Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 11

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Question 1
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An aeroplane flying at a constant speed, parallel to the horizontal ground, $$\sqrt {3}\ km$$ above it, is observed at an elevation of $$60^{o}$$ from a point on the ground. If, after five seconds, its elevation from the same point, is $$30^{o}$$, then the speed (in $$km/ hr$$) of the aeroplane, is

A
$$1500$$

B
$$750$$

C
$$720$$

D
$$1440$$

Solution

Let the observing point be $$O$$
The height at which the plane was flying above ground is constant at $$AP= QB= \sqrt{3}$$ km.
In the first case, distance of projection of plane from point of observation is $$\tan{60^0}=\frac{AP}{OP}\implies OP= 1$$ km.

In the first case, distance of projection of plane from point of observation is $$\tan{30^0}=\frac{BP}{OQ}\implies OP= 3$$ km.

So, a distance of $$3-1=2$$ km. is covered in $$5$$ seconds. So the speed of the plane is $$\dfrac{2\times3600}{5} = 1440$$ km/hr.

So option D is the correct answer.
Question 2
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A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of $$60^{\circ}$$ at the foot of the tower, and the angle of elevation of the top of the tower from A or B is $$30^{\circ}$$. The height of the tower is

A
$$\dfrac{2a}{\sqrt{3}}$$

B
$$2\sqrt{3}a$$

C
$$\dfrac{a}{\sqrt{3}}$$

D
$$a\sqrt{3}$$

Solution

Let us consider the height of the tower as $$h$$

We know that, $$h = AC \tan 30^\circ = BC \tan 30^\circ$$ ...(1)

We know that $$AC = BC$$ [$$\because$$ radius of circle]

Hence, we can say that $$\triangle ABC$$ is an isosceles triangle with $$AC=BC$$.

So, $$\angle ABC=\angle BAC$$ ...(2)

But $$\angle ACB=60^\circ$$ [given]

$$\Rightarrow \angle ABC+\angle BAC+\angle ACB=180^\circ$$

$$\Rightarrow \angle ABC+\angle ABC+60^\circ \,\,\,\,\,\,\,\,=180^\circ$$ [from (2)]

$$\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\times \angle ABC=120^\circ$$

$$\therefore\, \angle ABC=\angle BAC=60^\circ$$

Hence $$\triangle ABC$$ is an equilateral triangle.

$$\therefore\, AB = BC = CA = a$$ ...(3)

$$\therefore h = a \tan 30^\circ$$ [form equation (1)]

$$=\dfrac{a}{\sqrt{3}}$$
Question 3
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A bird is sitting on the top of a vertical pole $$20$$ m high and its elevation from a point $$O$$ on the ground is $$ 45^{\circ}$$. It flies off horizontally straight away from the point $$O$$. After one second, the elevation of the bird from $$O$$ is reduced to $$ 30^{\circ}$$. Then the speed (in $$m/s$$) of the bird is

A
$$ 40(\sqrt{2}-1)$$

B
$$ 40(\sqrt{3}-\sqrt{2})$$

C
$$ 20\sqrt{2}$$

D
$$ 20(\sqrt{3}-1)$$

Solution
Question 4
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Let $$10$$ vertical poles standing at equal distances on a straight line, subtend the same angle of elevation $$\alpha$$ at a point $$O$$ on this line and all the poles are on the same side of $$O$$. If the height of the longest pole is $$h$$ and the distance of the foot of the smallest pole from $$O$$ is $$a$$; then the distance between two consecutive poles, is

A
$$\displaystyle \frac{h \sin \alpha + a \cos \alpha}{9 \cos \alpha}$$

B
$$\displaystyle \frac{h \cos \alpha - a \sin \alpha}{9 \sin \alpha}$$

C
$$\displaystyle \frac{h \sin \alpha + a \cos \alpha}{9 \sin \alpha}$$

D
$$\displaystyle \frac{h \cos \alpha - a \sin \alpha}{9 \cos \alpha}$$

Solution

Since all $$10$$ poles are subtending equal angles at $$O$$. Let the distance between two consecutive poles $$=d$$.
Distance form $$O$$ to the smallest pole $$=a$$
Total base distance in right angled triangle $$=a+9d$$
$$\tan\alpha =\cfrac { h }{ a+9d } $$
$$\quad 9d\tan\alpha +a\tan\alpha =h$$
$$\quad d=\cfrac { h-a\tan\alpha }{ 9\tan\alpha } $$
$$\quad =\cfrac { h-a\dfrac{\sin\alpha }{\cos \alpha} }{ 9 \dfrac{\sin\alpha }{\cos \alpha} } $$
$$=\cfrac { h\cos\alpha -a\sin\alpha }{ 9\sin\alpha } $$
Question 5
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$$AB$$ is a vertical pole with $$B$$ at the ground level and $$A$$ at the top. A man finds that the angle of elevation of the point A from a certain point $$C$$ on the ground is $$60^{{o}}$$. He moves away from the pole along the line $$BC$$ to a point $$D$$ such that $$CD=7$$ m. From $$D$$ the angle of elevation of the point $$A$$ is $$45^{{o}}$$. Then the height of the pole is

A
$$\displaystyle \frac{7\sqrt{3}}{2}$$ $$(\sqrt{3}-1)\mathrm{m}$$

B
$$\displaystyle \frac{7\sqrt{3}}{2}\left (\frac{1}{\sqrt{3}+1}\right)\mathrm{m}$$

C
$$\displaystyle \frac{7\sqrt{3}}{2}\left (\frac{1}{\sqrt{3}-1}\right)\mathrm{m}$$

D
$$\displaystyle \frac{7\sqrt{3}}{2}(\sqrt{3}+1)\mathrm{m}$$

Solution

From question, we have
$$\displaystyle \tan 60^{0}=\frac{h}{BC} \ldots$$ ....(i) and
$$\displaystyle \tan 45^{0}=\frac{h}{7+BC} \ldots$$ .....(ii)
$$ \Rightarrow 7+BC=h\Rightarrow BC=h-7$$
From (i), we have
$$ \displaystyle \sqrt{3}=\frac{h}{h-7} \Rightarrow \sqrt{3}h-7\sqrt{3}=h\Rightarrow \sqrt{3}h-h=7\sqrt{3}$$
$$\Rightarrow h=\displaystyle \frac{7\sqrt{3}}{\sqrt{3}-1}= \displaystyle \frac{7\sqrt{3}(\sqrt{3}+1)}{2}\mathrm{m}$$
Question 6
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A tower $$T_{1}$$ of height $$60\ m$$ is located exactly opposite to a tower $$T_{2}$$ of height $$80\ m$$ on a straight road. From the top of $$T_{2}$$, if the angle of depression of the foot of $$T_{1}$$ is twice the angle of elevation of the top of $$T_{2}$$ from top of $$T_1,$$ then find the width of the road between the feet of the towers $$T_{1}$$ and $$T_{2}$$.$$\left(\text{Use }\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\right)$$

A
$$20\sqrt {2}$$

B
$$10\sqrt {2}$$

C
$$10\sqrt {3}$$

D
$$20\sqrt {3}$$

Solution

Let the width of the road between the feet of the towers $$ { T }_{ 1 }$$ and $${ T }_{ 2 }$$ be $$w.$$

In $$\triangle ABC,$$

$$\tan { \theta }=\dfrac {BC}{AC}$$
$$\Rightarrow \tan\theta=\dfrac{20}{w}\quad\quad\quad\dots(i)$$

Now, in $$\triangle BOD,$$
$$\tan { 2\theta } =\dfrac {BO}{OD}$$
$$\Rightarrow\tan { 2\theta } =\dfrac{80}{w}\quad\quad\quad\dots(ii)$$

Now, by using the given formula of $$\tan2\theta$$ in the question,

$$\begin{aligned}{}\tan 2\theta &= \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}\\\frac{{80}}{w}& = \frac{{2 \times \frac{{20}}{w}}}{{1 - {{\left( {\frac{{20}}{w}} \right)}^2}}}\\\left[ {1 - {{\left( {\frac{{20}}{w}} \right)}^2}} \right] \times \frac{{80}}{w} &= \frac{{40}}{w}\\1 - {\left( {\frac{{20}}{w}} \right)^2} &= \frac{1}{2}\\{\left( {\frac{{20}}{w}} \right)^2} &= \frac{1}{2}\\\frac{{20}}{w} &= \frac{1}{{\sqrt 2 }}\\w &= 20\sqrt 2 \ m\end{aligned}$$

Hence, the distance between the two towers is equal to $$20\sqrt2 \ m.$$
Question 7
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A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point $$A$$ on the path, he observes that the angle of elevation of the top of the pillar is $$\displaystyle { 30 }^{ \circ }$$. After walking for $$10$$ minutes from $$A$$ in the same direction, at a point $$B,$$ he observes that the angle of elevation of the top of the pillar is $$\displaystyle { 60 }^{ \circ }$$. Then the time taken (in minutes) by him, from $$B$$ to reach the pillar is:

A
$$6$$

B
$$10$$

C
$$20$$

D
$$5$$

Solution

In $$\triangle BCD,$$
$$\tan 60^\circ =\dfrac{h}{y}$$
$$\Rightarrow \sqrt3=\dfrac{h}{y}$$
$$\Rightarrow h={\sqrt3}y\quad\quad\quad\dots(i)$$

In $$\triangle ACD,$$
$$\begin{aligned}{}\tan 30^\circ &= \frac{h}{{x + y}}\\\frac{1}{{\sqrt 3 }} &= \frac{{\sqrt 3 y}}{{x + y}}\quad\quad\quad\dots[\text{By using }(i)]\\x + y& = 3y\\x& = 2y\\y& = \frac{x}{2}\end{aligned}$$

A man walks $$x$$ distance in $$10$$ minutes.
So, the time taken by man to walk $$\dfrac{x}{2}$$ distance will be $$5$$ minutes.
Question 8
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$$PQR$$ is a triangular park with $$PQ = PR = 200\ m$$. A T.V. tower stands at the mid-point of $$QR$$. If the angles of elevation of the top of the tower at $$P, Q$$ and $$R$$ are respectively $$45^{\circ}, 30^{\circ}$$ and $$30^{\circ}$$, then the height of the tower (in $$m$$) is

A
$$100\sqrt {3}$$

B
$$50\sqrt {2}$$

C
$$100$$

D
$$50$$

Solution

Let the height of the tower $$MN$$ be $$h$$

In $$\Delta QMN$$

tan 30° = $$\dfrac{MN}{QM}$$

$$\therefore QM=\sqrt{3}h=MR$$ ...$$(1)$$ {$$\because$$ $$M$$ is the point $$QM=MR$$}

Now in $$\Delta$$ $$MNP$$

$$\because \angle MPN = \angle NMP = 45^o $$

$$\therefore MN=PM$$ ($$\because$$ In a triangle sides opposite to equal angles are equal ) ...$$(2)$$

$$\therefore MN=PM=h$$

In $$\Delta$$PMQ

By using Equation (1)

$$PM$$ $$=$$ $$\sqrt{(200)^2-(\sqrt{3}h})^{2} \,\,\,\,$$ {By using Pythagoras theorem}

$$\therefore$$ From $$(2)$$

$$h=\sqrt{(200)^2-(\sqrt3h)^2} $$

$$\Rightarrow h=\sqrt{(40000)-3h^2} $$

$$\Rightarrow h^2=40000-3h^2 $$ {By squaring both the sides}

$$\Rightarrow 4h^2=40000 $$

$$\Rightarrow h^2 = 10000\,m$$

$$\Rightarrow h = 100\,m$$

Hence option C is correct.
Question 9
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A ladder rests against a wall so that its top touches the roof of the house. If the ladder makes an angle of $$\displaystyle { 60 }^{ \circ }$$ with the horizontal and height of the house be $$\displaystyle 6\sqrt { 3 } \text{ m}$$, then the length of the ladder is:

A
$$\displaystyle 12\sqrt { 3 } \text{ m}$$

B
$$12 \text{ m}$$

C
$$\displaystyle \frac { 12 }{ \sqrt { 3 } } \text{ m}$$

D
None of these

Solution

From the question, we can say
$$\sin60^o=\dfrac{BC}{AC}$$
$$\displaystyle $$ Length of ladder, $$\displaystyle AC=\frac { 6\sqrt { 3 } }{ \sin { { 60 }^{ \circ } } } =\dfrac{6\sqrt{3}}{\dfrac{\sqrt{3}}{2}}=12\text{ m}$$
Question 10
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A kite is flying at an inclination of $$60^\circ$$ with the horizontal. If the length of the thread is $$120\text{ m},$$ then the height at which kite is:

A
$$60\sqrt 3\text{ m}$$

B
$$60\text{ m}$$

C
$$\dfrac{60}{\sqrt 3}\text{ m}$$

D
$$120\text{ m}$$

Solution

Let at point $$A$$ kite is flying at a height $$AB$$ equal to $$h.$$

In $$\triangle ABC,$$
$$\begin{aligned}{}\sin 60^\circ &= \frac{{AB}}{{AC}}\\\\\frac{{\sqrt 3 }}{2} &= \frac{h}{{120}}\\\\h& = \frac{{120 \times \sqrt 3 }}{2}\\\\ &= 60\sqrt 3 \text{ m}\end{aligned}$$

Hence, the kite is flying at the height of $$60\sqrt3\text{ m}.$$