Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 33

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Question 1
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The angle of elevation of a ladder leaning against a wall is $$\displaystyle 60^{\circ}$$ and the foot of the ladder is $$4.6\ m$$ away from the wall The length of the ladder is

A
$$2.3\ m$$

B
$$4.6\ m$$

C
$$7.8\ m$$

D
$$9.2\ m$$

Solution

Consider the given figure. Let $$AC$$ denote the length of the ladder and $$AB$$ denote the wall. Now

$$cos60^{\circ}=\dfrac{BC}{AC}$$

$$AC=\dfrac{BC}{cos\ 60^{\circ}}$$

$$AC=BC.sec\ 60^{\circ}$$

$$=4.6\times 2$$
$$=9.2$$
Hence the length of the ladder is $$9.2\ m$$.
Question 2
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The angle of elevation of the top of a tower from a certain point is $$\displaystyle 30^{0}$$ If the observer moves 20m towards the tower the angle of elevation of the top of the tower increases by $$\displaystyle 15^{0}$$ The height of the tower is

A
$$17.3$$m

B
$$21.9$$m

C
$$27.3$$m

D
$$30$$m

Solution

Let AB be the tower and C and D be the point of observation
Given CD =20 m And $$\angle BCA=30^{0}$$ and $$ \angle BDA=30+15=45^{0}$$
Let height of tower is h
In triangle BAD
$$tan 45^{0}=\frac{AB}{AD}\Rightarrow 1=\frac{h}{AD}\Rightarrow AD=h$$
In triangle BAC
$$tan 30^{0}=\frac{AB}{AC}$$ ( AC=CD+AD)
$$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+h}$$
$$\Rightarrow \sqrt{3}h=20+h$$$$\Rightarrow \sqrt{3}h-h=20$$
$$\Rightarrow h(1.732-1)=20$$
$$\Rightarrow h=\frac{20}{0.732}=27.3$$
Question 3
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A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it If it takes $$12$$ minutes for the angle of depression to change from $$\displaystyle 30^{0}$$ to $$\displaystyle 45^{0}$$ how soon after this will the car reach the observation tower?

A
$$14$$ min. $$35$$ sec.

B
15 min. 49 sec.

C
16 min. 23 sec.

D
18 min. 5 sec.

Solution

Let AB be the tower and C and D be the point of positions of two cars
Given $$\angle ACB=45^{0}$$$$\angle ADB=30^{0}$$
Let AB=h ,CD=x and AC=y
In $$\Delta ABC$$
$$tan45^{0}=\frac{AB}{AD}$$
$$\Rightarrow 1=\frac{AB}{AD}$$
$$\Rightarrow AB=AD\Rightarrow y=h$$
In $$ABD$$
$$tan 30^{0}=\frac{AB}{AD}$$
$$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+y}$$
$$\Rightarrow x+y=\sqrt{3}h $$
$$\therefore x=(x+y)-y=(\sqrt{3})h-h=h(\sqrt{3}-1)$$
So car cover distance $$h(\sqrt{3}-1)$$ in 12 seconds
Then car cover distance h in=$$\left [ \frac{12\times h}{h(\sqrt{3}-1)} \right ]=\frac{12}{0.73}=16 min 23 sec$$
Question 4
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If the height of a pole is $$\displaystyle 2\sqrt{3}$$ metres and the length of its shadow is 2 metres then the angle of elevation of the sun is equal to

A
$$\displaystyle 30^{0}$$

B
$$\displaystyle 45^{0}$$

C
$$\displaystyle 60^{0}$$

D
$$\displaystyle 90^{0}$$

Solution

Height of a pole is $$2\sqrt { 3 } $$

length of its shadow is $$2$$ metre

Let ABC be a triangle where AB be $$2\sqrt { 3 } $$ height of a pole

AC be $$2$$ metre length of shadow

Angle of elevation of the sun will be
$$\tan { \theta = } \dfrac { AB }{ AC } =\dfrac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 } $$

$$\tan { { 60^ o } } =\sqrt { 3 } $$

Angle of elevation of the sun is equal to $$ { 60^o } $$
Question 5
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If a flag-staff 6 metres high placed on the top pf a tower throws a shadow of $$\displaystyle 2\sqrt{3}$$ metres along the ground then the angle that the sun makes with the ground is

A
$$\displaystyle 60^{0}$$

B
$$\displaystyle 30^{0}$$

C
$$\displaystyle 45^{0}$$

D
None of these

Solution

OA and AB be the shadows of the tower OP and the flag-staff PQ respectively on the grounds Let the sum makes an angle $$\displaystyle \theta $$ with the ground
$$\displaystyle AB=2\sqrt{3};OA=x;OP=h;PQ=6$$
In triangles OAP and OBQ We have
$$\displaystyle \tan \theta =\frac{h}{x}$$ and $$\displaystyle \tan \theta =\frac{h+6}{x+2\sqrt{3}}$$
$$\displaystyle \Rightarrow \frac{h}{x}=\frac{h+6}{x+2\sqrt{3}}$$
$$\displaystyle \Rightarrow x=\frac{h}{\sqrt{3}}$$
or $$\displaystyle \frac{h}{x}=\tan \theta =\sqrt{3}$$ so $$\displaystyle \theta=60^{0} $$
Question 6
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If the angles of elevaton of the top of a tower from two points distance s and t $$\displaystyle \left ( s> t \right )$$ from its foot are $$\displaystyle 30^{0}$$ and $$\displaystyle 60^{0}$$ respectively then the height of the tower is

A
$$\displaystyle \sqrt{s+t}$$

B
$$\displaystyle \sqrt{st}$$

C
$$\displaystyle \sqrt{s-t}$$

D
$$\displaystyle \sqrt{\frac{s}{t}}$$

Solution

$$AB$$ is the tower $$S$$ and $$T$$ be the points of observation such that $$AS = s$$ and $$AT = t$$
$$\displaystyle \frac{AB}{AT}=\tan 30^{0}\Rightarrow AB=\frac{t}{\sqrt{3}}$$
$$\displaystyle \frac{AB}{AS}=\tan 60^{0}\Rightarrow AB=s\sqrt{3}$$
$$\displaystyle \therefore AB^{2}=st$$ or $$\displaystyle AB=\sqrt{st}$$
Question 7
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A man standing on the bank of the river observes that the angle subtended by a tree on the opposite bank is $$\displaystyle 60^{0}$$. When he retires $$36$$ metres from the bank he finds the angle to be $$\displaystyle 30^{0}$$. The breadth of the river is

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

B
$$18\ m$$

C
$$12\ m$$

D
$$27\ m$$

Solution

Let the breadth of the river be AC = 'x' metres
Clearly $$\displaystyle \tan 30^{0}=\frac{AB}{AD}=\frac{h}{x+36}$$ and
$$\displaystyle \tan 60^{0}=\frac{AB}{AD}=\frac{h}{x}$$
$$\displaystyle \Rightarrow \frac{\tan 60^{0}}{\tan 30^{0}}=\frac{h}{x}\div \frac{h}{x+36}$$
$$\displaystyle \Rightarrow 3=\frac{x+36}{x}$$
$$\displaystyle \Rightarrow x=18m$$
Question 8
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A man on the top of a rock lying on a seashore observes a boat coming towards it. If it takes $$10$$ minutes for the angles of depression to change from $$\displaystyle 30^{\circ}$$ to $$\displaystyle 60^{\circ}$$ how soon will the boat reach the shore?

A
$$20$$ minutes

B
$$15$$ minutes

C
$$10$$ minutes

D
$$5$$ minutes

Solution

AB is the rock C and D be the two positions of the boat
Given:
$$\angle EAC=30^{\circ}$$ and $$\angle EAD=60^{\circ}$$
Now,
$$\angle ADB=EAD$$ $$[\because AE \parallel BC, AD~ \text{is transversal}, ~\angle ADB$$ and $$\angle EAD $$ are alternate interior angles $$]$$
$$\implies \displaystyle \angle ADB=60^{\circ}$$
$$\angle EAC=ACB$$ $$[\because AE \parallel BC, AC~ \text{is transversal}, ~\angle EAC$$ and $$\angle ACB $$ are alternate interior angles $$]$$
$$\implies \displaystyle \angle ACB=30^{\circ}$$
Let $$CD=x, ~DB=y$$
From $$\triangle ABC$$
$$\displaystyle \tan 30^{0}=\frac{AB}{x+y}=\frac{1}{\sqrt{3}}$$
$$\displaystyle \Rightarrow AB=\frac{x+y}{\sqrt{3}}$$
From $$\triangle ABD$$
$$\displaystyle \tan 60^{0}=\frac{AB}{y}\Rightarrow AB=\sqrt{3}y$$
$$\displaystyle \therefore \frac{x+y}{\sqrt{3}}=\sqrt{3}y$$ or $$\displaystyle \frac{x}{2}=y$$
The boat takes $$10$$ minutes to cover distance $$x,$$
so it will take $$5$$ minutes to cover $$\displaystyle \frac{x}{2}$$
$$\displaystyle \therefore $$ Time to reach $$\displaystyle B=\left ( 10+5 \right )$$ min $$= 15$$ Minutes from point $$C$$
Question 9
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The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of $$100\ m$$ from its base is $$\displaystyle 45^{o}$$ If the angle of elevation of the top of the complete pillar at the same point is to be $$\displaystyle 60^{o}$$ then the height of the incomplete pillar is to be increased by

A
$$\displaystyle 50\sqrt{2}\ m$$

B
$$100\ m$$

C
$$\displaystyle 100\left ( \sqrt{3-1} \right )\ m$$

D
$$\displaystyle 100\left ( \sqrt{3+1} \right )\ m$$

Solution

Let BC be the incomplete and BD be the complete pillar In Triangles ABC and ABD we have
Given $$BC=100 $$m,
And $$\angle CAB=45^{0}$$$$\angle DAB=60^{0}$$
Let $$CD= x$$ m
In $$\Delta ABC$$
$$tan 45^{o}=\frac{BC}{AB}$$
$$\Rightarrow 1=\frac{BC}{100}$$
$$\Rightarrow BC=100 m$$
In $$ABD$$
$$tan60^{0}=\frac{AD}{AB}$$
$$\Rightarrow \sqrt{3}=\frac{BD}{100}$$
$$\Rightarrow BD=100\sqrt{3}$$m
Then $$BD=BC+CD=100+x$$
$$\therefore 100+x=100\sqrt{3}$$
$$\Rightarrow x=100(\sqrt{3}-1)$$m
Question 10
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The length of the shadow of a person is x when the angle of elevation of the sun is $$\displaystyle 45^{0}$$. If the length of the shadow increased by $$\displaystyle \left ( \sqrt{3}-1 \right )x$$ then the angle of elevation becomes

A
$$\displaystyle 15^{0}$$

B
$$\displaystyle 18^{0}$$

C
$$\displaystyle 25^{0}$$

D
$$\displaystyle 30^{0}$$

Solution

Let $$AB$$ be the height of the person, $$AC$$ and $$AD$$ be his shadows at two instants.

$$\displaystyle AC=x$$
$$\displaystyle AD= AC+CD =x+\left ( \sqrt{3}-1 \right )x$$$$\displaystyle =\sqrt{3}x$$

and $$\displaystyle \angle ACB=45^{0}$$

$$\displaystyle \frac{AB}{AC}=\tan 45^{0}=1\\ \Rightarrow AB=AC=x$$

$$\displaystyle \tan \theta =\frac{AB}{AD}\\ =\dfrac{x}{\sqrt{3x}}\\ =\dfrac{1}{\sqrt{3}}$$

$$\displaystyle \Rightarrow \theta =30^{0}$$