Question Set D

CBSE 10th Maths Some Applications of Trigonometry Sample Question Papers Study Material - QB365

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Questions:

Generic 2

26 x 2 = 52

In the given figure, the height of the girl is 1.5 m and the height of the tree is 13.5 m.

(Note The figure is not to scale.)
If AB= 12√3 m, what is the angle of elevation of the top of the tree from her eyes?

An observer, 1.5 m tall, is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.

A kite is flying at a height of 90 m above the ground.The string attached to the kite is temporarily tied to a point on the ground.The inclination of the string with the ground is 60°.Find the length of the string assuming that there is no slack in the string.

If \(\cos { \left( A-B \right) } =\frac { \sqrt { 3 } }{ 2 } \) and \(\sin { \left( A+B \right) } =\frac { \sqrt { 3 } }{ 2 } \), find A and B, where (A + B) and (A - B) are acute angles.

If 4 cos \(\theta\) = 11 sin \(\theta\), find the value of \(\frac { 11\cos { \theta } -7\sin { \theta } }{ 11\cos { \theta } +7\sin { \theta } } \)

If \(\sqrt { 2 } \sin { \theta } =1\), find the value of \(\sec ^{ 2 }{ \theta } -cosec^{ 2 }\theta \)

A 4m tall tree casts a shadow of length m.If at the same time a flagpole casts a shadow 40m in length, then the find the length of the flagpole.

A circus artist is climbing a 18m long rope which is tightly stretched and tied from the top of a vertical pole to the ground level is 30⁰, then find the height of the pole.

the shadow of a tower standing on a level ground is found to be 30m longer when the sun's altitude is 30⁰ than when it is 60⁰.Find the height of the tower.

10.

From a point 100m above lake, the angle of elevation of stationary helicopter is 30⁰ and angle of depression of the helicopter in the lake is 60⁰.Find the height of the helicopter.

11.

From a parachute vertically above a straight road, the angles of depression of two accidental vehicles, at an instant is found to be 45⁰ and 60⁰.If the vehicles are 100m apart, find the height of the parachute.
Which precautions one should take to avoid accidents on roads?

12.

The angles of elevation and depression of the top and bottom of a lighthouse from the top of a building, 60m high, are 30⁰ and 60⁰ respectively.
Find:
(i)the difference between the heights of the lighthouse and the building.
(ii)distance between the lighthouse and the building.

13.

A boy of height 1.3m spot a balloon moving with the wind in a horizontal level at some height from the ground. The angle of elevation of the balloon from the eyes of the boy at any instant is 60^o. After 2seconds, the angle of elevation reduces to 30^o. If the speed of the wind at that moment is \(29\sqrt{3}\) m/s, then find the height of the balloon from the ground.

14.

From a balloon vertically above a straight road, the angles of depression of two cars at an instant is found o be 45⁰ and 60⁰ .If the cars are 100m apart, find the height of the balloon.

15.

A ladder 15 metres long just reaches the top of a vertical wall.If the ladder makes an angle of 60⁰ with the wall, find the height of the wall.

16.

A tower stands vertically on the ground. From a point on the ground which is 60m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60^o. Find the height of the tower.

17.

A kite is flying at a height of 45m above the ground. The string attached tot the kite is temporarily tied to a point on the ground is 60^o. Find the length of the string assuming that there is no slack in the string.

18.

A circus artist is climbing a rope 12m long which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground is 30^o

19.

An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.

20.

A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60^o with the wall, then find the height of the wall.

21.

An aeroplane when flying at a height of 3125 m from the ground passes vertically below another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 30^o and 60^o respectively. Find the distance between the two planes at that instant.

22.

If a pole 6 m high throws shadow of \(2\sqrt { 3 } \) m, then find the angle of elevation of the sun.

23.

Find the angle of the elevation of the sun if the length of the shadow of the tower of height 20 m is \(20\sqrt { 3 } \).

24.

When the length of the shadow of a pole of height 7 m is equal to 7 m then find the elevation of these source of light.

25.

An observer, 1.7 m tall, is \(20\sqrt { 3 } \) m away from a tower. The angle of elevation from the eye of observer to the top of tower is 30^o . Find the height of tower.

26.

A ladder, leaning against a wall, makes an angle of 60^o with the horizontal. If the foot of the ladder is 2.5 m away from the wall. find the length of the ladder.

Multiple Choice Question

8 x 1 = 8

27.

Seaweed is found under 80 m deep seafloor. To reach it, a diver makes a 45° dive from a boat. What is the distance travelled by the diver to reach the seafloor?

(a)

80 m

(b)

80.2 m

(c)

80√2 m

(d)

80√3 m

28.

If sin A \(=\frac{1}{2}\) then the value of cot A is

(a)

\(\sqrt{3}\)

(b)

\(\frac{1}{\sqrt{3}}\)

(c)

\(\frac{\sqrt{3}}{2}\)

(d)

29.

If cos A \(=\frac{4}{5}\) then the value of tan A is

(a)

3 / 5

(b)

3 / 4

(c)

4 / 3

(d)

5 / 3

30.

\(\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}\) =

(a)

tan 90°

(b)

(c)

sin 45°

(d)

31.

\(\frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}\)=

(a)

sin 60°

(b)

cos 60°

(c)

tan 60°

(d)

sin 30°

32.

If sin θ = 1/2 , then the value of sin θ (sin θ – cosec θ) is

(a)

√3/2

(b)

-√3/2

(c)

3/4

(d)

-3/4

33.

The figure shows the observation of point C from point A. The angle of depression from A is:

(a)

30°

(b)

60°

(c)

75°

(d)

45°

34.

In the above fig Q and α respectively are

(a)

Angle of Depression and Angle of Depression

(b)

Angle of Elevation and Angle of depression

(c)

Angle of Elevation and Angle of Elevation

(d)

Angle of depression and Angle of Elevation

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Answers Key:

Generic 2

26 x 2 = 52

According to the figure,

\(A B =E C=12 \sqrt{3} \mathrm{~m}\)
\(A E =B C=15 \mathrm{~m} \)
\(B D =135 \mathrm{~m}\)
\(\Rightarrow D C+B C =135 \mathrm{~m}\)
\(D C =135-1.5 \mathrm{~m} =12 \mathrm{~m}\)
In \(\triangle D E C\),
\(\tan \theta =\frac{D C}{E C} \)
\(=\frac{12}{12 \sqrt{3}} \text { [from Eqs. (i) and (ii)] } \)
\(\tan \theta =\frac{1}{\sqrt{3}}\)
\(\therefore \quad \theta =30^{\circ} {\left[\because \tan 30^{\circ}=\frac{1}{\sqrt{3}}\right] }\)
Hence, angle of elevation of top of the tree from her eyes is \(30^{\circ}\).

Let BE = 22 m be the height of the tower and AD = 1.5m be the height of the observer. The point D be the observer's eye. Draw DC \\ AB.
Then, AB = 20.5 m = DC and, EC = BE - BC = BE - AD = 22 -1.5 = 20.5 m [∵ BC = AD]
Let S be the angle of elevation make by observer's eye to the top of the tower i.e. ∠EDC = θ.

In right angled ΔDCE,
\(\tan \theta=\frac{P}{B}=\frac{C E}{D C}=\frac{20.5}{20.5}=1\)
45\(\unicode{xb0} \)

In right ΔABC, \({AB \over AC}=sin\ 60^0\)
\(\Rightarrow\ x={180\sqrt3\over 3}\)
\(\Rightarrow\ \ x=60\sqrt3\)
= 60 x 1.732
Hence length of string = 103.92m.

\(\cos { \left( A-B \right) } =\frac { \sqrt { 3 } }{ 2 } =\cos { { 30 }^{ ° } } \)
\(\Rightarrow\) A - B = 30° ...(i)
\(\sin { \left( A+B \right) } =\frac { \sqrt { 3 } }{ 2 } =\sin { { 60 }^{ ° } } \)
\(\Rightarrow\) A + B = 60° ... (ii)
Adding equations (i) and (ii),
2A = 90°
\(\therefore\) A = 45°
From eqn. (ii), B = 60° - A = 60° - 45° = 15°

Given : 4 cos \(\theta\) = 11 sin \(\theta\)
\(\Rightarrow \quad \cos { \theta } =\frac { 11 }{ 4 } \sin { \theta } \)
Now \(\frac { 11\cos { \theta } -7\sin { \theta } }{ 11\cos { \theta } +7\sin { \theta } } =\frac { 11\times \frac { 11 }{ 4 } \sin { \theta } -7\sin { \theta } }{ 11\times \frac { 11 }{ 4 } \sin { \theta } +7\sin { \theta } } \)
\(=\frac { \sin { \theta } \left( \frac { 121 }{ 4 } -7 \right) }{ \sin { \theta } \left( \frac { 121 }{ 4 } +7 \right) } \)
\(=\frac { 121-28 }{ 121+28 } =\frac { 93 }{ 149 } \)

Given, \(\sqrt { 2 } \sin { \theta } =1\)
\(\sin { \theta } =\frac { 1 }{ \sqrt { 2 } } =\sin { { 45 }^{ ° } } \)
\(\therefore \quad \theta ={ 45 }^{ ° }\)
Now \(\sec ^{ 2 }{ \theta } -cosec^{ 2 }\theta \) = \(\sec ^{ 2 }{ { 45 }^{ ° } } -cosec^{ 2 }{ 45 }^{ ° }\)
\(={ \left( \sqrt { 2 } \right) }^{ 2 }-{ \left( \sqrt { 2 } \right) }^{ 2 }\)
= 2 - 2
= 0

80m

25.95m

10.

200m

11.

\(50(3+\sqrt{3})m;\)To avoid accident everyone should follow road safety rules.

12.

(i)20m
(ii)34.64m

13.

Let AB be the position of boy, P and Q be the two positions of the balloon, such that AB=1.3 m, ㄥPAE=60^o, ㄥQAE=30^o and speed of the balloon =29\(\sqrt { 3 } \) m/s.
∴ ED=FC=AB=1.3 m
Distance covered by the balloon in 2 seconds
=2 x 29\(\sqrt { 3 } \)=58\(\sqrt { 3 } \)m=FE
Consider rt. ㄥed ΔAEQ
QE/AE=tan 30⁰
QE=AE x \(\frac { 1 }{ \sqrt { 3 } } \)
QE=(AF+FE) x \(\frac { 1 }{ \sqrt { 3 } } \)
PF=\(\left( \frac { PF }{ \sqrt { 3 } } +58\sqrt { 3 } \right) \times \frac { 1 }{ \sqrt { 3 } } \)
[∵ QE=PF]
PF=PF/3+58
PF\(\left( 1-\frac { 1 }{ 3 } \right) \)=58
PF=58 x 3/2
PF=87 m
Now PC=PF+FC
=87+1.3
=88.3 m
Hence, the height of the ballon is 88.3 m

14.

Let the height of the ballon at P be h meters. Let A and B be the two cars. Thus, AB=100 m
ㄥPAQ=45^o and ㄥPBQ=60^o
Consider right-angled ΔAQP
\(\frac { PQ }{ AQ } \)=tan^o
⇒ \(\frac { PQ }{ AQ } \)=1
⇒ PQ=AQ= h m
Now, consider right-angled ΔPBQ, we have
\(\frac { PQ }{ BQ } \)=tan 60^o
\(\frac { PQ }{ BQ } \)=\(\\ \\ \\ \\ \\ \\ \\ \sqrt { 3 } \\ \)
\(\frac { h }{ h-100 } =\sqrt { 3 } \)
⇒ h=\(\frac { \sqrt { 3 } (100) }{ \sqrt { 3 } -1 } \)
⇒ h=\(\frac { \sqrt { 3 } (100) }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } \)
⇒ h=\(\frac { \sqrt { 3 } (100)(\sqrt { 3 } +1) }{ 2 } \)
=50(3+\(\sqrt { 3 } \))
Hence, the height of the ballon is 50(3+\(\sqrt { 3 } \)) m.

15.

Here, length of the ladder is 15 m and angle of elevation is 90^o- 60^o i.e., 30^o
Let h m be the height of the wall Consider right-angled ΔABC
\(\frac { BC }{ AC } \)=sin 30^o
\(\frac { h }{ 12 } =\frac { 1 }{ 2 } \)
h=\(\frac { 15 }{ 2 } \)m
Hence, the height of the wall is 15/2 m or 7.5 m.

16.

Let AB be tower of height h m and C is a point on the ground such that BC=60 m

As ㄥACB=600 and AB=h m
Consider rt. angled ΔABC, we have
tan 600=\(\frac { AB }{ BC } \)
⇒ \(\sqrt { 3 } =\frac { h }{ 60 } \)
⇒ h=60\(\sqrt { 3 } \) m.
Hence, the height of the tower is 60 \(\sqrt { 3 } \) m.

17.

Let p is the position of the kite with height 45 m from the point Q on the ground. Let R is the other point on the ground to which kite is temporarily tied
AS ㄥPRQ=60^o and PQ=45 m (given)
Consider art. ΔPQR, we have
sin 60^o=\(\frac { PQ }{ PR } \)
\(\frac { \sqrt { 3 } }{ 2 } =\frac { 45 }{ PR } \)
PR=\(\frac { 90 }{ \sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \)
PR=\(\frac { 90\sqrt { 3 } }{ 3 } \)
=30\(\sqrt { 3 } \)
Hence, the length of the string is 30m.\(\sqrt { 3 } \) m.

18.

Let AB=h m be the height of the pole
AC=12 m is the length of rope
and ㄥACB=30^o
Consider rt. angled ΔABC, we have
sin 300=\(\frac { AB }{ AC } \)
\(\frac { 1 }{ 2 } =\frac { h }{ 12 } \)
h=\(\frac { 12 }{ 2 } \)=6 m

19.

\(45^{\circ}\)

20.

\({15\over2}{\sqrt {3}} \ m \)

21.

Let A and D are two aeroplanes such that BD = 3125 m
ㄥACB = 60^o, ㄥDCB = 30^o
ㄥACB = 60^o.ㄥDCB = 30^o
In rt. ΔABC, \(\frac { AB }{ BC } \) = tan 60^o
⇒ \(\frac { x+3125 }{ BC } =\sqrt { 3 } \)
⇒ BC=\(\frac { x+3125 }{ \sqrt { 3 } } \)
⇒ BC=\(\frac { x+3125 }{ \sqrt { 3 } } \)m
In rt. ΔDBC, \(\frac { DB }{ BC } \) = tan 30^o
⇒ \(\frac { BD }{ BC } =\frac { 1 }{ \sqrt { 3 } } \)
⇒ BD=\(\frac { BC }{ \sqrt { 3 } } \)
⇒ BD = \(\frac { 1 }{ \sqrt { 3 } } \times \frac { 3125+x }{ \sqrt { 3 } } \)
⇒ BD = 3125 = \(\frac { 3125+x }{ 3 } \)
Now, x = 6250 m
⇒ AD = 6250 m

22.

Let AB is pole and BC is its shadow
∴ AB = 6m, BC = 2\(\\ \\ \\ \\ \sqrt { 3 } \) m
In right ΔABC, \(\frac { AB }{ BC } \)=tanፀ
⇒ tanፀ = \(\frac { 6 }{ 2\sqrt { 3 } } \)
⇒ tanፀ = \(\sqrt { 3 } \)
⇒ ፀ = 60^o

23.

Let AB is tower and BC is its shadow
∴ AB = 20 m and BC = 20\(\sqrt { 3 } \)
In right ΔABC, \(\frac { AB }{ BC } \) = tanፀ
⇒ tanፀ = \(\frac { 20 }{ 20\sqrt { 3 } } \)
⇒ tanፀ = \(\frac { 1 }{ \sqrt { 3 } } \)
⇒ ፀ = 30^o

24.

Let AB is pole and BC is its shadow
∴ AB = 7 m, BC = 7 cm
In right ∆ABC, \(\frac { AB }{ BC } \) = tanፀ
tanፀ = 7/7 = 1 ⇒ ፀ = 45^o

25.

In right ΔABC, \(\frac { AB }{ BC } \) = tan30^o
⇒ \(\frac { AB }{ 20\sqrt { 3 } } =\frac { 1 }{ \sqrt { 3 } } \) ⇒ AB = 20 m
Height of tower AD = AB + BD
= 20 + 1.7 = 21.7 m