SSC CGL Previous Year Questions: Trigonometry - 4

# SSC CGL Previous Year Questions: Trigonometry - 4 - SSC CGL

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## 40 Questions MCQ Test SSC CGL (Tier - 1) - Previous Year Papers (Topic Wise) - SSC CGL Previous Year Questions: Trigonometry - 4

SSC CGL Previous Year Questions: Trigonometry - 4 for SSC CGL 2023 is part of SSC CGL (Tier - 1) - Previous Year Papers (Topic Wise) preparation. The SSC CGL Previous Year Questions: Trigonometry - 4 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL Previous Year Questions: Trigonometry - 4 MCQs are made for SSC CGL 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SSC CGL Previous Year Questions: Trigonometry - 4 below.
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SSC CGL Previous Year Questions: Trigonometry - 4 - Question 1

### (3π/5) radians is equal to   (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 1 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 2

### Evaluate: 3 cos 80° cosec 10° + 2 cos 59° cosec 31°    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 2

3 cos 80°. cosec 10° + 2 cos 59°. cosec 31°
= 3 cos (90° – 10°). cosec 10° + 2 cos (90° – 31°). cosec 31°
= 3 sin 10°. cosec 10° + 2 sin 31°. cosec 31° = 3 + 2 = 5
[∵ cos (90° – θ) = sinθ; sinθ.cosecθ = 1]

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 3

### If the angles of elevation of a balloon from two consecutive kilometre–stones along a road are 30° and 60° respectively, then the height of the balloon above the ground will be    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 3  SSC CGL Previous Year Questions: Trigonometry - 4 - Question 4

From 125 metre high towers, the angle of depression of a car is 45°. Then how far the car is from the tower?   (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 4 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 5

If tan α + cot α = 2, then the value of tan 7α + cot7α is    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 5

tan α + cot α = 2 ⇒ tan2 α – 2 tan α + 1 = 0
⇒ tan2 α – tan α – tan α + 1 = 0
⇒ tan α (tan α – 1) – 1 (tan α – 1) = 0 (tan α – 1) (tan α – 1) = 0
∴ tan α = 1 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 6

If x = a cos θ – b sin θ, y = b cos θ + a sin θ, then find the value of x2 + y2.    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 6

x = a cos θ – b sin θ
y = b cos θ + a sin θ
x2 + y2 = (a cos θ – b sin θ)2 + (b cos θ + a sin θ)2 = a2 cos2 θ + b2 sin2 θ + a2 sin2 θ + b2 cos2 θ
= a2 + b2

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 7

In a triangle, the angles are in the ratio 2 : 5 : 3. What is the value of the least angle in the radian?    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 7

Let angles are 2x, 5x and 3x.
2x + 5x + 3x = 180º (sum of interior angle of triangles is 180º)
10x = 18º
x = 18º
∴ Least angle in degree = 2x = 2 × 18 = 36º SSC CGL Previous Year Questions: Trigonometry - 4 - Question 8

The value of (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 8 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 9

The value of (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 9 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 10

The value of (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 10 [∵ cos (90º – θ) = sin θ and cot (90º – θ) = tan θ]

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 11

If cos4θ – sin4θ = 2/3, then the value of 1 – 2 sin2θ is    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 11 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 12

If cosθ + sinθ = √2 cos θ, then cosθ – sinθ is    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 12

cosθ + sinθ = √2 cos θ
On squaring both sides,
cos2θ + sin2θ + 2cos θ. sin θ = 2 cos2 θ
⇒ cos2θ – sin2θ = 2cosθ. cos θ
⇒ (cos θ + sin θ) (cos θ – sin θ) = 2 sin θ. cos θ
⇒ 2 cos θ (cosθ – sinθ) = 2sin θ. cos θ SSC CGL Previous Year Questions: Trigonometry - 4 - Question 13

The value of (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 13

Expression SSC CGL Previous Year Questions: Trigonometry - 4 - Question 14

If tan α = n tan β, and sin α = m sin β, then cos2 α is    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 14   SSC CGL Previous Year Questions: Trigonometry - 4 - Question 15

The tops of two poles of height 24 m and 36 m are connected by a wire. If the wire makes an angle of 60° with the horizontal, then the length of the wire is    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 15 BC = ED = 24 m
∴ AE = AD – ED = 36 m  – 24 m = 12 m
In ΔABE, SSC CGL Previous Year Questions: Trigonometry - 4 - Question 16

The value of (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 16

Expression SSC CGL Previous Year Questions: Trigonometry - 4 - Question 17

If tan θ = 3/4 and θ is acute, then cosec θ   (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 17 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 18

One of the four an gles of a rhombus is 60°. If the length of each side of the rhombus is, 8 cm, then the length of the longer diagonal is    (SSC CGL 2nd Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 18 ∴ ∠BAO = 30°
∠ABO = 60° ∴ OA = 43
∴ AC = 83 metre

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 19

If (1 + sinα) (1 + sinβ) (1 + sinγ) = (1 – sinα) (1 – sinβ) (1 – sinγ), then each side is equal to   (SSC CGL 2nd Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 19

(1 + sin α) (1 + sin β) (1 + sin γ) = (1 – sin α)  (1 – sin β) (1 – sin γ) = x
∴ x.x = (1 + sin α) (1– sin α) (1 + sin β)
(1 – sin β) (1 + sin γ) (1 – sin γ)
x2 = (1 – sin2 α) (1 – sin2 β) (1 – sin2 γ)
x2 = cos2 α. cos2β.cos2 γ
⇒ x = ± cos α. cos β . cos γ

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 20

The value of 152 (sin 30° + 2 cos2 45° + 3 sin 30° + 4 cos2 45° +...+17 sin 30° + 18 cos45°) is    (SSC CGL 2nd Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 20

152 (sin 30° + 2 cos2 45° + 3 sin 30° + 4 cos2 45° +...+17 sin 30° + 18 cos45°) Thus, Required number is a perfect square of an integer.

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 21

Maximum value of (2 sin θ + 3 cos θ) is    (SSC CGL 2nd Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 21

Maximum value of 2 sin θ + 3 cos θ SSC CGL Previous Year Questions: Trigonometry - 4 - Question 22

If sin ( A – B ) = 1/2 and cos (A + B) = 1/2 where A > B > 0 and A + B is an acute angle, then the value B is    (SSC CGL 2nd Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 22

sin (A – B) = 1/2 = sin 30° ⇒ A – B= 30°
cos (A + B) = 1/2 = cos 60° ⇒ A + B = 60°
∴ A + B + A – B = 30° + 60° = 90°
⇒ 2A = 90° ⇒ A = 45°
∴ A – B = 30° SSC CGL Previous Year Questions: Trigonometry - 4 - Question 23

Evaluate : tan 1° tan 2° tan 3° ........... tan 89°.    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 23

tan 89° = tan (90° – 1°) = cot 1°
tan 88° = tan (90° – 2°) = cot 2°
∴ Expression = tan 1°, cot 1°, tan 2°,  cot2° ... tan 45° = 1
[∵ tan θ. cot θ = 1]

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 24

If secθ + tanθ = 2 + √5,  then the value of sinθ + cosθ is?   (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 24  SSC CGL Previous Year Questions: Trigonometry - 4 - Question 25

If sinθ + cosecθ = 2, then sin9θ + cosec9θ is equal to ______.    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 25

sinθ + cosecθ = 2 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 26

The degree measure of 1 radian (taking π = (22 / 7)) is

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 26 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 27

The value of (sin2 25° + sin2 65°) is:    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 27

sin2 25° + sin2 65°
= sin2 25° + sin2 (90° – 25°)
= sin2 25° + cos2 25° = 1

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 28

The value of cos 1° cos 2° cos 3°............ cos 177° cos 178° cos 179° is:    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 28

cos 90° = 0
∴ cos 1°, cos 2° .... cos 179° = 0

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 29

The angle of elevation of a tower from a distance 100 m from its foot is 30°. Height of the tower is:    (SSC CGL 1st Sit. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 29 AB = h metre
∠ACB = 30°;
BC = 100 metre SSC CGL Previous Year Questions: Trigonometry - 4 - Question 30

If 2(cos2θ - sin2θ) = 1, (θ is positive acute angle), then cotθ is equal to?   (SSC CHSL 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 30

2 cos 2θ = 1
cos 2θ = 1/2
θ = 30°
cot 30° = √3

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 31

If x sin θ + y cos θ = and then the correct relation is   (SSC CHSL 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 31 Put x = sin θ
y = cos θ in the above equation, we have ⇒ 1 = 1
⇒ x = sin θ & y = cos θ is the solution of above equation. Now, on using x = sin θ & y = cos θ in SSC CGL Previous Year Questions: Trigonometry - 4 - Question 32

If 5 tan θ = 4, then (SSC CHSL 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 32 SSC CGL Previous Year Questions: Trigonometry - 4 - Question 33

The length of the shadow of a vertical tower on level ground increases by 10 metres when the altitude of the sun changes from 45° to 30°. Then the height of the tower is    (SSC CHSL 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 33  SSC CGL Previous Year Questions: Trigonometry - 4 - Question 34

2 cosec2  23° cot 2  67° – sin2  23° – sin2  67°– cot2 67° is equal to    (SSC CHSL 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 34 = 2 sec2 23° - 1 - tan2 23°
= (sec2 23°- 1) + (sec2 23°- tan2 23°)
= tan2 23° + 1 = sec2 23°

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 35

If cosec θ – cot θ = 7/2, the value of cosec θ is:    (SSC Sub. Ins. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 35

cosec θ – cot θ = 7/2 ... (i)
cosec2 θ – cot2 θ = 1
⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1  SSC CGL Previous Year Questions: Trigonometry - 4 - Question 36

If sin θ – cos θ = 1/2 the value of sin θ + cos θ is:    (SSC Sub. Ins. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 36

sin θ – cos θ = 1/2
sin θ + cos θ = x. SSC CGL Previous Year Questions: Trigonometry - 4 - Question 37

The minimum value of 4 tan2θ + 9 cot2θ is equal to    (SSC Sub. Ins. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 37

4 tan2 θ + 9 cot2 θ
⇒ (2 tan θ)2 + (3 cot θ)2
(2 tan θ)2 + (3 cot θ)2 – 12 + 12 = (2 tan θ – 3 cot θ)2 + 12
∴ Minimum value = 12 because (2 tan θ – 3 cot θ)2 ≠ 0

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 38

The value of tan 1° tan 2° tan 3° ... tan 89° is:    (SSC Sub. Ins. 2013)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 38

tan 1° tan 2° tan 3° ... tan 89°
= tan 1° tan 2°... tan45° ... tan (90° – 2 ) tan (90° – 1)
= tan 1° tan 2° ... 1 ... cot 2° cot 1°
= (tan 1° cot 1°) (tan 2° cot 2°) ... 1 = 1

SSC CGL Previous Year Questions: Trigonometry - 4 - Question 39

The simplified value of (1 + tan θ + sec θ) (1 + cot θ – cosec θ) is    (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 39

(1 + tan θ + sec θ) (1 + cot θ – cosec θ) SSC CGL Previous Year Questions: Trigonometry - 4 - Question 40

If sin2α = cos3α, then the value of (cot6α – cot2α) is    (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 4 - Question 40

If sin2α = cos3α
tan2α = cosα ...(1)
Now consider, cot6α – cot2α Substituting for tan2α with cos α from (1) above equation will be ## SSC CGL (Tier - 1) - Previous Year Papers (Topic Wise)

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