Some Applications of Trigonometry test

Result

Some Applications of Trigonometry test - 35

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

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

A
$$\displaystyle 18\sqrt{3}\ \text{m}$$

B
$$\displaystyle 18(\sqrt{3}+1)\ \text{m}$$

C
$$\displaystyle 18(\sqrt{3}-1)\ \text{m}$$

D
$$\displaystyle 6\sqrt{3}\ \text{m}$$

Solution

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 $$\displaystyle DE=BE\tan 45^{\circ}$$
$$=BE=18\sqrt{3}\ \text{m}$$
$$\displaystyle \therefore CD=CE+DE=AB+DE$$
$$=\left ( 18+18\sqrt{3} \right )\ \text{m}$$
$$\displaystyle = 18\left ( \sqrt{3}+1 \right )\ \text{m}$$
Question 2
1 / -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

A
$$\displaystyle b\tan \alpha \cot \beta $$

B
$$\displaystyle b\cot \alpha \cot \beta $$

C
$$\displaystyle b\tan \alpha \tan \beta $$

D
$$\displaystyle b\cot \alpha \tan \beta $$

Solution

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, $$\displaystyle \frac{x}{b}=\cot \beta $$)
Question 3
1 / -0

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

A
$$\displaystyle \frac{9-3\sqrt{3}}{2}km/h $$

B
$$\displaystyle \frac{9+3\sqrt{2}}{2}km/h $$

C
$$\displaystyle \frac{9\sqrt{3}}{2}km/h $$

D
$$\displaystyle 9\sqrt{3}km/h $$

Solution

$$\displaystyle \tan 60^{\circ}=\frac{150}{y}$$ 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
Question 4
1 / -0

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

A
$$\displaystyle 25\sqrt{2}m $$

B
$$\displaystyle 50\sqrt{2}m $$

C
25 m

D
50 m

Solution

$$\displaystyle \sin 45^{\circ}=\frac{h}{50}$$ or $$\displaystyle \frac{1}{\sqrt{2}}=\frac{h}{50}$$
or h=$$\displaystyle \frac{50}{\sqrt{2}}=\frac{2\times 25}{\sqrt{2}}=25\sqrt{2}$$
Question 5
1 / -0

The distance between two buildings is $$24\ \text{m}$$. The heights of the buildings are $$12\ \text{m}$$ and $$22\ \text{m}$$. Find the distance between the tops (in $$\text{m}$$).

A
$$20$$

B
$$26$$

C
$$30$$

D
$$32$$

Solution

From the diagram, $$ PB = CD = 12\ \text{m} $$
$$ AP = AB - PB = (22 - 12)\ \text{m} = 10\ \text{m} $$

Also $$ CP = BD = 24\ \text{ m} $$

Since $$ \triangle APC $$ is a right angled triangle,
$$ {AC}^{2} = {AP}^{2} + {PC}^{2} $$
$$ {AC}^{2} = {10}^{2} + {24}^{2} $$
$$ {AC}^{2} = 676 $$
$$ \therefore AC = 26\ \text{ m} $$

Hence, distance between their tops is $$ 26\ \text{m} $$.
Question 6
1 / -0

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

A
$$\displaystyle 10 \sqrt{3} $$

B
$$\displaystyle 10/ \sqrt{3} $$

C
$$\displaystyle 10 \sqrt{3}/3 $$

D
10

Solution

$$\displaystyle \tan 30^{\circ}=\frac{h}{10}=\frac{1}{\sqrt{3}}=\frac{h}{10}$$
$$or$$ $$h=\displaystyle \frac{10}{\sqrt{3}}$$
Question 7
1 / -0

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?

A
$$35\ ft$$

B
$$42\ ft$$

C
$$ 35\sqrt { 3 }\ ft$$

D
$$ 7+35\sqrt { 3 }\ ft$$

E
$$ 7+42\sqrt { 3 }\ ft$$

Solution

In $$\triangle BAC,$$
$$ \sin { 60° } =\dfrac { BC }{ AB } =\dfrac { BC }{ 70 } $$

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

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

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

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

Hence, the answer is $$\left( 7+35\sqrt { 3 } \right).$$
Question 8
1 / -0

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 _____.

A
Angle of elevation

B
Angle of depression

C
Line of sight

D
Horizontal level

Solution

If the given object is below the level of the observer, then the angle by which is called angle of depression.
Question 9
1 / -0

A tower subtends at angle $$\displaystyle\theta$$ at a point P on the same level as the foot of the tower and from a point h m above P the depression of the foot of the tower is $$\displaystyle\alpha$$ . The height of the tower is (in m)

A
$$h$$ $$\displaystyle\tan \theta \tan \alpha $$

B
$$h$$ $$\displaystyle\\cot \theta \cot \alpha $$

C
$$h$$ $$\displaystyle \tan \theta \cot \alpha $$

D
$$\displaystyle \cot \theta \tan \alpha $$

Solution

Here, $$\tan \theta=\dfrac{H}{x}\Rightarrow x=H\cot \theta.....(1)\\\tan \alpha=\dfrac{h}{x}\Rightarrow x=h\cot \alpha.....(2)$$

Equating (1) and (2)
$$H\cot \theta=h\cot \alpha\Rightarrow H=h\cot \alpha \tan \theta$$

Therefore, Answer is $$h\cot \alpha \tan \theta$$
Question 10
1 / -0

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 _____

A
$$30$$$$^o$$

B
$$45$$$$^o$$

C
$$60$$$$^o$$

D
$$90$$$$^o$$

Solution

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.