Some Applications of Trigonometry test

Result

Some Applications of Trigonometry test - 52

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A baloon is observed simultaneously from three points $$A, B$$ and $$C$$ on straight road directly under it. The angular elevation at $$B$$ is twice and at $$C$$ is thrice that of $$A$$. If the distance between $$A$$ and $$B$$ is $$200 \,m$$ and the distance between $$B$$ and $$C$$ is $$100 \,m$$, then the height of ballon is

A
$$50$$

B
$$50\sqrt{3}m$$

C
$$50\sqrt{2}m$$

D
$$None\ of\ these$$

Solution
Question 2
1 / -0

As observed from the top of a light house $$100$$m high from sea-level, the angles of depression of two ships are $$30^0$$ and $$45^0$$. If the ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. (Use $$\sqrt3 = 1.732$$)

A
$$60$$ m

B
$$73.2$$ m

C
$$75.32$$

D
$$78.6$$

Solution
Question 3
1 / -0

A man of height $$6$$ ft. observed the top of a tower at the top and the foot of a pole of height $$10 ft.$$ are $$45^o$$ and $$30^o$$ of elevation and depression respectively. The height of the tower is :

A
$$13.79 ft.$$

B
$$14.59 ft.$$

C
$$14.29 ft.$$

D
none of these

Solution

Let AB be the man and CD be the lower then $$\angle DAE=45^o$$ and $$\angle ACB =30^o$$
In $$\triangle ABC, \tan 30^o=\dfrac{AB}{BC}$$
$$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{6}{BC\Rightarrow BC=6\sqrt 3ft}$$
$$BC=AE=6\sqrt 3ft$$
In $$\triangle ADE, \tabn 45^o=\dfrac{DE}{AE}$$
$$\Rightarrow 1=\dfrac{DE}{6\sqrt 3}\Rightarrow DE=6\sqrt 3ft.$$
$$\therefore $$ Height of tower $$CD=CE+DE=AB+DE$$
$$=6+6\sqrt 3=6+6\times 1.732$$
$$=6+10.392$$
$$=16.391ft$$
Question 4
1 / -0

Length of the shadow of a $$20$$ meters high pole is $$20$$ meter at $$7.30^{o}am$$. The angle of elevation of the sun rays with the ground?

A
$$30^{o}$$

B
$$45^{o}$$

C
$$60^{o}$$

D
$$90^{o}$$

Solution

In the fig
AB is the tower and
BC is denote the shadow
Let $$\theta$$ be angle of elevation of the sun.
$$\Rightarrow \tan\theta =\dfrac{AB}{BC}$$
$$=\dfrac{20}{20}$$
$$\boxed{\tan \theta =1}$$
$$\therefore \boxed {\theta =45^o}$$ $$\because \tan 45=1$$
Question 5
1 / -0

The angle of elevation of the top of a tower at any point on the ground is $$\dfrac{\pi}{6}$$ and after moving $$20$$ meters towards the tower it becomes $${\pi}{3}$$.The height of the tower is equal to :

A
$$10 meters$$

B
$${10}{\sqrt3} meters$$

C
$$\dfrac{10}{\sqrt3} meters$$

D
$$5\sqrt3 meters$$

Solution

We can see from the diagram that $$CD=20\ m$$
Let $$BC=x\\AB=h$$

Now, in $$\triangle ABC$$
$$\tan60^O=\dfrac hx\Rightarrow x\sqrt3=h........................(1)\\\tan30^o=\dfrac h{x+20}\Rightarrow x+20=\sqrt3h\Rightarrow x+20=\sqrt3*x\sqrt3\Rightarrow x+20=3x\Rightarrow x=10$$

from (1)
$$h=x\sqrt3\Rightarrow h=10\sqrt3$$
Question 6
1 / -0

If the elevation of the sun is $$30^0$$, then the length of the shadow cast by a tower of $$150$$ ft height is :

A
$$75\sqrt3$$ ft.

B
$$200\sqrt3$$ ft.

C
$$150\sqrt3$$ ft.

D
none of these

Solution

Let the height of the tower $$AE=h=150ft.$$ and $$EC=x$$

We can see that, in $$\triangle AEC,$$

$$\tan30^o=\dfrac{AE}{EC}=\dfrac{h}{x}$$ {$$\tan30^o = \dfrac{1}{\sqrt3}$$}

$$\Rightarrow x=h\sqrt3=150\sqrt3ft.$$
Question 7
1 / -0

A boy standing on the ground, spots a balloon moving with the wind in a horizontal line at a constant height. The angle of elevation of the balloon from the boy at an instant is $$60^{\circ}$$. After 2 minutes, from the same point of observation, the angle of elevation reduces to $$30^{\circ}$$. If the speed of wind is $$29\sqrt{3}$$ m/min. then, find the height of the balloon from the ground level.

A
$$87\ m$$

B
$$78\ m$$

C
$$92\ m$$

D
$$75\ m$$

Solution
Question 8
1 / -0

The angle of elevation of the top of the tower at the eye of an observer is found to be $$45^{o}$$. Find the height of the observer, if he is standing at a distance of $$32.4\ m$$ away from the tower and the height of the tower is $$34\ m $$.

A
$$1.6m$$

B
$$2.4m$$

C
$$2.8m$$

D
$$3.8m$$
Question 9
1 / -0

As observed from the top of a $$75\ m$$ high lighthouse from the sea-level, the angles of depression of two ships arc $$30^{o}$$ and $$45^{o}$$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

A
$$75 \sqrt{3}-1m$$

B
$$40 \sqrt{4}-1m$$

C
$$69 \sqrt{6}-1m$$

D
$$72 \sqrt{3}-1m$$
Question 10
1 / -0

A man standing on a horizontal plane, observations the angle of elevation of the top of a tower to be $$\alpha$$. After walking a distance equal to double the height of the tower, the angle of elevation becomes $$2\alpha$$, then $$\alpha$$ is -

A
$$\dfrac{\pi}{8}$$

B
$$\dfrac{\pi}{12}$$

C
$$\dfrac{\pi}{6}$$

D
$$\dfrac{\pi}{2}$$

Solution

$$\tan 2\alpha=\dfrac{2\tan \alpha}{1-\tan^2\alpha}$$

$$\dfrac{h}{x-2h}=\dfrac{2h/x}{1-\left(\dfrac{h}{x}\right)^2}$$

$$\dfrac{h}{x-2h}=\dfrac{2k}{x(\dfrac{x^2-h^2}{x^2})}$$

$$\dfrac{x^2-h^2}{x}=2x-4h$$

$$x^2-h^2=2x^2-4hx$$

$$x^2+h^2-4h x=0$$

$$x^2-4hx+h^2=0$$

$$x=\dfrac{4h\pm\sqrt{12h^2}}{2}=h(2\pm\sqrt{3})$$

$$x$$ should be greater than $$2h$$

So $$x=(2+\sqrt{3})$$

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

$$\alpha=\dfrac{\pi}{12}$$

$$B$$ is correct.