Question Set B

CBSE 10th Maths Introduction to Trigonometry Model Questions Papers Study Material - QB365

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Questions:

2 Marks

Show that the points A(a,a), B(-a,a) and C(\(-a\sqrt3,a\sqrt3\)) form an equilateral triangle.

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.

The height of the lighthouse is h m. The angles of depression of two ships on opposite sides of this lighthouse are observed to be \({ 30 }^{ \circ }\ and \ \ { 45 }^{ \circ }\). Then, find the distance between the two ships.

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.

Find a relation between x and y such that the point (x , y) is equidistant from the points A (7, 1) and B (3, 5).

Prove that \(\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}\)

3 Marks

Two boats approach a lighthouse in mid - sea from opposite directions. The angles of elevations of the top of the lighthouse from two boats are 30^o and 45^o respectively. If the distance between two boats is 100 m, find the height of the lighthouse.

Point P divides the line segment joining the points A(2,1) and B(5,-8) such that \({AP\over AB}={1\over 3}\). If P lies on the 2x-y+k=0, find the value of k.

For what value of \(\lambda \), the three points (a,a), \((\lambda ,\sqrt { 3 } a)\) and (-a,-a) are the vertices of an equilateral triangle?

10.

The angle of elevation of a jet fighter from a point A on the ground is \({ 60 }^{ ° }\). After a flight of 10 s, the angle of elevation charge to \({ 30 }^{ ° }\) .

If the jet is flying at a speed of 432 Km/h, then find the constant height at which the jet is flying.

11.

Show that the points (a, a). (-a, - a) and \(\left( -\sqrt { 3 } a,\sqrt { 3 } a \right) \) are the vertices of an equilateral triangle. Also, find its area.

12.

Prove that \(\frac { \sin { A } +\cos { A } }{ \sin { A } -\cos { A } } +\frac { \sin { A } -\cos { A } }{ \sin { A } +\cos { A } } =\frac { 2 }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } .\)

13.

Prove that sec A (1 – sin A) (sec A + tan A) = 1.

Multiple Choice Question

14.

If altitude of the sun is 60°, the height of a tower which casts a shadow of length 30 m is:

(a)

30√3 cm

(b)

30/√3 m

(c)

15 m

(d)

15√2 m

15.

The angles of elevation of the top of a cliff from two points x and y metres from the base and in the same straight line with it are complementary. The height of the cliff is

(a)

\(x\sqrt { _{ ym } } m\)

(b)

\(\sqrt { x } y\quad m\)

(c)

\(\sqrt { XY } m\)

(d)

xy m

16.

The length of shadow of a tower on the plane ground is √3 times the height of the tower. The angle of elevation of sun is :

(a)

90^o

(b)

60^o

(c)

30^o

(d)

45^o

17.

A vertical tower is 20 m high. A man at some distance from the tower knows that the cosine of the angle of the elevation of the top of tower is 0.5. He is standing from the foot of the tower at a distance of:

(a)

30√3 m

(b)

20√3 m

(c)

20/√3 m

(d)

10/√3 m

18.

The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 60 m then, the height of the first tower is

(a)

139.5 m

(b)

142 m

(c)

135 m

(d)

140.83 m

19.

Consider a constellation of 3 stars A, B and C forming a right triangle with angle ABC = 90° and angle BAC = 30° . If the distance between star A and B is 3√3 x 1013 km, then how much time does light take to travel from star C to B with a speed of 3 x 108 m/s?

(a)

√3 x 10⁵ sec

(b)

10⁴ sec

(c)

√3 x 10⁴ sec

(d)

10⁵ sec

20.

A tree is broken by the wind. The top struck the ground at an angle of 30° and at a distance of 30 metres from the foot of the tree. The height of the tree in metres is

(a)

35√3

(b)

40√3

(c)

25√3

(d)

30√3

21.

A tower stands vertically on the ground from a point on the ground which is 25 m away from the foot of tower if the height of tower is 25√3 metres find the angle of elevation.

(a)

120°

(b)

90°

(c)

60°

(d)

30°

22.

A tower stands vertically on the ground from a point on the ground which is 15 m away from the foot of tower. If the height of tower is 15√3 meters find the angle of elevation

(a)

30°

(b)

60°

(c)

90°

(d)

120°

23.

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

(a)

(b)

40√3

(c)

20√3

(d)

24.

The angle of depression of a car, standing on the ground, from the top of a 75 m high tower, is 30°. The distance of the car from the base of the tower (in m.) is:

(a)

75√3

(b)

25√3

(c)

150

(d)

50√3

25.

A 20 m long ladder touches the wall at a height of 10 m. The angle which the ladder makes with the horizontal is

(a)

30°

(b)

45°

(c)

60°

(d)

90°

26.

Consider a ladder which makes an angle of 60° with a wall of height 10 m and its top just touches the top of the wall. If the ladder is now rotated in such a way that its top now touches the top of the opposite wall which has a height of 10/√3 m. What is the angle by which the ladder is rotated.

(a)

45°

(b)

60°

(c)

90°

(d)

30°

27.

A tree is broken by wind and its upper part touches the ground at a point 10 metres from the foot of the tree and makes an angle of 45° with the ground. The entire length of the tree is

(a)

20 m

(b)

10 (1 + √2) m

(c)

10√2 m

(d)

10 m

28.

A man has to clean a window at a height of 5 m on a building. He needs to reach a point 1.3m below the window to clean it. What should be the length of the ladder that he should use which, when inclined at an angle of 60° to the horizontal, would enable him to reach the required position?

(a)

2.46

(b)

2√3

(c)

2.46/ √3

(d)

2.46√3

29.

The angle of elevation from a point 30 feet from the base of a pole, of height h, as level ground to the top of the pole is 45° degree. Which equation can be used to find the height of the pole.

(a)

tan 45° = 30/h

(b)

tan 45° = h/30

(c)

sin 45° = h/30

(d)

cos 45° = h/30

30.

A man on a top of a tower observes a truck at an angle of depression α where tanα = 1/ √5 and sees that it is moving towards the base of the tower. Ten minutes later, the angle of depression of the truck is found to be β where tan β = √5 . If the truck is moving at a uniform speed, then how much more time it will take to reach the base of the tower.

(a)

150√5 sec

(b)

1500 sec

(c)

150 sec

(d)

150/ √5 sec

31.

The ——– is the line drawn from the eye of an observer to the point in the object viewed by the observer

(a)

Line of sight

(b)

Line of sight propagation

(c)

Line of symmetry

(d)

Line of incidence

32.

A kite is flying at a height of 75 metres from the ground level, attached to a string inclined at 60° to the horizontal. The length of the string to the nearest metre is

(a)

55 m

(b)

87 m

(c)

100 m

(d)

60 m

33.

Consider a ship with a right triangular mast. If the base of the mast is 10 m long, and the angle that the mast makes with the base is 60°, then what area of cloth is used to make the mast?

(a)

50 (√3 + 1) m²

(b)

50 √3 m²

(c)

50 m²

(d)

100 m²

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Answers Key:

2 Marks

\(AB=\sqrt { ({ -a-a) }^{ 2 }+({ -a-a) }^{ 2 } } =\sqrt { ({ -2a) }^{ 2 }+({ -2a) }^{ 2 } } =\sqrt { (4a^{ 2 }+4a^{ 2 } } =\sqrt { 8a^{ 2 } } =2\sqrt { 2a } \)
\(BC=\sqrt { ({ -a\sqrt { 3 } +a) }^{ 2 }+({ a\sqrt { 3 } +a) }^{ 2 } } =\sqrt { 3a^{ 2 }+a^{ 2 }-2\sqrt { 3 } a^{ 2 }+3a^{ 2 }+a^{ 2 }+2\sqrt { 3 } a^{ 2 } } =\sqrt { 8a^{ 2 } } =2\sqrt { 2a } \)
\(AC=\sqrt { ({ -a\sqrt { 3 } -a) }^{ 2 }+({ a\sqrt { 3 } -a) }^{ 2 } } =\sqrt { 3a^{ 2 }+a^{ 2 }+2\sqrt { 3 } a^{ 2 }+3a^{ 2 }+a^{ 2 }-2\sqrt { 3 } a^{ 2 } } =\sqrt { 8a^{ 2 } } =2\sqrt { 2a } \)
Since AB=BC=AC ∴△ABC is equilateral.

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

\(h\left( 1+\sqrt { 3 } \right) m\)

200m

Given point P(x, y) is equidistant from the points A(7, 1) and B(3, 5).
So, AP = BP
\(\Rightarrow\) AP² = BP²
\(\Rightarrow\) (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²
\(\left[\because \text { distance }=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\right]\)
\(\Rightarrow\) x² + 49 - 14x + y² + 1 - 2y
= x² + 9 - 6x + y² + 25 - 10y
\(\Rightarrow\)  -14x - 2y + 50 = -6x - 10y + 34
\(\Rightarrow\) -6x - 10y + 14x + 2y = 50 - 34
\(\Rightarrow\)   8x - 8y = 16
\(\Rightarrow\) x - y = 2
  [dividing by 8 on both sides]
Hence, the relation between x and y is x - y = 2.

\(L H S=\frac{\cot A-\cos A}{\cot A+\cos A}=\frac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}\)
\(=\frac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\frac{\left(\frac{1}{\sin A}-1\right)}{\left(\frac{1}{\sin A}+1\right)}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}=\operatorname{RHS}\)

3 Marks

AD is the lighthouse, Find AD=?
In right ΔABD,
h=x
\(\frac { h }{ x } \)=tan45^o
\(\frac { h }{ 100-x } \)=tan30^o ......(ii)
Solve for h and x.
⇒ \(\frac { h }{ 100-x } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \sqrt { 3 } h\)=100-x
⇒ \(\sqrt { 3 } \)x=100-x [Using eq (i)]
⇒ (\(\sqrt { 3 } \)+1)x=100 ⇒ x=\(\frac { 100 }{ \sqrt { 3 } +1 } \)
⇒ x=\(\frac { 100(\sqrt { 3 } -1) }{ (\sqrt { 3 } +1)(\sqrt { 3 } -1) } \)
⇒ x=\(\frac { 100(\sqrt { 3 } -1) }{ 2 } =50(\sqrt { 3 } -1)\)m
∴ h=height of lighthouse=\(50(\sqrt { 3 } -1)\)m

P is the point of intersection of line segment AB and line 2x-y+k=0.

Such that \(\frac {AP}{AB}=\frac{1}{3} \Rightarrow 3AP =AB\)
\(\Rightarrow \) 3AP = AP+PB ⇒ 2AP = PB
\(\Rightarrow \frac{AP}{AB} = \frac{1}{2} \Rightarrow\) AP : PB = 1:2
⇒ P divides the join of A(2,1) and B(5,-8) in the ratio 1:2.
∴ Coordinates of point P are \((\frac {5+4}{1+2}, \frac {2-8}{1+2})\) i.e, P(3,-2)
As points P lies on the 2x-y+k=0
∴ 6+2+k=0 ⇒ k=-8

Let A(a,a), B(λ, \(\sqrt{3}a\) ), C(-a,-a)
AB = \(\sqrt { { (\lambda -a) }^{ 2 }+({ \sqrt { 3 } a-a) }^{ 2 } } \)
BC \(=\sqrt { { (-a-\lambda ) }^{ 2 }+({ -a-\sqrt { 3 } a) }^{ 2 } } \)
As △ABC is equilateral triangle ⇒ AB = BC
\(\Rightarrow \ { (\lambda -a) }^{ 2 }+({ \sqrt { 3 } a-a) }^{ 2 }\ ={ (\lambda +a) }^{ 2 }+({ \sqrt { 3 } a+a) }^{ 2 }\)
\(\Rightarrow \ 0=4a\lambda +4\sqrt { 3 } { a }^{ 2 }\)
\(\Rightarrow \ \lambda \ = -\sqrt { 3 } a\)

10.

1039.2 m

11.

Let A(a,a), B(-a,-a) and C\(\left( -\sqrt { 3 } a,\sqrt { 3 } a \right) \) be the given points.
Then, AB = \(\sqrt { { \left( -a-a \right) }^{ 2 }+{ \left( -a-a \right) }^{ 2 } } \) [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
\(\sqrt { 4{ a }^{ 2 }+4{ a }^{ 2 } } =\quad 2\sqrt { 2 } \)a units
BC = \(\sqrt { { \left( -\sqrt { 3 } a+a \right) }^{ 2 }+{ \left( \sqrt { 3 } a+a \right) }^{ 2 } } \)
\(=\sqrt { { a }^{ 2 }{ \left( 1-\sqrt { 3 } \right) }^{ 2 }+{ a }^{ 2 }{ \left( \sqrt { 3 } +1 \right) }^{ 2 } } \\ =a\sqrt { 1+3-2\sqrt { 3 } +1+3+2\sqrt { 3 } } \\ =a\sqrt { 8 } =2\sqrt { 2 } a\) units
Clearly, we have AB = BC = AC
So, the \(\Delta \)ABC formed by the given points is an equilateral triangle.
Now, area of \(\Delta \)ABC = \(\frac { \sqrt { 3 } }{ 4 } \)(Side)² = \(\frac { \sqrt { 3 } }{ 4 } \) \(\times \) AB²
=\(\frac { \sqrt { 3 } }{ 4 } \)\(\times \) (\(2\sqrt { 2 } \)a)²
=\(2\sqrt { 3} \) a² sq units
Hence, area of an equilateral triangle is \(2\sqrt { 3} \) a² sq units.

12.

LHS = \(\frac { \sin { A } +\cos { A } }{ \sin { A } -\cos { A } } +\frac { \sin { A } -\cos { A } }{ \sin { A } +\cos { A } } \)
\(=\frac { { \left( \sin { A } +\cos { A } \right) }^{ 2 }+{ \left( \sin { A } -\cos { A } \right) }^{ 2 } }{ \left( \sin { A } -\cos { A } \right) \left( \sin { A } +\cos { A } \right) } \)
\(=\frac { \left[ \sin ^{ 2 }{ A } +2\cos { A } \sin { A } +\cos ^{ 2 }{ A } +\sin ^{ 2 }{ A } -2\cos { A } \sin { A } +\cos ^{ 2 }{ A } \right] }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } \)
\(=\frac { 2\sin ^{ 2 }{ A } +2\cos ^{ 2 }{ A } }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } =\frac { 2(\sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } ) }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } \)
\(=\frac { 2 }{ \sin ^{ 2 }{ A } -\cos ^{ 2 }{ A } } \left[ \because \sin ^{ 2 }{ A } +\cos ^{ 2 }{ A } =1 \right] \)
= RHS
Hence proved.

13.

LHS = sec A (1 – sin A)(sec A + tan A) \(=\left(\frac{1}{\cos A}\right)(1-\sin A)\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)\)
\(=\frac{(1-\sin \mathrm{A})(1+\sin \mathrm{A})}{\cos ^{2} \mathrm{~A}}=\frac{1-\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}\)
\(=\frac{\cos ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}=1=\mathrm{RHS}\)

Multiple Choice Question

14.

(a)
30√3 cm

15.

16.

17.

18.

(d)
140.83 m

19.

(b)
10⁴ sec

20.

(d)
30√3

21.

22.

(b)
60°

23.

24.

(a)
75√3

25.

(a)
30°

26.

27.

(b)
10 (1 + √2) m

28.

29.

30.

31.

(a)
Line of sight

32.

(b)
87 m

33.

(b)
50 √3 m²

CBSE 10th Maths Introduction to Trigonometry Model Questions Papers Study Material - QB365 Set B

QB365 Student Learing Managemnet System

Std: 10th Standard CBSE - Maths

chapter: Introduction to Trigonometry - Model Questions Papers

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