Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 34

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Question 1
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The top of a $$15$$ metre high tower makes an angle of elevation of $$\displaystyle 60^{0}$$ with the bottom of an electric pole and angle of elevation of $$\displaystyle 30^{0}$$ with the top of the pole. What is the height of the electric pole?

A
$$5$$ metres

B
$$8$$ metres

C
$$10$$ metres

D
$$12$$ metres

Solution

Let $$AB$$ be the tower and $$CD$$ be the electric pole
Then $$\displaystyle \angle ACB=60^{0},\angle EDB=30^{0}$$ and $$\displaystyle AB=15$$ m
Let $$\displaystyle CD=h.$$ Then $$\displaystyle BE=\left ( AB-AE \right )$$
$$\displaystyle =\left ( AB-CD \right )=\left ( 15-h \right )$$
We have $$\displaystyle \frac{AB}{AC}=\tan 60^{0}=\sqrt{3}$$
$$\displaystyle \Rightarrow AC=\frac{AB}{\sqrt{3}}=\frac{15}{\sqrt{3}}$$
And $$\displaystyle \frac{BE}{DE}=\tan 30^{0}=\frac{1}{\sqrt{3}}$$
$$\displaystyle \Rightarrow DE=\left ( BE\times \sqrt{3} \right )$$ $$\displaystyle =\sqrt{3}\left ( 15-h \right )$$
$$\displaystyle \Rightarrow 3h=\left ( 45-15 \right )$$
$$\Rightarrow h=10$$ m
Question 2
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From the top of the cliff $$150\ m$$ high the angles of depression of the top and bottom of a tower are observed to be $$\displaystyle 30^{\circ}$$ and $$\displaystyle 60^{\circ}$$ respectively. The height of the tower is

A
$$100\ m$$

B
$$\displaystyle 50\sqrt3\ {m}$$

C
$$\displaystyle 133\frac{1}{3}$$

D
$$\displaystyle 100\left ( \sqrt{3}-1 \right )$$

Solution

$$\displaystyle BF=\left ( 150-x \right )m.$$ Let $$\displaystyle AC=y$$
$$\displaystyle \tan 30^{0}=\frac{150-x}{y}$$
$$\displaystyle \tan 60^{0}=\frac{150}{y}$$
$$\displaystyle \Rightarrow y=\left ( 150-x \right )\sqrt{3}=\frac{150}{\sqrt{3}}$$
$$\displaystyle \therefore 150-x=50$$ or $$\displaystyle x=100\ m$$
Question 3
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A tree is broken by the wind and now its upper part touches the ground at a point $$10$$ m from the foot of the tree and makes an angle of $$\displaystyle 60^{0}$$ with the ground. The entire length of the tree is:

A
$$15$$ m

B
$$20$$ m

C
$$\displaystyle \left ( 10+\frac{20}{\sqrt3} \right )\sqrt{3}m$$

D
$$\displaystyle \left ( 10+\frac{\sqrt{3}}{2} \right )m$$

Solution

Let say the Length of the tree $$ = BT = BC + CT$$ and $$CT = CA$$ when the tree is broken.

Given: $$AB = 10m$$

In $$\triangle ABC$$,

$$\tan60^o = \dfrac{BC}{AB} $$ ............(1)

Also, $$sec\ 60^o = \dfrac{AC}{AB}$$ ...................(2)

$$ BT= $$ $$BC + AC$$

$$=$$ $$AB \displaystyle \tan 60^{0}+AB\ \text{sec} 60^{0}$$ {Using (1) and(2)}

$$= AB\sqrt3 + AB\sec60^o$$

$$\displaystyle =AB\left ( \sqrt{3}+ 2 \right ) =10\left ( 1+\frac{2}{\sqrt3} \right )\sqrt{3}m$$

$$\displaystyle =\left ( 10+\frac{20}{\sqrt3} \right )\sqrt{3}m$$
Question 4
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A flagstaff $$10\ m$$ high stands at the center of a equilateral triangle which is horizontal at the top of the flagstaff. Each side subtends at an angle of $$\displaystyle 60^{o}$$. The length of each side of the triangle is:

A
$$\displaystyle \dfrac{6\sqrt{3}}{3}$$

B
$$\displaystyle \dfrac{4\sqrt{6}}{3}$$

C
$$\displaystyle \dfrac{20}{\sqrt{3}}$$

D
$$\displaystyle \dfrac{6\sqrt{5}}{3}$$

Solution

$$\Rightarrow$$ Considering Right angle $$\triangle ADC$$ we have,
$$\Rightarrow$$ $$\tan 60^o=\dfrac{AD}{DC}$$
$$\Rightarrow$$ $$\sqrt{3}=\dfrac{10}{DC}$$

$$\Rightarrow$$ $$DC=\dfrac{10}{\sqrt{3}}$$
$$\Rightarrow$$ $$DC=\dfrac{10}{\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}$$
$$\therefore$$ $$DC=\dfrac{10\sqrt{3}}{3}$$
$$\Rightarrow$$ $$BC=2\times DC$$
$$\therefore$$ $$BC=2\times \dfrac{10\sqrt{3}}{3}$$
$$\therefore$$ $$BC=\dfrac{20\sqrt{3}}{3}$$
$$\therefore$$ The length of each side of the triangle is $$\dfrac{20}{\sqrt{3}}\,m$$
Question 5
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If the angle of elevation of an object from a point $$100 \text{ m}$$ above a lake is found to be $$\displaystyle 30^{\circ}$$ and the angle of depression of its image in the lake is $$\displaystyle 45^{\circ}$$, then the height of the object above the lake is

A
$$100\left ( 2-\sqrt{3} \right )\text{ m}$$

B
$$100\left ( 2+\sqrt{3} \right )\text{ m}$$

C
$$100\left ( \sqrt{3}-1 \right )\text{ m}$$

D
$$1000\left ( \sqrt{3}+1 \right )\text{ m}$$

Solution

Let $$AB$$ the surface of the lake $$O$$ be the point of observation $$P$$ be the object and $$P'$$ be its image.

Let the distance of object $$P$$ above lake be $$x.$$ So, it will be below the lake by same distance as they are mirror images.
$$PE=P'E=x$$
$$PD=x-100$$
$$P'D=x+100$$
$$OC=DE=100 \text{ m}$$

In $$\triangle OPD,$$
$$\tan 30^{\circ}=\dfrac{PD}{OD}$$
$$\Rightarrow\dfrac{1}{\sqrt3}=\dfrac{x-100}{OD}$$
$$\Rightarrow OD=\sqrt3 (x-100)\text{ m}$$

In $$\triangle OP'D,$$
$$\begin{aligned}{}\tan {45^\circ } &= \frac{{P'D}}{{OD}}\\OD& = P'D\\\sqrt 3 (x - 100) &= x + 100\\x& = \frac{{100(\sqrt 3 + 1)}}{{\sqrt 3 - 1}}\\ &= 100(2 + \sqrt 3 ){\text{ m}}\end{aligned}$$

Hence, the distance of the object above the lake is $$100(2+\sqrt3)\text{ m}.$$
Question 6
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The angle of elevation $$\displaystyle \theta $$ of a vertical tower from a point on the ground is such that its tangent is $$\displaystyle \frac{5}{12}$$. On walking $$192$$ meters towards the tower in the same straight line the tangent of angle of elevation $$\displaystyle \phi $$ is found to be $$\displaystyle \frac{3}{4}$$. Find the height of the tower

A
$$170$$ m

B
$$175$$ m

C
$$180$$ m

D
$$185$$ m

Solution

In $$\displaystyle \Delta PQR,\tan \theta =\frac{h}{192+x}=\frac{5}{12}$$
$$\displaystyle \Rightarrow x=\frac{1}{5}\left ( 12h-960 \right )$$ ...(i)
In $$\displaystyle \Delta PQS,\tan \phi =\frac{h}{x}=\frac{3}{4}$$
$$\displaystyle \Rightarrow x=\frac{4h}{3}$$ ...(ii)
From (i) and (ii), we get
$$\displaystyle \frac{4h}{3}=\frac{1}{5}\left ( 12h-960 \right )$$
$$\displaystyle \Rightarrow h=180$$
$$\displaystyle \therefore $$ The height of the tower is 180 m
Question 7
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From the top of a lighthouse, the angles of depression of two stations on opposite sides and $$a$$ distance apart are $$\displaystyle \alpha $$ and $$\displaystyle \beta $$. The height of the lighthouse is:

A
$$\displaystyle \frac{a}{\cot \alpha \cot \beta }$$

B
$$\displaystyle \frac{a}{\cot \alpha +\cot \beta }$$

C
$$\displaystyle \frac{a\cot \alpha \cot\beta }{\cot \alpha +\cot \beta }$$

D
$$\displaystyle \frac{a\tan \alpha \tan \beta }{\cot \alpha +\cot \beta }$$

Solution

Let Height of Light House be $$h=AD$$, as shown in the figure above.
and $$\angle ABD=\alpha\ (Given)$$
$$\angle ACD =\beta\ (Given)$$

Now, in $$\triangle ABD,\ BD=h\cot \alpha\quad \quad \left[\cot \theta=\dfrac {\text{Adjacent Side}}{\text{Opposite Side}} \right]$$

Also, in $$\triangle ADC, \ CD=h\cot\beta\quad \quad \left[\cot \theta=\dfrac {\text{Adjacent Side}}{\text{Opposite Side}} \right]$$

Given that, $$BC=a$$

$$\therefore \ a=BC=BD+CD\\$$
$$\Rightarrow a=h\cot\alpha+h\cot\beta\\$$
$$\Rightarrow a=h(\cot\alpha+\cot\beta)\\$$
$$\therefore \ h=\dfrac{a}{\cot\alpha+\cot\beta}\\$$

Hence, option B is correct.
Question 8
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A boat is rowed away from a cliff $$150$$ m high At the top of the the cliff the angle of depression of the boat change from $$\displaystyle 60^{0}$$ to $$\displaystyle 45^{0}$$ in $$2.5$$ minutes The speed of the boat (in m/sec) is

A
$$\displaystyle 1+\frac{1}{\sqrt{3}}$$

B
$$\displaystyle 1-\frac{1}{\sqrt{3}}$$

C
$$\displaystyle \sqrt{3}+3$$

D
$$\displaystyle 2\sqrt{3}-3$$

Solution

Let AB be cliff C and D be two position of the boat
$$\displaystyle AB=150m,\angle ACB=60^{0} and\angle ADB=45^{0}$$

$$\displaystyle AC=AB\cot 60^{0}$$

$$\displaystyle \Rightarrow AC=\frac{150}{\sqrt{3}}m$$

$$\displaystyle AD=AB\cot 45^{0}$$

$$\displaystyle \Rightarrow AD=150m$$

$$\displaystyle \therefore CD=150-\frac{150}{\sqrt{3}}$$

$$\displaystyle =\frac{150}{\sqrt{3}}\left ( \sqrt{3}-1 \right )$$ which is covered in 150sec.

$$\displaystyle \therefore $$ Speed of the boat in m/sec

$$\displaystyle =\dfrac{150\left ( \sqrt{3-1} \right )}{\sqrt{3}\times 150}$$

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

$$=\displaystyle 1-\frac{1}{\sqrt{3}}$$
Question 9
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An aeroplane flying at a height of $$300\ metre$$ above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on ground are $$\displaystyle 60^{\circ}$$ and $$\displaystyle 45^{\circ}$$ respectively. The height of the lower plane above the above the ground in meters is

A
$$\displaystyle 100\sqrt{3}$$

B
$$\displaystyle \frac{100}{\sqrt{3}}$$

C
$$50$$

D
$$\displaystyle 150\left ( \sqrt{3+1} \right ).$$

Solution

$$P$$ and $$Q$$ are positions of planes.
From $$\displaystyle \Delta OMP;MP=300\ m$$
$$\displaystyle OM=MP\cot 60^{0}=300\times \dfrac{1}{\sqrt{3}}\ m$$
$$\displaystyle =100\sqrt{3}$$
In $$\displaystyle \Delta OMQ,MQ=OM$$
Height of the lower plane,$$ MQ = OM = 100\sqrt{3}\ m$$
Question 10
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A flagstaff $$5$$ m high stands on a building $$25$$ m high. As observed from a point at a height of $$30$$ m, the flagstaff and the building subtend equal angles. The distance of the observer from the top of the flagstaff is

A
$$\displaystyle \frac{5\sqrt{3}}{2}$$

B
$$\displaystyle 5\sqrt{\frac{3}{2}}$$

C
$$\displaystyle 5\sqrt{\frac{2}{3}}$$

D
None of thses

Solution

OB is the bisector of DOAC of DAOC
$$\displaystyle \therefore \frac{OC}{OA}=\frac{BC}{AB}=\frac{x}{\sqrt{x^{2}+30^{2}}}=\frac{5}{25}$$
$$\displaystyle \Rightarrow x=\frac{30}{\sqrt{24}}=\frac{30}{2\sqrt{6}}$$
$$\displaystyle \therefore x=5\sqrt{\frac{3}{2}}$$