Some Applications of Trigonometry test

Result

Some Applications of Trigonometry test - 31

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 a cliff, the angle of depression of car on the ground is $$60^o$$. If the car is at a distance of 30 m from the cliff. Find the height of the cliff.

A
$$52\ m$$

B
$$51\ m$$

C
$$55\ m$$

D
$$56\ m$$

Solution

Given that, the angle of depression of car on the ground from the top of a cliff is $$60^0$$. Also, the car is at a distance of $$30\ m$$ from the cliff.
To find out: The height of the cliff.

Based on the given information, we can draw the figure shown above.
$$\therefore \angle PAC=\angle ACB=60^{0}$$

Now, in $$Δ ABC$$,
$$\tan 60^{0}=\dfrac{AB}{BC}\quad \quad \left[\because \ \tan \theta=\dfrac{P}{B}\right]$$

$$ \Rightarrow AB=30\times \tan 60^{0}$$
$$\Rightarrow AB=30\times \sqrt 3\quad \quad \left[\because \ \tan 60^0=\sqrt3 \right]$$
$$\Rightarrow AB=30\times 1.732$$
$$\therefore \ AB\approx52\ m$$

Hence, the height of the cliff is approximately $$52\ m$$.
Question 2
1 / -0

The angle of elevation of the top of an electric pole from a point $$15$$ feet away from the foot is $$30^o$$. Find the height of the pole.

A
$$5\sqrt 3$$ feet

B
$$8\sqrt 3$$ feet

C
$$6\sqrt 3$$ feet

D
$$7\sqrt 3$$ feet

Solution

Given, $$BC$$ is $$15$$ and angle of elevation on ground is $$30^{0}$$
Let the height of the pole be $$h$$
In right angled $$Δ ABC$$.
$$tan30^{0}=\dfrac{AB}{BC}$$
$$\dfrac{1}{\sqrt 3}=\dfrac{h}{15}$$
$$\therefore h=\dfrac{15}{\sqrt 3}$$
$$\therefore h=5\sqrt 3$$ feet
Question 3
1 / -0

The shadow of a pole standing on a horizontal plane is $$a$$ meters longer when the sun's elevation is $$\theta$$ than when it is $$\phi$$. The height of the pole will be: {Use $$\sin(A-B) = \sin A \cos B - \sin B \cos A$$}

A
$$a\displaystyle\frac{\cos\theta\cdot\cos\phi}{\cos(\theta-\phi)}$$ meter

B
$$a\displaystyle\frac{\sin\theta\sin\phi}{\sin(\phi-\theta)}$$ meter

C
$$a\displaystyle\frac{\sin\theta\cos\phi}{\sin(\theta-\phi)}$$ meter

D
$$a\displaystyle\frac{\sin\phi\cos\theta}{\cos(\theta-\phi)}$$ meter

Solution

Let the height of the pole standing on a horizontal plane be $$h$$
Let the length of the shadow be $$x$$ and $$(x+a)$$, when the sun's elevation is $$\phi $$ and $$\theta$$ respectively.
In $$\triangle APQ$$,

$$\dfrac{x}{h}=\cot \phi $$

$$\Rightarrow x=h \cot \phi $$ ....(1)

In $$\triangle BPQ$$,

$$\dfrac{x+a}{h}=\cot \theta$$

$$\Rightarrow (x+a)=h \cot \theta$$ ....(2)

$$\Rightarrow (x+a)-x=h \cot \theta-h \cot \phi$$ {using (1)}

$$\Rightarrow a=h(cot \theta-cot \phi)$$

$$\Rightarrow h=\cfrac { a }{ \left( \dfrac { \cos \theta }{ \sin \theta } -\dfrac { \cos \phi }{ \sin \phi } \right) } $$

$$\Rightarrow h=a\cfrac { \sin \theta \sin \phi }{ \left( \sin \phi \cos \theta -\cos \phi \sin \theta \right) } $$

$$\Rightarrow h=a\cfrac { \sin \theta \sin \phi }{ \sin\left( \phi - \theta \right) } $$
Question 4
1 / -0

The angle of elevation of the top of a tower from a point on the ground, which is $$30$$ m away from the foot of the tower is $$30$$. The height of the tower is :

A
$$\displaystyle 8\sqrt { 3 } $$ m

B
$$\displaystyle 9\sqrt { 3 } $$ m

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

D
$$\displaystyle 12\sqrt { 3 } $$ m
Question 5
1 / -0

The shadow of a vertical tower on level ground increases by $$20$$m, when the altitude of the sun changes from angle of elevation $$60^o$$ to $$45^o$$. Find the height of the tower.

A
$$47.32$$m

B
$$47$$m

C
$$48$$m

D
$$49.56$$m

Solution

Let the height of the tower be $$h$$
In right $$ΔOBC$$
$$tan60^{0}=\dfrac{h}{x}$$
$$\Rightarrow \sqrt 3=\dfrac{h}{x}$$
$$\Rightarrow h=\sqrt {3 x}$$ ....................(1)
In right $$ΔOAC$$.
$$tan45^{0}=\dfrac{h}{AB+BC}$$
$$\Rightarrow tan45^{0}=\dfrac{h}{20+x}$$
$$=>h=20+x$$ ..............(2)
From (1) & (2) we get
$$\sqrt {3 x}=20+x$$
$$\Rightarrow x=\dfrac{20}{\sqrt {3-1}}$$
$$\Rightarrow x=\dfrac{20\sqrt {3+1}}{2}$$
$$\Rightarrow x=10(\sqrt 3+1)$$
$$=27.32$$ metres. [since $$\sqrt 3=1.732$$]
$$\therefore h=20+x$$
$$=20+27.32$$
$$=47.32\ m$$
Thus the height of the tower is $$47.32\ m$$
Question 6
1 / -0

A tree, $$15$$m high, is broken by the wind in such a way that its top touches the ground and makes an angle $$30^o$$ with the ground. At what height from the bottom is the tree broken by the wind?

A
$$5$$m

B
$$6$$m

C
$$7$$m

D
$$8$$m

Solution

Let AB be the height of tree $$=15m$$.
The tree is broken at point C. The broken part of tree which touches the ground be CO.
$$AB=15$$, then let $$AC=x$$ and $$BC=15−x$$.
In right $$Δ OCA$$
$$sin 30^{0}=\dfrac{AC}{OC}$$
$$=>\dfrac{1}{2}=\dfrac{x}{15-x}$$
$$=>2x=15-x$$
$$=>3x=15$$
$$=>x=5$$m
The tree is broken at height of $$5m$$
Question 7
1 / -0

A person is $$x$$ m away from a tree which is $$10$$m high and angle of elevation of the top of the tree is $$30^o$$, then the value of $$x$$ is

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

B
$$10\sqrt 3$$

C
$$5\sqrt 3$$

D
None

Solution

Here, $$\tan30^o=\dfrac{10}{x}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{10}{x}\Rightarrow x=10\sqrt3$$

Therefore, Answer is $$10\sqrt3$$
Question 8
1 / -0

The angle of elevation of the top of a vertical tower from two points. $$30 \text{ m}$$ apart, and on the same straight line passing through the base of tower, are $$ 30^{\circ} $$ and $$ 60^{\circ} $$ respectively. The height of the tower is

A
$$10 \text{ m}$$

B
$$15 \text{ m}$$

C
$$15\sqrt{3}\text{ m} $$

D
$$30\text{ m}$$

Solution

Let $$AB$$ is the tower of height $$h$$, and $$BC$$ equal to $$x.$$

In $$\triangle ABC,$$
$$\tan 60^{\circ}=\dfrac{AB}{BC}$$

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

$$\Rightarrow x=\dfrac{h}{\sqrt3}\quad\quad\quad\dots(i)$$

In $$\triangle ABD ,$$
$$\tan 30^\circ=\dfrac{AB}{BD}$$
$$\Rightarrow\dfrac{1}{\sqrt{3}} =\dfrac{h}{(x+30)}$$
$$\Rightarrow x+30 =\sqrt3 h$$
$$\Rightarrow\sqrt3 h-30\quad\quad\quad\dots(ii)$$

From equation $$(i)$$ and $$(ii),$$
$$\begin{aligned}{}\frac{h}{{\sqrt 3 }} &= \sqrt 3 h - 30\\h &= 3h - 30\sqrt 3 \\2h &= 30\sqrt 3 \\h &= 15\sqrt 3 \text{ m} \end{aligned}$$

So, height of the tower is $$15\sqrt{3} \text{ m} .$$
Question 9
1 / -0

The shadow of a flagstaff is three times as long as the shadow of the flagstaff when the sun rays meet the ground at angle of $$60^o$$. Find the angle between the sun rays and ground at the time of the longer shadow.

A
$$\theta=30^o$$

B
$$\theta=60^o$$

C
$$\theta=45^o$$

D
$$\theta=15^o$$

Solution

Let the length of the shadow be $$x$$ when it makes an angle of $$60^{0}$$ at point C at the ground and $$3x$$ at an angle of $$\theta$$ at point D.
Let height of flagstaff be $$h$$
In right angled $$ ΔABC$$.
$$tan 60^{0}=\dfrac{h}{x}$$
$$\rightarrow h=x tan60^{0}$$ ..............(1)
In $$ΔBDA$$
$$tanθ=\dfrac{h}{3x}$$
$$\rightarrow h=3 x tan\theta$$ ...............(2)
From (1) & (2) we get,
$$x tan60^{0}=3 x tanθ$$
$$\rightarrow tanθ=\dfrac{tan60}{3}$$
$$\rightarrow tan \theta=\dfrac{\sqrt 3}{3}$$
$$\rightarrow tan\theta=\dfrac{\sqrt 3\times \sqrt 3}{3\times \sqrt 3}$$
$$\rightarrow tan \theta=\dfrac{1}{\sqrt 3}$$
$$\rightarrow θ=30^{0}$$
Question 10
1 / -0

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height $$h$$. At a point on the plane, the angle of elevation of the bottom of the flagstaff is $$\displaystyle \alpha $$ and that of the top of the flagstaff is $$\displaystyle \beta $$ Then the
height of the tower is :

A
$$\displaystyle \frac { h }{ \tan { \alpha } } $$

B
$$\displaystyle \frac { h\tan { \alpha } }{ \tan { \beta } -\tan { \alpha } } $$

C
$$\displaystyle \frac { \tan { \beta } }{ h } $$

D
$$\displaystyle \frac { \tan { \alpha } }{ \tan { \beta } -\tan { \alpha } } $$

Solution

Let $$BC$$ be the tower and $$CD$$ be the flagstaff.
The angle of elevation of the bottom of the flagstaff is $$\alpha $$ and that of the top of flagstaff is $$\beta$$.
Let $$h$$ be the height of the tower
In $$\triangle ABC$$, we have
$$\Rightarrow \tan \alpha=\dfrac{BC}{AB}$$ ....{i}
In $$\triangle ABD$$
$$\Rightarrow \tan \beta=\dfrac{BD}{AB}$$
$$\Rightarrow \dfrac{BC+h}{AB}=\dfrac{BD}{AB}$$ ....{ii}
Now dividing {ii} by {i}, we get
$$\dfrac{BC+h}{BC}=\dfrac{\tan \beta}{\tan \alpha}$$
$$\Rightarrow (BC+h)\tan \alpha=BC \tan \beta$$
$$\Rightarrow BC(\tan \beta-\tan \alpha)=h \tan \alpha$$
$$\Rightarrow BC=\dfrac{h \tan \alpha}{\tan \beta- \tan \alpha}$$