Ex. 1Multiple choice
The value of (sin30° + cos30°) – (sin60° + cos60°) is
- (A)–1
- (B)0
- (C)1
- (D)2
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Solution : Answer (B) tan 30°
Chapter 8 – Introduction To Trigonometry And Its Applications
Class 10 Mathematics · 52 questions · 23 with answers
Solved examples
Ex. 2Short answer
The value of is cot 60°
Ex. 3Multiple choice
The value of (sin 45° + cos 45°) is 1 3
- (A)
- (B)2
- (C)
- (D)1 2 2
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Solution : Answer (B)
Ex. 1Short answerExercise 8.1
The value of sinθ + cosθ is always greater than 1.
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Solution : False. The value of (sinθ + cosθ) for θ = 0° is 1.
Ex. 2Short answerExercise 8.1
The value of tanθ (θ < 90°) increases as θ increases.
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Solution : True. In Fig. 8.2, B is moved closer to C along BC. It is observed that (i) θ increases (as θ1 > θ, θ2 > θ1, ...) and (ii) BC decreases (B1C < BC, B2C < B1C, ...) Thus the perpendicular AC remains fixed and the base BC decreases. Hence tanθ increases as θ increases.
Ex. 3Short answerExercise 8.1
tanθ increases faster than sinθ as θ increases.
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Solution : True We know that sinθ increases as θ increases but cosθ decreases as θ increases. sin θ We have tan θ = cos θ Now as θ increases, sinθ increases but cosθ decreases. Therefore, in case of tanθ, the numerator increases and the denominator decreases. But in case of sinθ which can be sin θ seen as , only the numerator increases but the denominator remains fixed at 1. Hence tanθ increases faster than sinθ as θ increases.
Ex. 4Short answerExercise 8.1
The value of sinθ is a + , where ‘a’ is a positive number.
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Solution : False.
Ex. 1Short answerExercise 8.2
Prove that sin6θ + cos6θ + 3sin2θ cos2θ = 1
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Solution : We know that sin2θ + cos2θ = 1 Therefore, (sin2θ + cos2θ)3 = 1 or, (sin2θ)3 + (cos2θ)3 + 3sin2θ cos2θ (sin2θ + cos2θ) = 1 or, sin6θ + cos6 θ + 3sin2θ cos2θ = 1
Ex. 2Short answerExercise 8.2
Prove that (sin4θ – cos4θ +1) cosec2θ = 2
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Solution : L.H.S. = (sin4θ – cos4θ +1) cosec2θ = [(sin2θ – cos2θ) (sin2θ + cos2θ) + 1] cosec2θ = (sin2θ – cos2θ + 1) cosec2θ [Because sin 2θ + cos2θ =1] = 2sin2θ cosec2θ [Because 1– cos 2θ = sin2θ ] = 2 = RHS
Ex. 3Short answerExercise 8.2
Given that α + β = 90°, show that cos α cosecβ – cos α sin β = sin α
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Solution : cos α cosecβ – cos α sin β = cos α cosec (90° − α ) – cos α sin (90° − α ) [Given α + β = 90°] = cos α sec α – cos α cos α = 1 − cos2 α = sin α
Ex. 4Short answerExercise 8.2
If sin θ + cos θ = 3 , then prove that tan θ + cot θ = 1
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Solution : sin θ + cos θ = 3 (Given) or (sin θ + cos θ)2 = 3 or sin2 θ + cos2θ + 2sinθ cosθ = 3 2sinθ cosθ = 2 [sin2θ + cos2θ = 1] or sin θ cos θ = 1 = sin2θ + cos2θ sin 2 θ + cos2 θ or 1= sin θ cos θ Therefore, tanθ + cotθ = 1 INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 95
Ex. 1Short answerExercise 8.3
A spherical balloon of radius r subtends an angle θ at the eye of an observer. If the angle of elevation of its centre is φ, find the height of the centre of the balloon.
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Solution : In Fig. 8.3, O is the centre of balloon, whose radius OP = r and ∠PAQ = θ. Also, ∠OAB = φ. Let the height of the centre of the balloon be h. Thus OB = h. θ r Now, from ∆OAP, sin = , where OA = d (1) 2 d Also from ∆OAB, sin = . (2) sin φ d h = = From (1) and (2), we get θ r r 2 d θ or h = r sin φ cosec .
Ex. 2Short answerExercise 8.3
From a balloon vertically above a straight road, the angles of depression of two cars at an instant are found to be 45° and 60°. If the cars are 100 m apart, find the height of the balloon. INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 97
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Solution : Let the height of the balloon at P be h meters (see Fig. 8.4). Let A and B be the two cars. Thus AB = 100 m. From ∆PAQ, AQ = PQ = h PQ h Now from ∆PBQ, = tan 60° = 3 or = 3 BQ h – 100 or h= 3 (h –100) 100 3 Therefore, h = = 50 (3 + 3 ) 3 –1 i.e., the height of the balloon is 50 (3 + 3 ) m.
Ex. 3Short answerExercise 8.3
The angle of elevation of a cloud from a point h metres above the surface of a lake is θ and the angle of depression of its reflection in the lake is φ. tan φ + tan θ Prove that the height of the cloud above the lake is h . tan φ − tan θ
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Solution : Let P be the cloud and Q be its reflection in the lake (see Fig. 8.5). Let A be the point of observation such that AB = h. Let the height of the cloud above the lake be x. Let AL = d. x−h Now from ∆PAL, = tan θ (1) x+h From ∆QAL, = tanφ (2) From (1) and (2), we get x + h tan φ = x – h tan θ 2 x tan φ+ tan θ or = 2h tan φ − tan θ tan φ + tan θ Therefore, x = h . tan φ− tan θ INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 99
Questions
Q1Multiple choice
1
- (A)
- (B)
- (C)3
- (D)1
Q2Multiple choice
3 Solution : Answer (D)
If cos A = , then the value of tan A is
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(B)
3 4 5
- (A)
- (B)
- (C)
- (D)
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(B)
4 3 3 2. If sin A = , then the value of cot A is 1 3
- (A)3
- (B)
- (C)
- (D)1 3 2 3. The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is (A) –1 (B) 0 (C) 1 (D) 4. Given that sinθ = , then cosθ is equal to b b b2 – a2 a (A) (B) (C) (D) b2 – a2 a b b2 – a2 5. If cos (α + β) = 0, then sin (α – β) can be reduced to (A) cos β (B) cos 2β (C) sin α (D) sin 2α
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(B)
The value of (tan1° tan2° tan3° ... tan89°) is
- (A)0
- (B)1
- (C)2
- (D)
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(B) 1
If cos 9α = sinα and 9α < 90° , then the value of tan5α is
- (A)
- (B)3
- (C)1
- (D)0
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(C) 1
If ∆ABC is right angled at C, then the value of cos (A+B) is 1 3
- (A)0
- (B)1
- (C)
- (D)2 2
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(A) 0
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is
- (A)1
- (B)
- (C)2
- (D)3
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(A) 1
1 10. Given that sinα = and cosβ = , then the value of (α + β) is
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(B)
2
- (A)0°
- (B)30°
- (C)60°
- (D)90° INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 91 sin 2 22° + sin 2 68° 11. The value of the expression 2 2 + sin 2 63° + cos 63° sin 27° is cos 22° + cos 68° (A) 3 (B) 2 (C) 1 (D) 0 4sin θ − cos θ 12. If 4 tanθ = 3, then 4sin θ+ cos θ is equal to 2 1 1 3 (A) (B) (C) (D)
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(A) 0°
3 2 4 13. If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is 3 1 1
- (A)1
- (B)
- (C)
- (D)4 2 4 14. sin (45° + θ) – cos (45° – θ) is equal to (A) 2cosθ (B) 0 (C) 2sinθ (D) 1 15. A pole 6 m high casts a shadow 2 3 m long on the ground, then the Sun’s elevation is (A\) 60° (B) 45° (C) 30° (D) 90°
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(B)
1 We know that a− ≥ 0 or a + ≥ 2 , but sinθ is not greater than 1. a a Alternatively, there exists the following three posibilities : Case 1. If a < 1, then a + >1 Case 2. If a = 1, then a + >1 Case 3. If a > 1, then a + >1 However, sin θ cannot be greater than 1. INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS 93
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(B)
=1 cot 43°
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True
The value of the expression (cos2 23° – sin2 67°) is positive.
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False
The value of the expression (sin 80° – cos 80°) is negative.
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False [sin 80° – sin 10º = positive : as θ increases, value of sin θ increases ]
(1– cos 2 θ) sec 2 θ = tan θ
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True
If cosA + cos2A = 1, then sin2A + sin4A = 1.
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True
(tan θ + 2) (2 tan θ + 1) = 5 tan θ + sec2θ.
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False
If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.
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False
If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.
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False
The value of 2sinθ can be a + , where a is a positive number, and a ≠ 1. a 2 + b2
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False
cos θ = , where a and b are two distinct numbers such that ab > 0. 2 ab
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False
The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.
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False
If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains unchanged.
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True
1 + cos θ + sin θ = 2cosecθ tan A tan A
1 + sec A − 1 − sec A = 2cosec A
12 3. If tan A = , then sinA cosA =
25 4. (sin α + cos α) (tan α + cot α) = sec α + cosec α
( 3 +1) (3 – cot 30°) = tan 60° – 2 sin 60° cot 2 α
1 + = cosec α 1+ cosec α
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
Find the angle of elevation of the sun when the shadow of a pole h metres high is 3 h metres long.
If 3 tan θ = 1, then find the value of sin2θ – cos2 θ.
A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.
Simplify (1 + tan2θ) (1 – sinθ) (1 + sinθ)
If 2sin2θ – cos2θ = 2, then find the value of θ. cos2 (45° + θ) + cos 2 (45° – θ)
Show that =1 tan (60° + θ) tan (30° − θ)
An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Show that tan4θ + tan2θ = sec4θ – sec2θ.
If cosecθ + cotθ = p, then prove that cosθ = . p 2 +1
Prove that sec 2 θ+ cosec 2 θ = tan θ + cot θ
The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
If 1 + sin2θ = 3sinθ cosθ , then prove that tanθ = 1 or .
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is st .
The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Prove that the height of the h tan α tower is tan β − tan α . l 2 +1
If tanθ + secθ = l, then prove that secθ = . 2l
If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q (p2 – 1) = 2p.
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = a2 + b2 – c2 .
+ sec θ – tan θ 1 – sin θ 12. Prove that 1 + sec θ + tan θ = cos θ 13. The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower. 14. From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects. 15. A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal. p cos β – cos α Show that q = sin α – sin β . 16. The angle of elevation of the top of a vertical tower from a point on the ground is 60o . From another point 10 m vertically above the first, its angle of elevation is 45o. Find the height of the tower. 17. A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α and β, respectively. Prove that the height of the other house is h ( 1 + tan α cot β ) metres. 18. The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60o and 30o, respectively. Find the height of the balloon above the ground.