Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 60

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Question 1
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A spherical balloon of radius $$r$$ subtends an angle $$\angle$$ at the eye of an observer, while the angle of elevation of its centre is $$\beta$$. What is the height of the centre of the balloon (neglecting the height of the observer)?

A
$$\cfrac { r\sin { \beta } }{ \sin { \left( \cfrac { \alpha }{ 2 } \right) } } $$

B
$$\cfrac { r\sin { \beta } }{ \sin { \left( \cfrac { \alpha }{ 4 } \right) } } $$

C
$$\quad \cfrac { r\sin { \left( \cfrac { \beta }{ 2 } \right) } }{ r\sin { \alpha } } $$

D
$$\cfrac { r\sin { \alpha } }{ \sin { \left( \cfrac { \beta }{ 2 } \right) } } $$

Solution

$$\rightarrow$$ Let $$'O'$$ be the centre of the balloon, $$P$$ be the eye of the observer, and $$\angle APB$$ be the angle subtended by the balloon at the eye of the observer.

$$\therefore \angle APB = \alpha$$

$$\angle APO = \angle BPO = \dfrac{\alpha} { 2}$$

In $$\triangle OAP \ \sin \dfrac {\alpha}{2} = \dfrac {OA}{OP}$$

$$\rightarrow \sin \dfrac {\alpha}{2} = \dfrac {r}{OP}$$ or $$OP = r \text{cosec} \dfrac {\alpha}{2}$$

In $$OPL$$
$$\sin \beta = \dfrac {OL}{OP}$$ or $$PL = OP\sin \beta $$

$$\therefore OL = r\text{cosec} \dfrac {\alpha}{2} \sin \beta$$

$$ = \dfrac {r\sin \beta}{\sin \left (\dfrac {\alpha}{2}\right )}$$.
Question 2
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The angles of elevation of the top of a tower from the top and foot of a pole are respectively $$30^o$$ and $$45^o$$. If $${ h }_{ T }$$ is the height of the tower and $${ h }_{ p }$$ is the height of the pole, then which of the following are correct?
1. $$\dfrac { 2{ h }_{ p }{ h }_{ T } }{ 3+\sqrt { 3 } } ={ h }_{ p }^{ 2 }$$

2. $$\dfrac { { h }_{ T }-{ h }_{ p } }{ \sqrt { 3 } +1 } =\dfrac { { { h }_{ p } } }{ 2 } $$

3. $$\dfrac { 2({ h }_{ p }+{ h }_{ T }) }{ { h }_{ p } } =4+\sqrt { 3 } $$

Select the correct answer using the code given below.

A
1 and 3 only

B
2 and 3 only

C
1 and 2 only

D
1, 2 and 3
Question 3
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The angle of elevation of a cloud from a point $$h$$ $$m$$ above the level of water in a lake is $$\alpha$$ and the angle of depression of its reflection in the lake is $$\beta$$. Then height of the cloud above the water level.

A
$$\cfrac { h\sin { \left( \beta -\alpha \right) } }{ \sin { \left( \beta +\alpha \right) } } $$

B
$$\cfrac { h\sin { \left( \beta -\alpha \right) } }{ \sin { \left( \beta -\alpha \right) } } $$

C
$$\cfrac { h\sin { \left( \alpha +\beta \right) } }{ \sin { \left( \alpha -\beta \right) } } $$

D
none of these

Solution

In $$\Delta ACP-$$

$$\tan\alpha=\dfrac{PC}{AC}=\dfrac{H-h}{x}$$

$$\implies x=\dfrac{H-h}{\tan\alpha}$$

Similarly, in $$\Delta ACP'-$$

$$\tan\beta=\dfrac{CP'}{AC}=\dfrac{H+h}{x}$$

$$\implies x=\dfrac{H+h}{\tan\beta}$$

$$\implies \dfrac{H-h}{\tan\alpha}=\dfrac{H+h}{\tan\beta}$$

$$\implies H=h\dfrac{\tan\alpha+\tan\beta}{\tan\beta-\tan\alpha}$$

$$\implies H=h\dfrac{\sin\alpha\cos\beta+\sin\beta\cos\alpha}{\sin\beta\cos\alpha-\sin\alpha\cos\beta}$$

$$\implies H=\dfrac{h\sin(\alpha+\beta)}{\sin(\beta-\alpha)}$$
Question 4
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A pole has to be erected at a point on the boundary of a circular park of diameter $$13$$ meters in such a way that the difference of its distance from two diametrically opposite fixed gates $$A$$ and $$B$$ on the boundary is $$7$$ meters. It is possible to do so? If yes, at what distances from the two gates should the pole be erected?

A
$$6.71\ m$$

B
$$0.71\ m$$

C
$$1.51\ m$$

D
$$5.51\ m$$

Solution

Let $$P$$ be point at which pole erected diameter$$=AB=13m$$
Radius$$=r=\cfrac{13}{2}m$$
$$AP=l_{1}, BP=l_{2}$$
$$l_{1}=r\theta$$
$$l_{2}=r(\pi-\theta)$$
$$l_{1}-l_{2}=7$$
$$\Rightarrow r(2\theta-\pi)=7$$
$$\Rightarrow 2\theta=\cfrac{14}{13}+\cfrac{22}{7}$$
$$\Rightarrow \theta=\cfrac{192}{91}$$
$$\pi-\theta=\cfrac{22}{7}-\cfrac{192}{91}=\cfrac{94}{91}$$
$$\therefore$$ The given condition is possible.
Distance of pole from A$$=r\theta=\cfrac{96}{7}m$$
Distance of pole from B$$=r(\pi-\theta)=\cfrac{47}{7}m=6.71m$$
Question 5
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A monkey is sitting at a height of $$5$$ m on a tree. There is a fruit on the ground at a distance of $$10$$ m. from the foot of the tree. The monkey gets down and goes straight towards the fruit. If instead it would have climbed up to the top of the tree and from there jumped at the fruit, the distance travelled would have been the same as before. Find the height of the tree.

A
$$15m$$

B
$$7.5m$$

C
$$10m$$

D
$$12.5m$$

Solution

Let $$AB$$ be the height of the tree, $$D$$ is the point it is sitting at and the fruit is at point $$C$$
Given, $$DA+AC = DB + BC$$
$$\Rightarrow AC + AD = (5 + 10)m = 15m$$ (Given)
$$AC =15-AD$$
In $$\triangle ABC$$, applying pythagoras theorem,
$$AC^{2} = AB^{2} + BC^{2}$$
$$(15-AD)^{2} = (AD + 5)^{2} + 10^{2}$$
$$225 + (AD)^{2} - 30AD = (AD)^{2} + 25 + 10AD + 100$$
$$\Rightarrow 100 = 40AD$$
$$\Rightarrow AD = \dfrac {100}{40} = 2.5\ m$$
$$\therefore AB = AD + DB$$
$$= (2.5 + 5) m$$
$$AB = 7.5\ m$$
Therefore, height of tree is $$7.5\ m$$
So, the answer is option $$(b)$$.
Question 6
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A man in a boat rowing away from lighthouse $$100$$ m high takes $$2$$ minutes to changes the angle of elevation of the top of lighthouse from $$60^0$$ to $$45^0$$.If the speed of the boat is $$\dfrac{5(\sqrt3-1)}{m\sqrt3}m/s$$.Find $$m$$

A
2

B
3

C
5

D
6

Solution
Question 7
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The shadow of a tower standing on the level ground to be $$60\text{ m}$$ shorter when the sun's altitude changes from $$30^\circ$$ to $$60^\circ$$, find the height of the tower.

A
$$90\sqrt{30}\text{ m}$$

B
$$30\sqrt{3}\text{ m}$$

C
$$90\text{ m}$$

D
None of the above

Solution

In $$\triangle ABC,$$
$$\tan 60^\circ=\cfrac {AB}{CB}$$
$$\Rightarrow \sqrt {3}=\cfrac {AB}{CB}$$
$$\Rightarrow CB=\cfrac {AB}{\sqrt {3}}\quad\quad\dots(i)$$

In $$\triangle ABD,$$
$$\tan30^\circ=\cfrac {AB}{DB}$$
$$\Rightarrow \cfrac {1}{\sqrt {3}}=\cfrac {AB}{DB}$$
$$\Rightarrow DB=\sqrt {3}AB$$
$$\Rightarrow DC+CB=\sqrt {3}AB$$
$$\Rightarrow 60+CB=\sqrt {3}AB$$
$$\Rightarrow CB=\sqrt {3}AB-60\quad\quad\dots(ii)$$

On comparing $$(i)$$ & $$(ii),$$
$$\cfrac {AB}{\sqrt {3}}=\sqrt {3}AB-60$$
$$\Rightarrow \cfrac {AB}{\sqrt {3}}-\sqrt {3}AB=-60$$
$$\Rightarrow \cfrac {AB-3AB}{\sqrt {3}}=-60$$
$$\Rightarrow 2AB=60\sqrt {3}$$
$$\Rightarrow AB=30\sqrt {3}$$

So, height of the tower is equal to $$30\sqrt{3}\text{ m}$$.
Question 8
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A boy standing on a horizontal plane finds a bird flying at a distance of $$100 \,m$$ from him at an elevation of $$30^o$$ A girl standing on the roof of $$20 \,m$$ high building finds the angle of elevation of same bird to be $$45^o$$. Both the boy and the girl are on opposite side of the bird. Find the distance of the bird from the girl.

A
$$\dfrac{30}{\sqrt2}$$

B
$$30\sqrt2$$

C
$$10\sqrt2$$

D
$$\dfrac{10}{\sqrt2}$$

Solution
Question 9
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There are two statins $$P.Q$$ due north, due south of a tower of height $$15$$ meters. The angle of depression of $$P$$ and $$Q$$ as seen from top a tower are $$\cot^{-1}\dfrac{12}{5},\sin^{-1}\dfrac{3}{5}$$. The distance between $$P$$ and $$Q$$ is ..

A
$$48$$

B
$$56$$

C
$$65$$

D
$$25$$

Solution
Question 10
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A 10 meters high tower is standing at the centre of an equilateral triangle and each side of the triangle subtends an angle of $${60^o}$$ at the top of the tower. Then the length of each side of the triangle is

A
$$5m$$

B
$$5\sqrt 6 m$$

C
$$4\sqrt 6 $$

D
$$4m$$

Solution