SSC CGL Previous Year Questions: Trigonometry - 5 - SSC CGL MCQ

# SSC CGL Previous Year Questions: Trigonometry - 5 - SSC CGL MCQ

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## 41 Questions MCQ Test SSC CGL (Tier - 1) - Previous Year Papers (Topic Wise) - SSC CGL Previous Year Questions: Trigonometry - 5

SSC CGL Previous Year Questions: Trigonometry - 5 for SSC CGL 2024 is part of SSC CGL (Tier - 1) - Previous Year Papers (Topic Wise) preparation. The SSC CGL Previous Year Questions: Trigonometry - 5 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL Previous Year Questions: Trigonometry - 5 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SSC CGL Previous Year Questions: Trigonometry - 5 below.
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SSC CGL Previous Year Questions: Trigonometry - 5 - Question 1

### The angles of elevation of the top of a tower standing on a horizontal plane from two points on a line passing through the foot of the tower at a distance 9 ft and 16 ft respectively are complementary angles. Then the height of the tower is    (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 1

In ΔABC
tan α = h/9 ... (1)
In ΔABD
tan β = h/16

α + β = 90° (given)
β = 90 – α
Since, tan β = h/16

From eqn. (1) and (2)

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 2

### is equal to   (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 2

Dividing Numerator and Denominator by cos θ

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 3

### then the value of  is   (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 3

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 4

In a triangle ABC, AB = AC, BA is produced to D in such a manner that AC = AD. The circular measure of ∠BCD is    (SSC CGL 2nd Sit. 2012)

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

AB = AC
∴ ∠ABC = ∠ACB ...(1)

[opposite angles of equal sides are equal]
AC = AD and BA = AC
∴ ∠ACD = ∠ADC and ∠ABC = ∠ACB ...(2)
In a triangle,
∠ABC + ∠ADC + ∠DCB = 180° (Angle sum property)
∠ABC + ∠ADC + ∠ACB + ∠ACD = 180°
2∠ACB + 2∠ACD = 180°
[From eqn. (1) & (2)]
∴ ∠BCD = 90° or π / 2

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 5

The radian measure of 63°14'51" is   (SSC CGL 2nd Sit. 2012)

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

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 6

If sec θ + tan θ = √3, then the positive value of sin θ is    (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 6

sec2 θ – tan2 θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 7

If 2y cos θ = x sin θ and 2x sec θ – y cosec θ = 3, then the relation between x and y is    (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 7

2y cos θ = x sin θ

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 8

If sinθ + sin2θ = 1, then the value of cos12θ + 3cos10θ + 3cos8θ + cos6θ – 1 is    (SSC CGL 1st  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 8

sinθ + sin2θ = 1
⇒ sinθ = 1 – Sin2θ
⇒ sinθ = cos2θ
∴ cos12θ + 3cos10θ + 3cos8θ + cos6θ – 1
= (cos4θ + cos2θ)3 – 1
= (sin2θ + cos2θ)3 – 1 = 1 – 1 = 0

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 9

When the angle of elevation of the sum increases from 30° to 60°, the shadow of a post is diminished by 5 metres. then the height of the post is    (SSC CGL 1st  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 9

AB = Pole = h metre
BD = x metre
From ΔABC,

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 10

If 0 ≤ 2 ≤ π/2, 2y cosθ = x sin θ and 2x secθ – y cosecθ = 3, then the value of x2 + 4y2 is    (SSC CGL 1st  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 10

2y cos θ = xsin θ
⇒ x sec θ = 2y cosec θ
Also, 2x sec θ – y cosec θ = 3
⇒ 4y cosec θ – y cosec θ = 3
⇒ 3y cosec θ = 3
⇒ y cosec θ = 1
⇒ y = sin θ
∴ x sec θ = 2y cosec θ
= 2sin θ . cosec θ = 2
⇒ x = 2cos θ
∴ x2 + 4y2 = 4cos2θ + 4sin2θ = 4

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 11

If x = cosecθ – sinθ and y = secθ – cosθ, then the value of x2y2 (x2 + y2 + 3) is     (SSC CGL 1st  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 11

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 12

If tan (x + y) tan (x – y) = 1, then the value of tan (2x/3) is   (SSC CGL 1st  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 12

tan (x + y). tan (x – y) = 1
⇒ tan (x + y) = cot (x – y) = tan (90°– x + y)
⇒ x + y = 90° – x + y ⇒ 2x = 90°

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 13

The least value of (4sec2θ + 9 cosec2θ) is    (SSC CGL 1st  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 13

4 sec2 θ + 9cosec2 θ
= 4 (1 + tan2 θ) + 9 (1 + cot2 θ)
= 4 + 4 tan2 θ + 9 + 9cot2 θ
= 4 tan2 θ + 9cot2θ +  12 – 12 + 13
= (2tan2 θ – 3Cot2 θ)2 + 25
{∵ least value of 2 tan2 θ – 3cot2 θ = 0}
∴ the minimum value is 25.

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 14

If 5 tanθ = 4, then the value of     (SSC CGL 1st  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 14

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 15

If x, y are positive acute angles, x + y < 90° and sin (2x – 20°) = cos (2y + 20°), then the value of sec (x + y) is    (SSC CGL 2nd  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 15

sin (2x – 20°) = cos (2y + 20°)
⇒ sin (2x – 20°) = sin (90° – 2y – 20°) = sin (70° – 2y)
⇒ 2x – 20° = 70° – 2y ⇒ 2 (x + y) = 90° ⇒ x + y = 45°
∴ sec (x + y) = sec 45° = √2

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 16

If α is a positive acute angle and 2sinα + 15cos2α = 7, then the value of cotα is:   (SSC CGL 2nd  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 16

2sinα + 15cos2α = 7
⇒ 2 sinα + 15 (1 – sin2α) = 7
⇒ 2 sinα + 15 – 15 sin2α = 7
⇒ 15 sin2α – 2 sinα – 8 = 0
⇒ 15 sin2α – 12 sinα + 10 sinα – 8 = 0
⇒ 3 sinα (5 sinα – 4) + 2 (5 sinα – 4) = 0
⇒ (3 sinα + 2) ( 5 sinα – 4) = 0
⇒ 5 sinα – 4 = 0
⇒  sinα = 4/5
∴ cosecα = 5/4

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 17

If (a2 – b2) sinθ + 2ab cosθ = a2 + b2, then the value of tanθ is    (SSC CGL 2nd  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 17

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 18

If tan θ – cot θ = a and cos θ – sin θ = b, then the value of (a2 + 4) (b2 – 1)2 is:   (SSC CGL 2nd  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 18

Put θ = 45°
a = tan 45° – cot 45°,  b = sin 45° – cos 45°
a = 1 – 1 = 0

Put in equation
(a2 + 4) (b2 –1)2 = (0 + 4) (0 – 1)2 = 4

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 19

If a3 – b3 = 56 and a – b = 2, then the value of (a2 + b2) is:    (SSC CGL 2nd  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 19

(a – b)3 = a3 – b3 – 3ab (a – b)
⇒ 8 = 56 – 3ab (2)
⇒ 6ab = 56 – 8 = 48
⇒ 2ab = 16 ...(i)
∴ a2 + b2 = (a – b)2 + 2ab
= 4 + 16 = 20

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 20

The simplified value of    (SSC CGL 2nd  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 20

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 21

The least value of 4 cosec2α + 9sin2α is:   (SSC CGL 2nd  Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 21

4 cosec2α + 9sin2α
= 4 cosec2α + 4 sin2α + 5 sin2α
= 4 [(cosecα – sinα)+ 2] + 5 sin2α
= 12  [∵ cosecα – sinα ≥ 1]

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 22

If tan(x + y) tan (x – y) = 1, then the value of tan x is:    (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 22

tan (x + y) tan (x – y) = 1
⇒ tan (x + y) = cot (x – y) = tan (90° – (x – y))
⇒ x + y = 90°  – ( x – y)
⇒ 2x = 90°
⇒  x = 45°
∴ tan x = tan 45° = 1

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 23

A boy standing in the middle of a field, observes a flying bird in the north at an angle of elevation of 30° and after 2 minutes, he observes the same bird in the south at an angle of elevation of 60°. If the bird flies all along in a straight line at a height of 50√3 m, then its speed in km/h is:    (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 23

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 24

In a right-angled triangle ABC, ÐB is the right angle and AC = 2√5 cm. If AB – BC = 2 cm, then the value of (cos2A – cos2C) is:   (SSC CGL 2nd Sit.  2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 24

Let BC = x
∴ AB = x + 2
∴ AB2 + BC2 = AC2
⇒ (x + 2)2 + x2 = (2√5) 2 ⇒ x2 + 4x + 4 + x2 = 20
⇒ 2x2 + 4x – 16 = 0
⇒ x2 + 2x – 8 = 0 ⇒ x2 + 4x – 2x – 8 = 0
⇒ x (x + 4)  – 2 (x + 4) = 0
⇒ (x – 2) (x + 4) = 0
⇒ x = 2 = BC
∴ AB = 2 + 2 = 4 cm

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 25

If cot A + cosec A = 3 and A is an acute angle, then the value of cos A is:    (SSC CGL 2nd Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 25

cot A +  cosec A = 3
cosec2 A – cot2 A = 1
(cosec A – cot A) (cosec A + cot A) = 1
cosec A – cot A = 1/3 ... (i)
cosec A + cot A = 3 ... (ii)

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 26

If cos A + cos2 A = 1, then sin2 A + sin4 A is equal to   (SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 26

cos A = 1 – cos2 A = sin2A
∴ sin2A + sin4A = sin2A + cos2A = 1

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 27

A tree is broken by the wind. If the top of the tree struck the ground at an angle of 30° and at a distance of 30 m from the root, then the height of the tree is    (SSC CGL 1st Sit.2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 27

AB = tree
BC = broken part
∴ BC = CD

∴ AB = AC + BC
= 10√3 + 20√3 = 30√3 metre

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 28

sec4θ – sec2θ is equal to   (SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 28

sec4θ – sec2θ = sec2 θ (sec2θ – 1)
= (1 + tan2θ) (1 + tan2θ – 1) = tan2θ + tan4θ

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 29

If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, sin θ ≠ 0, cos θ ≠ 0, then x2 + y2 is        (SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 29

x sin3θ + y cos3θ = sinθ.cosθ
=> (x sinθ). sin2θ + (y cosθ) cos2θ = sinθ.cosθ
=> xsinθ . sin2θ + xsinθ . cos2θ = sinθ . cosθ
=> x sinθ (sin2θ + cos2θ) = sinθ . cosθ
=> x = cosθ
now, x sinθ = y cosθ
=> cosθ . sinθ = y cosθ
=> y = sinθ
x2 + y2 = cos2θ + sin2θ = 1

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 30

If sin θ + cos θ = √2 cos (90° – θ), then cot θ is    (SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 30

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 31

If 2 – cos 2θ = 3 sin θ cos θ, sin θ ≠ cos θ then tan θ is   (SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 31

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 32

(SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 32

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 33

The least value of 2 sin2 θ + 3 cos2 θ is    (SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 33

2sin2 θ + 3 cos2 θ = 2 sin2 θ + 2cos2 θ + cos2 θ
= 2 (sin2θ + cos2θ) + cos2θ = 2 + cos2θ
∴ Least value = 2 + 0 = 2 [∵ cos2 θ ≥ 0]

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 34

If A and B are complementary angles, then the value of sin A cos B + cos A sin B – tan A tan B + sec2 A – cot2 B is    (SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 34

A + B = 90° ⇒  A = 90° – B
⇒ sin A = sin (90° – B) = cos B
Similarly,
⇒ cos A = sin B and tan A = cot B
∴ sin A . cos B + cos A. sin B – tan A. tanB + sec2 A– cot2 B = cos2 B+ sin2 B – cot B. tan B + sec2A – tan2A
= 1 – 1 +1 = 1
[∵ tan B. cot B = 1, sec2 A - tan2 A= 1]

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 35

P and Q are two points observed from the top of a building 10√3 m high. If the angles of depression of the points are complementary and PQ = 20 m, then the distance of P from the building is   (SSC CGL 1st Sit. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 35

In this type of question, just put
h2 = x(x + 20)
⇒ (10√3)2 = x (x + 20)
300 = x (x + 20)
10 (30) = x (x + 20)
x + 20 = 30

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 36

value of cos θ is:    (SSC CHSL 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 36

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 37

then value of cot θ is:   (SSC CHSL 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 37

Dividing numerator and denominator by cos θ.

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 38

then the value of sin 2θ1 + tan 3θ2 is equal to (Assume that 0 < θ1 - θ2 < θ1 + θ2 < 90°)    (SSC Sub. Ins. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 38

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 39

The value of (sec θ – cos θ) (cosec θ – sin θ) (tan θ + cot θ) is   (SSC Sub. Ins. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 39

(sec θ – cos θ) (cosec θ – sin θ) (tan θ + cot θ)

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 40

tan 7° tan 23° tan 60° tan 67° tan 83° is equal to   (SSC Sub. Ins. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 40

tan 7° tan 23° tan 60° tan 67° tan 83°
⇒ tan (90° – 83°) tan (90° – 67°) tan 60° tan 67° tan 83°
⇒ cot 83° cot 67° tan 60° tan 67° tan 83°
[∵ tan (90° – θ) = cotθ]

SSC CGL Previous Year Questions: Trigonometry - 5 - Question 41

A ladder is resting against a wall at height of 10m. If the ladder is inclined with the ground at an angle of 30°, then the distance of the foot of the ladder from the wall is    (SSC Sub. Ins. 2012)

Detailed Solution for SSC CGL Previous Year Questions: Trigonometry - 5 - Question 41

Let ‘x’ be the distance of foot of ladder

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