \,m$ from its base. If the tree is broken at a height of

$5 \,m$ from the ground hen the actual height of the tree is

0%
$25 \,m$
0%
$13 \,m$
0%
$18 \,m$
0%
$17 \,m$

Explanation

Let

$AB$ is the unbroken part of the tree and

$BC$ is the broken part of it.
Here, triangle ABC is right-angled triangle.

$\Rightarrow(BC)^{2}=(AB)^{2}+(AC)^{2}$

$\Rightarrow(BC)^{2}=5^{2}+12^{2}$

$\Rightarrow(BC)^{2}=25+144$

$\Rightarrow(BC)^{2}=169$

$\Rightarrow(BC)^{2}=13m$
Now actual height of tree

$=AB+BC$

$\Rightarrow 5+13$

$\Rightarrow 18m$
1670093_1790132_ans_5ac62df6e736491da6e714a6ddd4be2e.png

The ratio of the length of a tree and its shadow is

$1: \frac{1}{\sqrt{3}}$ . The angle of the sun's elevation is

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

Find the angle of elevation of the top of a tower, whose height is

$100 m$ , at a point whose distance from the base of the tower is

$100 m$ .

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

Choose the correct alternative answer for the following question.
When we see at a higher level, from the horizontal line, angle formed is .............

0%
angle of elevation
0%
angle of depression
0%
0
0%
straight angle

If the shadow of 10 m high tree is

$10 \sqrt{3} \ m$ , then find the angle of elevation of the sun.

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

Explanation

We can draw the figure shown above, based on the given infomation.

From

$\triangle ABC$ ,

$AB =$ height of tree

$= 10\ m$

$BC =$ shadow of tree

We know that,

$\tan \theta=\dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$

Hence,

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

$\Rightarrow \tan \theta = \dfrac {10}{10\sqrt {3}}$

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

$\Rightarrow \tan \theta = \tan 30^{\circ}\quad \quad \left[\because \ \tan 30^o=\dfrac{1}{\sqrt3}\right]$

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

Hence, the angle of the elevation of the Sun is

$30^\circ$ .

2110147_1868689_ans_575eba726d574b269bbe5f04a99d6ebd.png

The shadow length of a vertical pillar is same as the height of pillar, then angle of elevation of sun is___.

0%
$45^{\circ}$
0%
$30^{\circ}$
0%
$60^{\circ}$
0%
$50^{\circ}$

Explanation

Let

$BC$ be a tower with height

$h\ m$ .

Then according to question, its shadow length

$AB$ will be

$h\ m$ .

Let angle of elevation

$= \theta$

In right angled

$\Delta ABC$ , we have

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

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

$\Rightarrow \tan\theta=1$

$\Rightarrow \tan \theta = \tan 45^{\circ}$

$\Rightarrow \theta = 45^{\circ}$

Hence, the angle of elevation is

$45^{\circ}$ .

1856135_1875861_ans_40f743f2862146e1a7bade1ac4d2c541.png

If ratio of length of a vertical rod and length of its shadow is

$1 : \sqrt{3}$ , then angle of elevation of sun is :

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

Explanation

Let length of vertical rod is

$BC$ and length of shadow is

$AB$ .
Let angle of elevation of sun is

$\theta$ then,
From right

$\Delta ABC$ ,

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

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

$\theta = 30^{0}$
Hence, angle of elevation of sun

$= 30^{0}$
So, correct choice is (A).
1857122_1875933_ans_1aaba89c23f84575aab5f27ddec93aa5.png

If angle of elevation of sum is

$45^{0}$ then the length of the shadow casts by a

$12 m$ high tree is

0%
$6\sqrt{3}m$
0%
$12\sqrt{3}m$
0%
$\frac{12}{\sqrt{3}}m$
0%
$12m$

Explanation

Let $BC$ is a tree of height $12 m$ and shadow of tree $AB = x$ m and angle of elevation of sun $= 45^{0}$

i.e., $\angle CAB = 45^{0}$

From right angled $\Delta ABC$ ,

$\tan 45^{0} =\dfrac{BC}{AB}$

$1 =\dfrac{12}{x}$

$x = 12 m$

Hence, length of shadow of tree = $12 m$

So, correct choice is (D).

1857843_1876907_ans_410f97269b2a47f780baa27d1ff8af9c.png

A kite is flying at a height of

$30 m$ from the ground. The length of string from the kite to the ground is

$60 m$ . Assuming that there is no slack in the string, the angle of elevation of the kite at the
ground is :

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

Explanation

Let $A$ is a kite which is $30$ meter high from the ground which is flying with string of $60 m$ long.

Let $\angle ACB = \theta$

From right angled

$\sin \theta =\dfrac{AB}{AC}$

$\sin \theta =\dfrac{30}{60}$

$\sin \theta =\dfrac{1}{2}$

$sin \theta = \sin 30^{0}$

$\theta = 30^{0}$

So, correct choice is (B)

1857817_1876902_ans_b84b27fb7a4841eeb82648032ad946b2.png

If the angle of elevation of a point at the horizontal slope of a hill is

$60^{0}$ . If one has to walk

$300 m$ to reach the top of the hill from the point then the distance of the point from the foot of the hill is:

0%
$300\sqrt{3}m$
0%
$150m$
0%
$150\sqrt{3}m$
0%
$\frac{150}{\sqrt{3}}m$

Explanation

Let $BC$ be the height of the hill and $AB = x$ be the distance of point from the foot.

Also, the length between $A$ to $C$ i.e., $AC = 300 m$ and $\angle BAC = 60^{0}$

From right angled $\Delta ABC$ ,

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

$\dfrac{1}{2} =\dfrac{x}{300}$

$x =\dfrac{300}{2}$

$\therefore x = 150 m$

Hence, distance of point from base is $150 m$ .

So, the correct choice is (B).

1857836_1876905_ans_52145851ff834ab4a8d41d7840a2e9da.png

If the height of a pole is

$6 m$ and the length of its shadow is

$2\sqrt{3} m$ , then the angle of elevation of sun is equal to

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

Explanation

Let

$AB$ is a pole of height

$6 m$

Length of a shadow of pole

$BC = 2\sqrt{3} m$

Let angle of elevation of sun

$\angle ACB = \theta$

From right angled

$\Delta ABC$ ,

$\tan \theta =\frac{AB}{BC}$

$\tan \theta =\frac{6}{2\sqrt{3}}$

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

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

$\theta = 60^{0}$

So, correct choice is (A).

1857808_1876896_ans_94b163738bb74b7e9d0ef07c143aab43.png

$25\ m$ ladder is placed against a vertical wall of a building. The foot of the ladder is

$7\ m$ from the base of the building. If the top of the ladder slips

$4\ m$ , then the foot of the ladder will slide by:

0%
$5\ m$
0%
$8\ m$
0%
$9\ m$
0%
$15\ m$

Explanation

The ladder forms a right angled triangle with the wall and the base with length of the ladder being the length of hypotenuse of this triangle.

Hypotenuse

$= 25$ m

Base

$=7$ m

Height

$=25^2-7^2=24$ m

Now the ladder slides. In the new position, the hypotenuse of the new right angled triangle will be the same.

The top slides down by a distance

$4m$ . Hence, the height is

$24-4=20$ m. Let the base slide by a distance

$x$ . Therefore, hypotenuse

$= 25$ m

Base

$=(7+x)$ m

Height

$=20$ m

By Pythagoras Theorem, we get

$25^2-20^2 = (7+x)^2$

$\therefore (7+x) = 15$

Hence,

$x=8$

The correct answer is Option B.

From the top of cliff

$300\ meters$ high, the top of tower was observed at an angle of depression

${30}^{\circ}$ and from the foot of the tower the top of the cliff was observed at an angle of elevation

${45}^{\circ}$ . The height of the tower is

0%
$50\left( 3-\sqrt { 3 } \right) m$
0%
$200\left( 3-\sqrt { 3 } \right) m$
0%
$100\left( 3-\sqrt { 3 } \right) m$
0%
$None\ of\ these$

Explanation

Let

$AB$ be the cliff with

$AB=300\ m$

Let

$DE$ be the tower with height

$h\$

Let

$BE=x\ m$

In right angles

$\triangle ABE$ ,

$\displaystyle \frac { AB }{ BE } =\tan { { 45 }^{ 0 } }$

$\Rightarrow \dfrac { 300 }{ x } =1$

$\Rightarrow x=300\ m$

In right angled

$\triangle ACD$

$\displaystyle \dfrac { AC }{ CD } =\tan { { 30 }^{ 0 } }$

$\Rightarrow \dfrac { 300-h }{ x } =\dfrac { 1 }{ \sqrt { 3 } }$

$\displaystyle \Rightarrow \frac { 300-h }{ 300 } =\frac { 1 }{ \sqrt { 3 }}$

$\Rightarrow 300\sqrt { 3 } -\sqrt { 3 } h=300$

$\Rightarrow 300\sqrt { 3 } -300=\sqrt { 3 } h$

$\displaystyle \Rightarrow h=\frac { 300\left( \sqrt { 3 } -1 \right) }{ \sqrt { 3 }}$

By rationalize the denominator , we get

$\Rightarrow h=100\sqrt { 3 } \left( \sqrt { 3 } -1 \right) m\quad \\ \Rightarrow h=100\left( 3-\sqrt { 3 } \right)\ m$

From the top of a tree on one side of a street the angles of elevation and depression of the top and foot of a tower on the opposite side are respectively found to be

$\alpha$ and

$\beta$ . If

$h$ is the height of the tree, then the height of the tower is:

0%
$\displaystyle \frac{h\sin(\alpha+\beta)}{\cos\alpha\sin\beta}$
0%
$\displaystyle \frac{h\sin(\alpha+\beta)}{\sin\alpha\cos\beta}$
0%
$\displaystyle \frac{h\cos(\alpha-\beta)}{\cos\alpha\cos\beta}$
0%
$\displaystyle \frac{h\cos(\alpha+\beta)}{\cos\alpha\cos\beta}$

Explanation

Let

$h_{t}+h=H$ be the total height of tower.
Let

$x$ be the distance between tower and tree.

$\displaystyle \tan \beta= \frac{h}{x}$

$\displaystyle \tan \alpha= \frac{h_{t}}{x}$

Now,

$\displaystyle \tan \alpha= \frac{h_{t}}{h \cot \beta}$

$\displaystyle h_{t}=h \dfrac{\tan \alpha}{\tan \beta}$

Total height of the tower,

$\displaystyle H= h+h \dfrac{\tan \alpha}{tan \beta}$

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

Substituting for

$\tan \alpha$ and

$\tan \beta$
we get,

$\displaystyle H= h \left ( \frac{\sin \beta \cos \alpha +\sin \alpha \cos \beta}{\sin \beta \cos \alpha } \right )$
Hence, option 'A' is correct.

A lamp post standing at a point

$A$ on a circular path of radius

$r$ subtends an angle

$\alpha$ at some point

$B$ on the path, and

$AB$ subtends an angle of

$45^{\circ}$ at any other point on the path, the height of the lamp post is

0%
$\sqrt{2}r\cot\alpha$
0%
$\displaystyle \frac{r}{\sqrt{2}}\tan\alpha$
0%
$\sqrt{2}r\tan\alpha$
0%
$\displaystyle \frac{r}{\sqrt{2}}\cot\alpha$

Explanation

Let

$AM$ is the lamp post at point

$A$ of height

$h.$

$AB$ subtends an angle of

$45^\circ$ at any point on the circle. Therefore it will subtend an angle of

$90^\circ$ at the center.

So, we can say that

$\triangle OAB$ is a right-angled triangle with a right angle at

$O.$

By applying pythagoras theorem,

$\begin{aligned}{}O{A^2} + O{B^2}& = A{B^2}\\{r^2} + {r^2} &= A{B^2}\\A{B^2}& = 2{r^2}\\AB& = r\sqrt 2 \end{aligned}$

$\triangle AMB,$

$\tan \alpha = \dfrac {AM}{AB}$

$\Rightarrow \tan \alpha = \dfrac {h}{r \sqrt 2}$

$\Rightarrow h = r \sqrt 2 \tan \alpha$

A tower subtends an angle

$\alpha$ at a point on the same level as the root of the tower and at a second point,

$b$ meters above the first, the angle of depression of the foot of the tower is

$\beta$ . The height of the tower is

0%
$b\cot { \alpha } \tan { \beta }$
0%
$b\tan { \alpha } \tan { \beta }$
0%
$b\tan { \alpha } \cot { \beta }$
0%
None of these

Explanation

From the figure, in

$\triangle ABD$

$\dfrac { AB }{ BD } =\tan { \alpha }$

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

$\Rightarrow h=x\tan { \alpha } \quad\quad\quad\dots(i)$

In right angled

$\triangle BCE$

$\dfrac { BE }{ EC } =\tan { \beta }$

$\Rightarrow \dfrac { b }{ x } =\tan { \beta }$

$\Rightarrow x=\dfrac { b }{ \tan { \beta } } \quad\quad\quad\dots(ii)$

Now, from equation

$(i)$ and

$(ii),$

$h=\dfrac { b }{ \tan { \beta } } \times \tan { \alpha }$

$=b\tan { \alpha } \cot { \beta }$

Hence, height of the tower is equal to

$b\tan\alpha\cot\beta.$

A person walking along a canal observes that two objects are in the same line which is inclined at an angle

$\alpha$ to the canal. He walks a distance

$C$ for the and observes that the objects subtended their greatest angle

$\beta$ then the distance between the object is

0%
$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\beta-\cos\alpha}$
0%
$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\alpha-\cos\beta}$
0%
$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\alpha+\cos\beta}$
0%
$\displaystyle \frac{2C\cos\alpha\cos\beta}{\sin\alpha-\sin\beta}$

Explanation

$\tan \alpha = \dfrac {h_1}{a} - (1), \tan \alpha = \dfrac {h_2}{a + b} - (2), \tan \beta = \dfrac {h_1}{a - c} - (3)$

$\tan \beta = \dfrac {h_2}{a + b - c} - (4)$

$\tan \beta (a + b -c) = \tan \alpha (a + b)$ (From (1) & (3))

$\tan \alpha a =\tan \beta (a - c)$ (From (2) and (4))

$a = \dfrac {\tan \beta c}{\tan \beta - \tan \alpha}$

The angular depression of the top and the foot of a tower as seen from the top of a second tower which is

$150m$ high and standing on the same level as the first are

$\alpha$ and

$\beta$ respectively. If

$\tan\alpha=\dfrac{4}{3}$ and

$\tan\beta=\dfrac{5}{2}$ , the distance between their tops is:

0%
$100m$
0%
$120m$
0%
$110m$
0%
$130m$

Explanation

Let

$AB$ be the length of

$1^{st}$ tower.

Let

$BC$ be the length of

$2^{nd}$ tower.

Here given that CD

$=150 m$

$BE=CD=150m$

From

$\triangle BCE$

$\tan \beta=\cfrac{BE}{CE}$

or,

$\cfrac{5}{2}=\cfrac{150}{CE}$

or,

$CE=150\times \cfrac{2}{5}=60$

$\tan\alpha=\cfrac{4}{3}$

$\sec\alpha=\sqrt {1+\cfrac{16}{9}}\\\ \ \ \ \ \ \ \ =\sqrt {\cfrac{25}{9}}\\\ \ \ \ \ \ \ =\cfrac{5}{3}$ [only positive value taken].

Thus,

$\cos\alpha=\cfrac{3}{5}$

$\cos\alpha=\cfrac{CE}{AC}$

or,

$\cfrac{60}{AC}=\cfrac{3}{5}$

or,

$AC=\cfrac{5}{3} \times 60=100$

$\therefore$ Distance between their tops

$=100m$

Vertical pole

$PO$ stands at the centre

$O$ of a square

$ABCD$ . If

$AC$ subtends an angle at the top

$P$ of the pole, then the angle

$90^{0}$ subtended by a side of the square at

$p$ is

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

At the foot of a mountain the elevation of its peak is found to be

$\displaystyle \frac{\pi }{4}.$ After ascending

$h$ metres toward the mountain up a slope of

$\displaystyle \frac{\pi }{6}$ inclination, the elevation is found to be

$\displaystyle \frac{\pi }{3}.$ Height of the mountain is

0%
$\displaystyle \frac{h}{2}(\sqrt{3}+1)m$
0%
$h(\sqrt{3}+1)m$
0%
$\displaystyle \frac{h}{2}(\sqrt{3}-1)m$
0%
$h(\sqrt{3}-1)m$

Explanation

Let 'A' be the top of hill and 'P' be a point on it's foot.
We have

$\angle APB=\dfrac{\pi }{4}, \angle QPB=\dfrac{\pi }{6}, \angle AQR=\dfrac{\pi }{3}, PQ=h\ m$
In

$\Delta APB,$

$PB=AB \cot \dfrac{\pi }{4}=AB$
In

$\Delta PQ_{1}Q$

$QQ_{1}=PQ \sin \displaystyle \frac{\pi }{6}=\dfrac{h\sqrt{3}}{2}$
In

$\Delta AQR, \tan \displaystyle \dfrac{\pi }{3}=\dfrac{AR}{QR}=\dfrac{AB-QQ_{1}}{PB-PQ_{1}}$

$\Rightarrow \displaystyle \dfrac{AB-\dfrac{h}{2}}{AB-\dfrac{h\sqrt{3}}{2}}=\sqrt{3}$

$\Rightarrow AB=\displaystyle \dfrac{h}{(\sqrt{3}-1)}=\dfrac{h(\sqrt{3}+1)}{2}\ m$

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 13 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Some Applications Of Trigonometry Quiz 13 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.