Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 37

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Weekly Quiz Competition

Question 1
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An aeroplane at an altitude of $$200$$ m observes the angle of depression of two opposite points on the two banks of a river to be 45$$^o$$ and 60$$^o$$. Then, the width of the river is _____.

A
$$\dfrac{400}{\sqrt3}$$ m

B
$$400$$$$\sqrt3$$ m

C
$$\left(200 - \displaystyle{\frac{200}{\sqrt3}}\right)$$ m

D
$$\left(200 + \displaystyle{\frac{200}{\sqrt3}}\right)$$ m

Solution

From the given figure, A is the aeroplane, and B and C
are the two points of the two banks of the river.
AD = height of aeroplane's altitude $$= 200$$ m

$$\because XAB = 45^o\quad \therefore \angle ABD=45^o$$ {as they are alternate interior angles}

and, Similarily,

$$\because \angle$$YAC = 60$$^o$$ $$\therefore$$ $$\angle$$ACD = 60$$^o$$

Now, In the right-angled triangle ABD

$$\tan 45^o= \dfrac{AD}{BD}$$

$$\Rightarrow 1=\dfrac{200}{BD}$$

$$\therefore BD=200$$ ....(1)

Also,
In the right-angled triangle ADC

$$\tan60^o=\dfrac{AD}{DC}$$

$$\Rightarrow \sqrt{3}=\dfrac{200}{DC}$$

$$\therefore DC=\dfrac{200}{\sqrt{3}}$$ ....(2)

Now as we know Width of the river $$= BD + DC$$

$$\therefore From \,(1) \,and \,(2)$$

$$ BD + DC$$ $$=\left(200 + \displaystyle{\frac{200}{\sqrt3}}\right)$$ m

So, option D is correct.
Question 2
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The shadow of a vertical tower on the level ground increases by $$10$$ m when the altitude of the sun changes from 45$$^o$$ to 30$$^o$$. The height of the tower is _____ .

A
$$5($$$$\sqrt3$$ $$+ 1)$$

B
$$10($$$$\sqrt3$$ $$- 1)$$

C
$$9$$ m

D
$$13$$ m

Solution

From the given figure,
AB = tower = h,
AC and AD are shadow at 45$$^o$$ and 30$$^o$$.
$$\therefore$$ $$\angle$$C = 45$$^o$$ and $$\angle$$D = 30$$^o$$.
Also, $$AC = x$$ m and $$AD = (x + 10)$$ m

$$\therefore$$ cot 30$$^o$$ = $$\displaystyle{\frac{x + 10}{h} \Rightarrow}$$ x + 10 = h$$\sqrt3$$

Also, cot 45$$^o$$ = $$\displaystyle{\frac{x}{h} \Rightarrow 1 = \frac{x}{h} \Rightarrow}x = h $$

$$\therefore$$ x + 10 = h$$\sqrt3$$ & $$x = h$$

$$\Rightarrow$$ h + 10 = h$$\sqrt3$$

$$\Rightarrow$$ h($$\sqrt3$$ - 1) = 10 $$\Rightarrow$$ h = $$\displaystyle{\frac{10}{\sqrt - 2} = \frac{10}{\sqrt3 - 1} \times \frac{\sqrt3 + 1}{\sqrt3 + 1} = \frac{10(\sqrt3 + 1)}{3 - 1}}$$ = 5($$\sqrt3 $$+ 1) m.

So, option A is correct.
Question 3
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The string of a kite is $$100$$ m long and it makes an angle of 60$$^o$$ with the horizontal. If there is no slack in the string, the height of the kite above the ground is _____

A
$$100$$$$ \sqrt3$$ m

B
$$50$$$$ \sqrt3$$ m

C
$$50$$$$ \sqrt2$$ m

D
$$100$$ m

Solution

Let $$AB$$ be the string of the kite $$B$$ and $$AC$$ be the horizontal line.
$$\therefore$$BC$$\perp$$ AC, AB $$=$$ 100 m, $$\angle$$ $$=$$ 60$$^o$$
$$\sin 60^o= \displaystyle{\frac{BC}{AB} \Rightarrow \frac{\sqrt3}{2} = \frac{h}{100} \Rightarrow} h = 50\sqrt 3$$ m

$$\therefore$$ height of the kite above ground is 50$$\sqrt3$$ m

So, option B is correct.
Question 4
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An observer standing $$50$$ m away from a building notices that the angles of elevation of the top and bottom of a flag post on the building are 60$$^o$$ and 45$$^o$$ respectively. The height of the flag post is _____.

A
$$50$$$$\sqrt3$$ m

B
$$50$$($$\sqrt3$$ + 1) m

C
$$50$$($$\sqrt3$$ - 1) m

D
$$\dfrac{50}{\sqrt3}$$

Solution

Let $$AB$$ be the tower, $$BC$$ be the flag post and $$D$$ be the observer's position.
Then $$DA = 50$$ m, $$\angle ADB = 45^o$$ and $$\angle ADC = 60^o$$

Now, tan45$$^o$$ = $$\displaystyle{\frac{AB}{DA} \Rightarrow 1 = \frac{AB}{50} \Rightarrow}$$ AB = 50 m.

And, tan60$$^o$$ = $$\displaystyle{\frac{AC}{OA} \Rightarrow \frac{AC}{50} = \sqrt3 \Rightarrow \Rightarrow}$$ AC = 50$$\sqrt3$$ m

$$\therefore$$ Height of the flag post $$= AC - AB$$
$$= 50\sqrt3 - 50$$
$$=50(\sqrt3-1)$$m

So, option C is correct.
Question 5
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From the top of a hill the angle of depression of two consecutive mile stones on east are $$30^o$$ and $$45^o$$. The height of the hill is _____ km.

A
$$(\sqrt3 +1)$$

B
$$(\sqrt3 - 1)$$

C
$$\displaystyle{\frac{1}{2}(\sqrt3} + 1)$$

D
$$\displaystyle{\frac{1}{2}(\sqrt3} - 1)$$

Solution

From the given figure, AB is height of the hill $$= h$$ m
AD and AC are distance between the base of hill and two milestone being $$x$$ km & $$x + 1$$ km respectively.

Also $$\angle D$$ = 45$$^o$$ = $$\displaystyle{\frac{AB}{AD} \Rightarrow 1 = \frac{h}{x} \Rightarrow}$$ $$h = x$$ ...eq. (1)

tan 30$$^o$$ = $$\displaystyle{\frac{1}{\sqrt3} \Rightarrow \frac{h}{h + 1} \Rightarrow}$$ $$\sqrt3h = h + 1$$

$$\Rightarrow$$($$\sqrt3 - 1)h$$ = 1$$\Rightarrow$$ h = $$\displaystyle{\frac{1}{\sqrt3 - 1} = \frac{\sqrt3 + 1}{(\sqrt3 - 1)(\sqrt3 + 1)} = \frac{1}{2}(\sqrt3 + 1)}$$

$$\therefore$$ Height of the hill is $$\displaystyle{\frac{1}{2}(\sqrt3}$$ + 1) km

So, option C is correct.
Question 6
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On the same side of the tower 300 m high, The angles of depression of two objects are $$45^o$$ and $$60^o$$ respectively. The distance between two objects is _____ .(take $$\sqrt3$$ = 1.73)

A
$$127$$ m

B
$$117$$ m

C
$$137$$ m

D
$$147$$ m

Solution

Let say, AB = height of tower $$= 300$$ m, C and D be the positions of two objects such that $$\angle$$ACB = 45$$^o$$ and $$\angle$$ADB = 60$$^o$$

Now, cot 45$$^o$$ = $$\displaystyle{\frac{AC}{AB} \Rightarrow 1 = \frac{AC}{300} \Rightarrow}$$ AC = 300 m

and, cot 60$$^o$$ = $$\displaystyle{\frac{AD}{AB} \Rightarrow \frac{AD}{300} = \frac{1}{\sqrt3} \Rightarrow AD = \frac{300}{\sqrt3}}$$

$$\therefore$$ AD = $$\displaystyle{\frac{300}{\sqrt3} = \frac{300 \times \sqrt3}{\sqrt3 \times \sqrt3} = 100\sqrt3}$$ m

$$\therefore CD = AC - AD = (300 - 100\sqrt3$$)
$$= 300 - (100 \times 1.73)$$
$$= 300 - 173 $$
$$= 127$$ m

So, option A is correct.
Question 7
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A pole stands vertically on the ground if the angle of elevation at the top of the pole from a point 60 m away from the pole is 30$$^o$$, then the height of the pole is:
(Use $$\sqrt3$$ = 1.73)

A
$$34.2\ m$$

B
$$34.3\ m$$

C
$$34.4\ m$$

D
$$34.6\ m$$

Solution

Suppose AB represents the pole,
O is angle of elevation at the top of the pole from a point $$60\ m$$ away from the pole.

Then, OA = 60 m, $$\angle$$O = 30$$^o$$

$$\therefore \ tan \ {30}^o = \displaystyle{\frac{AB}{OA} \Rightarrow \frac{1}{\sqrt3} = \frac{AB}{60} \Rightarrow AB = \frac{60}{\sqrt3}} = 20\sqrt3$$.

$$\therefore \ AB = 20 \times 1.73 = 34.6 \ m$$ $$[\sqrt{3}=1.73]$$

$$\therefore\ $$ Height of the pole is 34.6 m
Question 8
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If the length of the shadow of a pole on ground is twice the length of the pole, then the angle of elevation of the sun is _____

A
30$$^o$$

B
45$$^o$$

C
60$$^o$$

D
None of these

Solution

Referring figure, if $$AB$$ is pole and $$AC$$ is length of the shadow of pole,
then $$AB = x$$, $$AC = 2x$$
and let $$\angle\; ACB=\theta$$

Then, $$\tan$$$$\theta$$ = $$\displaystyle{\frac{AB}{AC} \Rightarrow \tan \theta = \frac{x}{2x} = \frac{1}{2}}$$

$$\therefore$$ Value of $$\theta$$ is neither 30$$^o$$, 60$$^o$$ nor 45$$^o$$.

So, option D is correct.
Question 9
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A ladder leans a wall making an angle of 60$$^o$$ with the ground. The foot a the ladder is 2 m away from the wall. The length of the ladder is:

A
$$8$$ m

B
$$6$$ m

C
$$4$$ m

D
$$2$$ m

Solution

Let AC be the ladder and AB be the wall.
Then, BC is the distance between base of the ladder & wall = 2 m.
Also, $$\angle$$C = 60$$^o$$.
Now, In $$\triangle$$ABC, cos 60$$^o$$ = $$\displaystyle{\frac{BC}{AC}}$$

$$\therefore$$ AC = $$\displaystyle{\frac{BC}{\cos 60^o} = \frac{2}{\frac 12}}$$ = 4 m

$$\therefore$$ Length of the ladder is 4 m.
Question 10
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An aeroplane is at a vertical height of 675 m from the bank of a river. The angle of elevation of the aeroplane from a point just on the opposite bank of that river is found to be 60$$^o$$. The width of the river is _____ m.

A
225$$\sqrt3$$

B
215$$\sqrt3$$

C
205$$\sqrt3$$

D
None of these

Solution

Based on the given information, we can draw the figure shown above.
Let $$A$$ be the aeroplane at a vertical height of $$675\ m$$ above the bank of a river the angle of elevation of the aeroplane from a point just on the opposite bank is found to be $$60^o$$. Let that point be $$C$$.

We know that, $$\tan \theta=\dfrac{\text{Opposite Side}}{\text{Adjacent Side}}$$

Hence, $$\tan 60^o = \displaystyle\frac{AB}{BC}$$

$$ BC = \dfrac{AB}{\tan 60^o} $$

$$= \dfrac{675}{\sqrt3} \quad [\because \ \tan 60°=\sqrt3]$$

$$= \dfrac{675 \times \sqrt3}{\sqrt3 \times \sqrt3}$$

$$ = 225\sqrt3$$

Thus, width of the river is $$225\sqrt3\ m$$.