Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 51

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Question 1
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The lower window of a house is at $$2\ m$$ above the ground and its upper window is $$4\ m$$ vertically above the lower window. At certain instant the angle of elevation of balloon from these windows are observed to be $$60^0$$ and $$30^0$$ respectively. Then the height of the balloon above ground is

A
$$12\ m$$

B
$$14\ m$$

C
$$8\ m$$

D
$$16\ m$$

Solution

Let the height of the first window be $$EG=2\ m$$
and, the height of the second window be $$CG=(2+4)\ m=6\ m$$
Let, $$AB$$ be $$x$$ and $$FG$$ be $$y$$

From $$\Delta ABC$$
$$\implies \tan 30°=\cfrac {AB} {BC}$$
$$\implies \cfrac 1 {\sqrt3}= \cfrac x y$$
$$\implies y=\sqrt{3}x$$

From $$\Delta ADE$$
$$\implies \tan 60°=\cfrac {AD} {DE}$$
$$\implies \sqrt{3}= \cfrac {x+4} y$$
$$\implies \sqrt{3}y=x+4$$
$$\implies \sqrt{3}\times\sqrt{3}x=x+4$$
$$\implies 3x=x+4$$
$$\implies 2x=4$$
$$\implies x=2$$

Hence, the height of the balloon from the ground is$$=(2+4+2)\ m=8\ m$$
Question 2
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The angle of elevation of the top of a tower from the top and bottom of a building of height $$a$$ are $${0}^{o}$$ and $${45}^{o}$$ respectively. If the tower and the building stand at the same level, then height of tower is:

A
$$\dfrac { a\left( 3+\sqrt { 3 } \right) }{ 2 } $$

B
$$a(\sqrt {3}+1)$$

C
$$a\sqrt {3}$$

D
$$a(\sqrt {3}-1)$$
Question 3
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A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle $$30^o$$ with it. The distance between the foot of the tree to the point where the top touches the ground is $$8\ m.$$ Find the height of the tree.

A
$$\dfrac{12}{\sqrt3}\ m$$

B
$${24}{\sqrt3}\ m$$

C
$$\dfrac{8}{\sqrt3}\ m$$

D
$${8}{\sqrt3}\ m$$

Solution

From the above figure,
$$\tan30^{\circ}=\dfrac{b}{8}$$
$$\Rightarrow \dfrac1{\sqrt3} = \dfrac b8$$
$$\Rightarrow b=\dfrac{8}{\sqrt3}$$
Also,
$$\cos30^{\circ}=\dfrac{8}{a}$$
$$\Rightarrow \dfrac{\sqrt3}2 = \dfrac 8a$$
$$\Rightarrow a=\dfrac{16}{\sqrt3}$$

Length of the tree $$ = a + b$$
$$=\dfrac{8}{\sqrt3}+\dfrac{16}{\sqrt3}$$
$$=\dfrac{24}{\sqrt3}$$
$$=8\sqrt3\ m$$
Question 4
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A tower $${T_1}$$ of height 60m is located exactly opposite to a tower $${T_2}$$ of height 80m on a straight road. From the top of $${T_1}$$ , if the angle of depression of the foot of $${T_2}$$ , is twice the angle of elevation of the top of $${T_1}$$ , then the width (in m) of the road between the feet of the towers $${T_1}$$ and $${T_1}$$ is:

A
$$10\sqrt 2 $$

B
$$10\sqrt 3 $$

C
$$20\sqrt 3 $$

D
$$20\sqrt 2 $$

Solution

Given In figure we first find $$\tan \theta$$
$$\Rightarrow \tan \theta =\dfrac{20}{x}$$ -----------(1)
$$\Rightarrow \tan 2 \theta =\dfrac{60}{x}$$ --------------(2)
$$\Rightarrow\dfrac{\tan \theta}{\tan 2 \theta}=\dfrac{\dfrac{20}{x}}{\dfrac{60}{x}}=\dfrac{1}{3}$$
$$\Rightarrow \dfrac{{\tan \theta}{\dfrac{2 \tan \theta}{1-\tan 2 \theta}} =\dfrac{1}{3}$$
$$\because \tan 2\theta=\dfrac{2\tan \theta}{1\tan ^2\theta}$$
$$\Rightarrow 1-\tan^2\theta=\dfrac{2}{3}$$
$$\Rightarrow 1-\dfrac{2}{3}=\tan^2\theta$$
$$\Rightarrow \tan \theta =\dfrac{1}{\sqrt 3}$$
Put in eqn (1)
$$\Rightarrow \dfrac{1}{\sqrt 3}=\dfrac{20}{x}$$
$$\therefore x=20\sqrt 3$$
$$\therefore $$ Width of the rod between the feel of there $$T_1$$ and $$T_2$$ is $$20\sqrt 3.$$
Question 5
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An electrician has to repair an electric fault on a pole of height $$5$$m. She needs to reach a point $$1.3$$m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of $$60^0$$ to the horizontal, would enable her to reach the required position? Also, how far the foot of the pole should she place the foot of the ladder? ( You may take $$\sqrt3=1.73$$).

A
$$\text{length of ladder}=7.4m$$ , $$\text{distance}=2.14m$$

B
$$\text{length of ladder}=4.27m$$ , $$\text{distance}=3.7m$$

C
$$\text{length of ladder}=4.27m$$ , $$\text{distance}=2.14m$$

D
$$\text{length of ladder}=4.7m$$ , $$\text{distance}=2.14m$$

Solution

$$\sin60°=\cfrac { 3.7 }{ l } $$
$$\cfrac { \sqrt { 3 } }{ 2 } =\cfrac { 3.7 }{ l } $$
$$l=4.277 m$$
$$\tan60°=\cfrac { 3.7 }{ x } $$
$$x=2.138$$
$$x\approx 2.14m$$
Length of ladder $$=4.27m$$
Distance $$=2.14m$$
Question 6
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From the top of a building $$15\text{ m}$$ high the angle of elevation of the top a tower is found to be $$30^\circ$$. From the bottom of the same building, the angle of elevation of the tower id found to be $$60^\circ$$. Find the height of the tower and the distance between the tower and the building.

A
$$h= 22.5\text{ m} , D=38.97\text{ m}$$

B
$$h=22.5\text{ m}, D=12.97\text{ m}$$

C
$$h= 38.97\text{ m}, D=22.5\text{ m}$$

D
$$h=12.97\text{ m}, D= 22.5\text{ m}$$

Solution

Let the height of the tower be $$h$$ and the distance between tower and building be $$d.$$
This means the height from the top of the building to the top of the tower is $$h-15 \text{ m}.$$

Now, from the top of building,
$$\tan 30^\circ$$ $$=\dfrac{h-15}{d}\quad\quad\quad\dots(i)$$

And, from the bottom of the building,
$$\tan 60^\circ$$ $$=\dfrac{h}{d}\quad\quad\quad\dots(ii)$$

Dividing $$(i)$$ by $$(ii)$$,

$$\dfrac{h-15}{h}$$ $$=\dfrac{\tan 30^\circ}{\tan 60^\circ}$$

$$\Rightarrow$$ $$h$$ $$= 22.5 \text{ m}$$

Now, using equation $$(ii)$$,

$$\tan 60^\circ$$ $$=\dfrac{h}{d}$$

$$\Rightarrow$$ $$d=\dfrac{h}{\tan 60^0}$$

$$\Rightarrow$$ $$d$$ $$=12.97\text{ m}$$

So. the correct option is $$B$$.
Question 7
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The angle of elevation of the top of two points at a distance of $$9$$ m and $$16$$ m from the base of the tower and in the same straight line in the same direction with it are complimentary. Then the height of the tower is

A
$$12$$m

B
$$15$$ m

C
$$20$$ m

D
$$25$$ m

Solution

$$Given$$ , $$BC=9m$$ ,$$BD=16m$$ and $$\angle ACB + \angle ADB = 90 $$
Let $$AB=x$$ , $$\angle ACB = \theta $$ and $$\angle ADB=90- \theta$$
In $$\Delta ABC$$
$$\frac{x}{9}$$ $$=$$ $$tan \theta$$ ......1
In $$\Delta ABD$$
$$\frac{x}{16}$$ $$=$$ $$tan (90-\theta)$$
$$\frac{x}{16}$$ $$=$$ $$cot \theta$$ .....2
Multiply 1 and 2
We get ,$$\frac{x}{9}$$ $$\times$$ $$\frac{x}{16}$$ $$=1$$
$$x^2=144$$
$$x=12$$
So, the height of the tower is 12m.
Question 8
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The angle of elevation of a jet plane from a point $$'A'$$ on the ground is $$60^{o}$$. After a flight of $$15\ seconds$$, the angle of elevation changes to $$30^{o}$$. If the jet plane is flying at a constant height of $$1500\sqrt{3}\ meters$$. Find the speed of jet plane $$(\sqrt{3}=1.732)$$.

A
$$200m/s$$

B
$$600m/s$$

C
$$300m/s$$

D
$$400m/s$$

Solution

Let D and E be the initial and final positions of the plane respectively.

Consider the $$\triangle ABD$$

$$\tan 60=\dfrac{BD}{AB}$$ ............ $$\because tan(\Theta ) = \dfrac {Opposite}{Adjacent}$$

$$\sqrt3=\dfrac{1500\sqrt3}{AB}$$

$$AB=1500$$

Consider the $$\triangle ACE$$

$$\tan 30=\dfrac{CE}{AC}$$

$$\dfrac{1}{\sqrt3}=\dfrac{1500\sqrt3}{AC}$$

$$AC=4500$$

$$BC=AC-AB=4500-1500=3000m$$

$$DE=BC=3000 \,m$$

i.e. the jet plane travels a distance of $$3000 \, m$$ in $$15$$ sec
Therefore,
Speed of the jet plane$$=\dfrac{3000}{15}=200 \, m/s$$
Question 9
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A boy is flying a kite with a string of length $$100$$ m. If the string is tight and the angle of elevation of the kite is $$26^032'$$, find the height of the kite correct to one decimal place (ignore the height of the boy).

A
$$43.87$$ m

B
$$53.87$$ m

C
$$34.78$$ m

D
$$42.87$$ m
Question 10
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From the point on a bridge across a river, the angles of depressions of the banks on opposite sides of the river are $$30^\circ$$ and $$45^\circ$$, respectively. If the bridge is at a height of $$3\ m$$ from the banks, find the width of the river.

A
$$3(\sqrt { 3 } -1)$$ m

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

C
$$(3+\sqrt { 3 } )$$ m

D
$$(3-\sqrt { 3 } )$$ m

Solution

Let width of river =AB

And bridge is at height of $$3m$$ from banks

So, $$DP=3m$$

Angel of depression of banks on the opposite sides of river are $$={{30}^{0}},{{45}^{0}}$$

So,$$\angle QPA={{30}^{0}} $$

$$ \angle RPB={{45}^{0}} $$

We need to find AB=?

Since , PD height so it will be perpendicular at AB

$$\angle PDA=\angle PDB={{90}^{0}}$$

And line QR is parallel to line AB

$$ \angle PAD=\angle QPA={{30}^{0}} $$   (Alternate angle)
Similarly,

$$\angle QPB=\angle PBD={{45}^{0}}$$          (Alternate angle)

Now, in triangle PAD ,

$$ \tan 30=\dfrac{PD}{AD} $$

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

$$ AD=3\sqrt{3} $$

Now in triangle PBD ,

   $$ \tan {{45}^{0}}=\dfrac{PD}{DB} $$
$$ 1=\dfrac{3}{DB} $$

$$ DB=3 $$

AB=AD+DB

$$ =3\sqrt{3}+3 $$

$$ =3\left( \sqrt{3}+1 \right) $$

Hence width of river is $$3\left( \sqrt{3}+1 \right)$$.