Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 15

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Question 1
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A bridge above the river makes an angle of $${45}^{o}$$ with the bank of river. If length of bridge above the river is $$150\ m$$ then breadth of river will be

A
$$75\ m$$

B
$$50\sqrt {2}\ m$$

C
$$150\ m$$

D
$$75\sqrt {2}\ m$$

Solution

Let $$AC$$ be the bridge and $$AB$$ be the width of river.
In $$\Delta ABC$$
$$\dfrac{AB}{AC}=\sin 45^{\circ}$$
$$\dfrac{AB}{150}=\dfrac{1}{\sqrt2}$$
$$AB=\dfrac{150\sqrt2}{2}=75\sqrt2$$
therefore width of the river is $$75\sqrt2\;m$$.
Question 2
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$$AB$$ is vertical pole with $$B$$ at the ground level and $$A$$ at the top. A man find 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
$$\dfrac { 7\sqrt { 3 } }{ 2 } \dfrac { 1 }{ \sqrt { 3 } -1 } m$$

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

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

D
$$\dfrac { 7\sqrt { 3 } }{ 2 } \dfrac { 1 }{ \sqrt { 3 } +1 } m$$

Solution

Then in triangle $$ABC$$
$$tan60^{\circ}=\dfrac{AB}{BC}$$
$$x=\frac{h}{\sqrt3}$$ ----- (i)
in $$\Delta ABD$$
$$tan45^{\circ}=\dfrac {AB}{BD}$$
$$1=\dfrac{h}{x+7}$$
$$h=x+7$$
$$h=\dfrac{h}{\sqrt3}+7$$ from (i)
$$h=\dfrac{7\sqrt3}{\sqrt3-1} $$
$$h=\dfrac{7\sqrt3}{2}(\sqrt3+1)$$
Therefore the height of the pole is $$\dfrac{{7\sqrt 3 }}{2}(\sqrt 3 + 1)\;m$$
Question 3
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As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. An equation to model the motion is y=20cos($$\frac {\pi}{4} (t-3))+23$$. Predict your height above the ground at a time of 1 seconds.

A
$$20.86 ft$$

B
$$23 ft$$

C
$$8.14 ft$$

D
$$18.96$$

Solution

$$y=20\ cos (\cfrac{\pi}{4}(t-3))+23$$
At $$t=1$$ second
Height above the ground,
$$y=20\ cos (\cfrac{\pi}{4}(1-3))+23$$
$$y=20\ cos (\cfrac{-2\pi}{4})+23$$
$$\implies y=23 ft$$. [ Since , $$\ cos (\cfrac{-\pi}{4})=0$$ ]
Question 4
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A flagstaff stands on the middle of a square tower. A man on the ground, opposite to the middle of one face and distant from it $$100$$ m, just see the flag ; on his receding another $$100$$ m, the tangents of the elevation of the top of the tower and the top of the flagstaff are found to be $$\dfrac {1}{2}$$ and $$\dfrac {5}{9}$$. Find the height of the flagstaff, the ground being horizontal

A
$$20$$

B
$$25$$

C
$$30$$

D
$$35$$

Solution

$$\dfrac{h_2}{b}=\dfrac{h_1}{100}$$

$$\dfrac{h_1}{200}=\dfrac{1}{2}$$

$$\Rightarrow h_1=100$$

$$\dfrac{h_1+h_2}{200+b}=\dfrac{5}{9}$$

$$\dfrac{100+b}{200+b}=\dfrac{5}{9}$$

$$b=\dfrac{100}{4}$$

$$\Rightarrow b=h_2=25$$
Question 5
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If a ladder $$13 m $$ is placed against a wait such that its roots at a distance from the wall, then the height of the top of the ladder from the ground :

A
$$ 10 m $$

B
$$ 11 m $$

C
$$ 12 m $$

D
None of these

Solution

If a ladder is $$13$$ m long is placed at a distance of $$12$$ m from a wall, then the weight of the wall is:
$$\begin{array}{l} A.T.Q, \\ h=\sqrt { { { \left( { 13 } \right) }^{ 2 } }-{ { \left( { 12 } \right) }^{ 2 } } } \\ =\sqrt { 169-144 } \\ =\sqrt { 25 } \\ =5\, \, m \end{array}$$
Question 6
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From the top of a building $$h$$ metres, the angle of depression of an object on the ground is $$\alpha$$, the distance of the object from the foot of the building is

A
$$h\cot\alpha$$

B
$$h\tan\alpha$$

C
$$h\cos\alpha$$

D
$$h\sin\alpha$$

Solution

$$\tan \alpha = \dfrac{h}{d}$$

$$\Rightarrow d = \dfrac{h}{\tan \alpha}$$

$$\Rightarrow d = h \cot \alpha $$
Question 7
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$$A$$ person walking $$50$$ metres towards a chimney in a horizontal line. The angle of elevation of its top changes from $$30^{\circ}$$ to $$45^{\circ}$$. Height of the chimney (in metres) is

A
$$25(3+\sqrt{3})\ m$$

B
$$50(\sqrt{3}+1)\ m$$

C
$$25(\sqrt{3}+1)\ m$$

D
$$25(\sqrt{3}-1)\ m$$

Solution

$$\tan 30 ^{\circ} = \dfrac{h} {x+50}$$

$$\dfrac{1}{\sqrt 3} = \dfrac{h} {x+50}$$ ... (1)

and $$\tan 45 ^{\circ} = \dfrac{h} {x}$$

$$\Rightarrow 1 = \dfrac{h} {x}$$

$$\Rightarrow x= h$$ ...(ii)

From (i) and (ii), we get
$$\dfrac{x+50}{\sqrt{3}} = x$$

$${x+50} = \sqrt 3x$$

$$50= \sqrt 3x -x$$

$$50= (\sqrt 3 -1)x$$

$$\dfrac{50}{\sqrt 3 -1}=x$$

$$x=\dfrac{50}{\sqrt 3 -1}$$

Rationalize the denominator, we get
$$\displaystyle x= \dfrac{50}{\sqrt{3}-1}\times \dfrac{(\sqrt{3}+1)} {(\sqrt{3}+1)}$$

$$x=\dfrac{50\times {(\sqrt 3+1)}}{{\sqrt 3}^2-1^2}$$

$$x=\dfrac{50\times {(\sqrt 3+1)}}{3-1}$$

$$x=\dfrac{50\times {(\sqrt 3+1)}}{2}$$

$$\Rightarrow x= 25(\sqrt{3}+1)\ m $$

$$\therefore h = x = 25 (\sqrt{3}+1)\ m$$
Question 8
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The angle of elevation of an object from a point $$P$$ on the level ground is $$\alpha$$. Moving $$d$$ meters on the ground towards the object, the angle of elevation is found to be $$\beta$$, then the height (in meters) of the object is

A
$$ d\tan\alpha$$

B
$$ d\cot\beta$$

C
$$\dfrac d{\cot\alpha+\cot\beta}$$

D
$$\dfrac d{\cot\alpha-\cot\beta}$$

Solution

Let the height of the object $$AB$$ be $$h$$ metres.

Given, $$\angle ACB = \alpha, \angle ADB = \beta,\ CD = d$$ metres

Let $$DB = x$$ metres.

Then, in right-angled $$\triangle ABD,$$

$$ \cot β=\dfrac{x}{h}$$

$$⇒ x = h \cot \beta$$ $$...(i)$$

In right-angled $$\triangle ACB,$$

$$\cot \alpha =\dfrac{x+d}{h}$$

$$⇒ x + d = h \cot \alpha$$ $$...(ii)$$

Let us subtract the equations: $$(ii)\ – (i)$$

$$\Rightarrow d = h \cot \alpha \ – h \cot \beta$$

$$\Rightarrow d=h(\cot \alpha \ -\cot \beta)$$

$$\Rightarrow h = \dfrac{d}{ \cot \alpha - \cot \beta }$$
Question 9
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The flag staff of height $$10$$ metres is placed on the top of a tower of height $$30$$ metres. At the top of a tower of height $$40$$ metres, the flag staff and the tower subtend equal angles then the distance between the two towers (in metres) is

A
$$10\sqrt{2}$$

B
$$20\sqrt{2}$$

C
$$30\sqrt{2}$$

D
$$40\sqrt{2}$$

Solution

$$\tan 2 \theta = \dfrac{40}{d}$$ ...(i)

$$\Rightarrow \dfrac{2\tan \theta}{1-\tan^{2} \theta} = \dfrac{40}{d}$$

$$\tan \theta = \dfrac{10}{d}$$ ...(ii)

Substituting (ii) in (i), we get
$$\dfrac{2\dfrac{10}{d}}{1-\dfrac{(10)^{2}}{d^{2}}}= \dfrac{40}{d}$$

$$\Rightarrow \displaystyle \frac{d}{d^{2}-10^{2}}= \frac{2}{d}$$

$$\Rightarrow d^{2} = 2d^{2} - 2 \times 10^{2}$$
$$\Rightarrow d^{2} = 2 \times 10^{2}$$
$$\Rightarrow d = 10 \sqrt{2}\ m$$
Question 10
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A man observes the angle of elevation of a balloon to be $$30^{0}$$ at a point $$A$$. He then walks towards the balloon and at a certain place $$B$$, he finds the angle of elevation to be $$60^{0}$$. He further walks in the direction of the balloon and finds it to be directly over him at a height of $$\dfrac12\ km$$, then the distance$$AB $$ is:

A
$$\displaystyle \frac{1}{\sqrt{2}}\ km$$

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

C
$$\displaystyle \frac{1}{\sqrt{4}}\ km$$

D
$$\displaystyle \frac{1}{\sqrt{5}}\ km$$