Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 39

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Question 1
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After covering a distance of $$10$$ meters from the flagpole, Mike lies on the ground, and measures an angle of $${70}^{o}$$ from the ground to the base of the ball at the top of the flagpole. Find the approximate height of the flagpole from the ground to the base of the ball at the top of the flagpole.

A
$$3m$$

B
$$9m$$

C
$$27m$$

D
$$29m$$

Solution

A triangle gets formed as seen in the figure.
We can write $$tan(70^o) = \cfrac{x}{10}$$ where $$x$$ is the height of the flagpole from the ground to the base of the ball at the top of the flagpole.
$$\therefore 2.747 = \cfrac{x}{10}$$
$$\therefore x = 27.47$$ m $$\approx 27$$ m
Question 2
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An aeroplane leaving from Bismarck travels on a bearing of $$120^o$$, as shown in the figure. If the plane is $$295$$ miles directly east of Pierre, how far apart are Bismarck and Pierre? select your answer to the nearest mile.

A
$$170$$

B
$$160$$

C
$$150$$

D
$$140$$

Solution

The angle which the plane makes with the line joining Bismarck and Pierre is $$60^o$$
So, $$\tan(60^o) = \cfrac{295}{x}$$
where, $$x$$ is the distance between Bismarck and Pierre.
$$\therefore x = \cfrac{295}{\sqrt{3}} = 170.31 $$ miles, the nearest integer being $$170$$ miles
Question 3
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In the figure, the length of $$\overline{QR}$$ is $$4$$, and $$T$$ is the midpoint of $$\overline{QS}$$. What is the length of $$\overline{RS}$$?

A
$$4$$

B
$$2\sqrt{3}$$

C
$$2\sqrt{13}$$

D
$$4\sqrt{13}$$

Solution

In right angled triangle $$\Delta RQT, QR = 4$$ and $$\angle RTQ = 30^o$$,
$$\therefore RT = \dfrac{4}{\sin(30^o)} = 8$$

Also, $$QT = \dfrac{4}{\tan(30^o)}$$

$$\Rightarrow QT = 4\sqrt{3}$$

$$\Rightarrow QS = 2 \times QT = 8\sqrt{3}$$

Since $$\angle SQR = 90^o, (RS)^2 = (QR)^2 + (QS)^2$$

$$\therefore (RS)^2 = 16 + (8\sqrt{3})^2$$

$$(RS)^2 = 16 + 192 = 208$$

$$RS = 4\sqrt{13}$$ units
Question 4
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A vertical pole subtends an angle $$\tan^{-1}\left (\dfrac {1}{2}\right )$$ at a point P on the ground. If the angles subtended by the upper half and the lower half of the pole at P are respectively $$\alpha$$ and $$\beta$$ then $$(\tan \alpha, \tan \beta) =$$

A
$$\left (\dfrac {1}{4}, \dfrac {1}{5}\right )$$

B
$$\left (\dfrac {1}{5}, \dfrac {2}{9}\right )$$

C
$$\left (\dfrac {2}{9}, \dfrac {1}{4}\right )$$

D
$$\left (\dfrac {1}{4}, \dfrac {2}{9}\right )$$

Solution

From the diagram,
$$\tan(\alpha +\beta) = \dfrac12= \dfrac{h}{d}$$, $$\tan\beta = \dfrac{h}{2d}=\dfrac14$$

$$\therefore \dfrac{\tan{\alpha}+\frac14}{1-\frac{\tan{\alpha}}{4}} = \dfrac12$$
$$\Rightarrow \tan{\alpha}= \dfrac29$$
Question 5
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The angle of elevation of a stationary cloud from a point $$2500$$ m above a lake is $$15^0$$ and from the same point the angle of depression of its reflection in the lake is $$45^o$$. The height (in meters) of the cloud above the lake, given that $$\cot\,15^o=2+\sqrt{3}$$, is

A
$$2500$$

B
$$2500\sqrt{2}$$

C
$$2500\sqrt{3}$$

D
$$5000$$

Solution

Let $$EB=x$$m, then
$$\tan 15^0=\dfrac{h}{x}$$

$$\Rightarrow x=h \cot 15^0$$ ...(1)
and from triangle EBD, we can get $$\tan 45^o=\dfrac{h+5000}{x}$$

$$\rightarrow x\cdot 1=h+5000$$ ...(2)

From (1) and (2)

$$\Rightarrow h \cot 15^0=h+5000$$
$$\Rightarrow h (2+\sqrt 3)=h+5000$$

$$\Rightarrow h=\dfrac{5000}{\sqrt{3}+1} = \dfrac{5000(\sqrt{3}-1)}{3-1}=2500(\sqrt{3}-1)m$$

$$\therefore$$ height of cloud is $$2500+2500(\sqrt{3}-1)$$ $$=2500\sqrt{3}$$m
Question 6
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There is a need of ramp in a company. The ramp shown in the figure must be placed at an angle of $$30^o$$, eight feet from the bottom step. Calculate the approximate length of the ramp.

A
$$67$$

B
$$116$$

C
$$128$$

D
$$134$$

Solution

Distance of ramp bottom from bottom step $$8$$ feet $$= 8 \times 12 = 96$$ inches
Distance of ramp bottom from building bottom $$= 96+20 = 116$$ inches

Therefore, $$ \cos (30) = \dfrac { \sqrt { 3 } }{ 2 }$$ $$=\dfrac { 116 }{ x }$$

$$\Rightarrow x = 134$$ inches
Question 7
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An aeroplane leaving from Bismarck travels on a bearing of $$120^o$$, as shown in the figure. If the plane flew at an average rate of $$400$$ miles per hour, how many minutes (approx) had it been in the air when it was $$295$$ miles from Pierre?

A
$$51$$

B
$$56$$

C
$$29$$

D
$$67$$

Solution

The angle that the plane makes with the line joining Bismarck and Pierre is $$60^o$$
We thus have $$\sin (60^o) = \cfrac{295}{x}$$ where $$x$$ is the distance flown by the plane.
That distance becomes $$x = \cfrac{295 \times 2}{\sqrt{3}}$$
$$x = 340.63 $$ mi
Since the speed is $$400$$ miles per hour, the time becomes $$340.63/400 \times 60 = 51.0945 $$ minutes
Question 8
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The horizontal distance between two towers is $$60\text{ m}$$ and the angle of depression of the top of the first tower as seen from the top of the second is $$30^o$$. If the height of the second tower be $$150\text{ m}$$, then the height of the first tower is

A
$$(150 + 60\sqrt{3}) \text{ m}$$

B
$$(150 - 60\sqrt{3}) \text{ m}$$

C
$$(150 + 20 \sqrt{3}) \text{ m}$$

D
None of these

Solution

As the Angle of Depression is given from the top of Second Tower
Hence, Second Tower is Larger than the First,so we can clearly see that, Height of First Tower must be less than the Height of the Second Tower $$(=150 \ m)$$

We can see that, $$BCDE$$ is a Rectangle
so, $$BC=ED=60$$
$$BE=CD=h\ (Let)$$

Now, in $$\triangle AED$$
$$\tan60^o=\dfrac{AE}{ED}\Rightarrow \sqrt3=\dfrac{AE}{60}\Rightarrow AE=60\sqrt3$$

So, Height of First Tower is $$h=BE=AB-AE=150-60\sqrt3$$
Question 9
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A boat sights the top of a $$40$$- foot lighthouse at an angle of elevation of $$25$$ degrees. Calculate the distance of the boat from the lighthouse (the horizontal distance) to the nearest tenth of a foot.

A
$$16.9$$ feet

B
$$18.7$$ feet

C
$$44.1$$ feet

D
$$85.8$$ feet

E
$$94.6$$ feet

Solution

Given, a boat sights the top of a $$40$$ foot light house at an angle of elevation $$25$$ degree
Then the lighthouse (horizontal ) the angle of elevation $$=90-25=65$$ degree
So, in right angle $$\triangle ABC$$,
$$\tan 65^{0}=\dfrac{BC}{40}$$
$$\Rightarrow BC=40\tan 65^{0}$$
$$\Rightarrow BC=40\times 2.1445$$
$$\Rightarrow BC=85.78$$ foot
Then horizontal distance of lighthouse to boat $$=85.78$$ foot.
Question 10
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A ramp is needed for a cat to help her walk up to the bed. The inclination of the ramp should be $$35^o$$ angle with the bedroom floor. How long must the ramp be to reach the top of the bed that is exactly three feet off the ground?

A
$$\displaystyle\frac{\sin 35^o}{3}$$

B
$$\displaystyle\frac{\sin 55^o}{3}$$

C
$$\displaystyle\frac{3}{\sin 55^o}$$

D
$$\displaystyle\frac{3}{\sin 35^o}$$

Solution

Let $$AB$$ is the height of the bed from the ground and $$BC$$ is the ramp that is $$35^{\circ}$$ angle with the bedroom floor.
In the right-angled triangle $$ABC$$ with angle $$A=$$ $$90^{\circ}$$
$$\Rightarrow \sin35^{\circ}=\dfrac{AB}{BC}$$
$$\therefore BC=\dfrac{3}{\sin35^{\circ}}$$