Problem Set 6 Geometry 10th Std Maths Part 2 Answers Chapter 6 Trigonometry
i. sin θ.cosec θ = ?
(A) 1
(B) 0
(C)
(D)
(A)
ii. cosec 45° = ?
(A)
(B)
(C)
(D)
Answer:
(B)
iii. 1 + tan2θ = ?
(A) cot2θ
(B) cosec2θ
(C) sec2θ
(D) tan2θ
Answer:
(C)
iv. When we see at a higher level, from the horizontal line, angle formed is ______
(A) angle of elevation.
(B) angle of depression.
(C) 0
(D) straight angle.
Answer:
(A)
Question 2
Maharashtra Board Solution
If sin θ = , find the value of cos θ using trigonometric identity.
Solution & Step-by-Step Answer:
sin θ = … [Given] We know that, sin2 θ + cos2 θ = 1 …[Taking square root of both sides]
Question 3
Maharashtra Board Solution
If tan θ = 2, find the values of other trigonometric ratios.
Solution & Step-by-Step Answer:
tan θ = 2 …[Given] We know that, 1 + tan2 θ = sec7 θ ∴ 1 + (2)7 = sec7 θ ∴ 1 + 4 = sec7 θ ∴ sec7 θ = 5 ∴ sec θ = …[Taking square root of both sides]
Question 4
Maharashtra Board Solution
If sec θ = , find the values of other trigonometric ratios.
Solution & Step-by-Step Answer:
sec θ = … [Given] We know that, 1 + tan2 θ = sec2 θ ∴ sin θ = , cos θ = , tan θ = , cot θ = , cosec θ =
Question 5
Maharashtra Board Solution
Prove the following: i. sec θ (1 – sin θ) (sec θ + tan θ) = 1 ii. (sec θ + tan θ) (1 – sin θ) = cos θ iii. sec2 θ + cosec2 θ = sec2 θ × cosec2 θ iv. cot2 θ – tan2 θ = cosec2 θ – sec2 θ v. tan4 θ + tan2 θ = sec4 θ – sec2 θ vi. = 2 sec2 θ vii. sec6 x – tan6 x = 1 + 3 sec2 x × tan2 x Proof: i. L.H.S. = sec θ (1 – sin θ) (sec θ + tan θ) ∴ sec θ (1 – sin θ) (sec θ + tan θ) = 1
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ii. L.H.S. = (sec θ + tan θ) (1 – sin θ)
∴ (sec θ + tan θ) (1 – sin θ) = cos θ
iii. L.H.S. = sec2θ + cosec2θ
∴ sec2θ + cosec2θ = sec2θ × cosec2θ
iv. L.H.S. = cot2θ – tan2θ
= (cosec2θ – 1) – (sec2θ – 1)
[∵ tan2θ = sec2θ – 1,
cot2θ = cosec2θ – 1]
= cosec2θ – 1 – sec2θ + 1
cosec2θ – sec2θ
= R.H.S.
∴ cot2θ – tan2θ = cosec2θ – sec2θ
v. L.H.S. = tan4θ + tan2θ
= tan2θ (tan2θ + 1)
= tan2θ. sec2θ
…[∵ 1 + tan2θ = sec2θ]
= (sec2θ – 1) sec2θ
…[∵ tan2θ = sec2θ – 1]
= sec4θ – sec2θ
= R.H.S.
∴ tan4θ + tan2θ = sec4θ – sec2θ
vii. L.H.S. = sec6x – tan6x
= (sec2x)3– tan6x
= (1 + tan2x)3– tan6x …[∵ 1 + tan2θ = sec2θ]
= 1 + 3tan2x + 3(tan2x)2+ (tan2x)3– tan6x …[∵ (a + b)3= a3+ 3a2b + 3ab2+ b3]
= 1 + 3 tan2x (1 + tan2x) + tan6x – tan6x
= 1 + 3 tan2x sec2x …[∵ 1 + tan2θ = sec2θ]
= R.H.S.
∴ sec3x – tan6x = 1 + 3sec2x.tan2x
x. We know that,
sin2θ + cos2θ = 1
∴ 1 – sin2θ = cos2θ
∴ (1 – sin θ) (1 + sin θ) = cos θ. cos θ
Question 6. A boy standing at a distance of 48 metres from a building observes the top of the building and makes an angle of elevation of 30°. Find the height of the building.
Solution & Step-by-Step Answer:
Let AB represent the height of the building and point C represent the position of the boy. Angle of elevation = ∠ACB = 30° BC = 48 m In right angled ∆ABC, tan 30° = … [By definition] ∴ The height of the building is 16 m.
Question 7
Maharashtra Board Solution
From the top of the lighthouse, an observer looks at a ship and finds the angle of depression to be 30°. If the height of the lighthouse is 100 metres, then find how far the ship is from the lighthouse.
Solution & Step-by-Step Answer:
Let AB represent the height of lighthouse and point C represent the position of the ship. Angle of depression ∠PAC 30° AB = 100m. Now, ray AP || seg BC ∴ ∠ACB = ∠PAC … [Alternate angles] ∴ ∠ACB = 30° AB = 100m In right angled ∆ABC, tan 30° = …[By definition] ∴ ∴ BC = 100m ∴ The ship is 1oom far from the lighthouse.
Question 8
Maharashtra Board Solution
Two buildings are in front of each other on a road of width 15 metres. From the top of the first building, having a height of 12 metre, the angle of elevation of the top of the second building is 30°. What is the height of the second building?
Solution & Step-by-Step Answer:
Let AB and CD represent the heights of the two buildings, and BD represent the width of the road. Draw seg AM ⊥ seg CD Angle of elevation = ∠CAM = 30° AB = 12m BD = 15m In ꠸ ABDM, ∠B = ∠D = 90° ∠M 90° …[segAM ⊥ segCD] ∠A 90° …[Remaining angle of ꠸ABDM] ꠸ABDM is a rectangle …[Each angle is 90°] ∴ The height of the second building is 20.65 m.
Question 9
Maharashtra Board Solution
A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2 m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)
Solution & Step-by-Step Answer:
Let AB represent the length of the ladder and AE represent the height of the platform. Draw seg AC ⊥ seg BD. Angle of elevation = ∠BAC = 70° AB = 20 m AE = 2m In right angled ∆ABC, sin 70° = …..[By definition] ∴ 0.94 = ∴ BC = 0.94 × 20 = 18.80 m In ꠸ACDE, ∠E = ∠D = 90° ∠C = 90° … [seg AC ⊥ seg BD] ∴ ∠A = 90° … [Remaining angle of ꠸ACDE] ∴ ꠸ACDE is a rectangle. … [Each angle is 90°] ∴ CD = AE = 2 m … [Opposite sides of a rectangle] Now, BD = BC + CD … [B – C – D] = 18.80 + 2 = 20.80 m ∴ The maximum height from the ground upto which the ladder can reach is 20.80 metres.
Question 10
Maharashtra Board Solution
While landing at an airport, a pilot made an angle of depression of 20°. Average speed of the plane was 200 km/hr. The plane reached the ground after 54 seconds. Find the height at which the plane was when it started landing, (sin 20° = 0.342)
Solution & Step-by-Step Answer:
Let AC represent the initial height and point A represent the initial position of the plane. Let point B represent the position where plane lands. Angle of depression = ∠EAB = 20° Now, seg AE || seg BC ∴ ∠ABC = ∠EAB … [Alternate angles] ∴ ∠ABC = 20° Speed of the plane = 200 km/hr = 200 × m/sec = m/sec ∴ Distance travelled in 54 sec = speed × time = × 54 = 3000 m ∴ AB = 3000 m In right angled ∆ABC, sin 20° = ….[By definition] ∴ 0.342 = ∴ AC = 0.342 × 3000 = 1026 m ∴ The plane was at a height of 1026 m when it started landing.
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