Basic Trigonometry Concepts 2 – Questions and Solved Examples for CAT

Friday, July 13th, 2018


Basic Trigonometry Concepts 2 – Questions and Solved Examples for CAT

In the earlier part of trigonometry series, we had seen basic concepts and some questions. In this part we will see some additional Examples to understand how to deal with CAT questions

Example 1: 3sinx + 4cosx + r is always greater than or equal to 0. What is the smallest value 'r' can to take?

a.) 5

b.) -5

c.) 4

d.) 3

Solution:  3sinx + 4cosx ≥ -r

⇒ 5(3/5sinx+4/5cosx) ≥ -r                  (Since 3,4 and 5 are Pythagorean triplets,)

⇒ 3/5 = cosA

⇒ 4/5 = sin A

⇒ 5(sinx cosA + sinA cosx) ≥ -r

⇒ 5(sin(x + A)) ≥ -r

(Since Now, 5sin (x + A) ≥ -r

We know, -1 ≤ sin (angle) ≤ 1

⇒ 5sin (x + A) ≥ -5

⇒ rmin = 5

Answer Choice (A)

Correct Answer: 5

Example 2:  Sin2014x + Cos2014x = 1, x in the range of [-5π, 5π], how many values can x take?

a.) 0

b.) 10

c.) 21

d.) 11

Solution: We know that Sin2x + Cos2x = 1 for all values of x.

Now, If Sin x or Cos x is equal to –1 or 1, then Sin2014x + Cos2014x will be equal to 1.

Sin x is equal to –1 or 1 when x = –4.5π or –3.5π or –2.5π or –1.5π or –0.5π or 0.5π or 1.5π or 2.5π or 3.5π or 4.5π.    (Since x is in the range of -5π, 5π)

Cos x is equal to –1 or 1 when x = –5π or –4π or –3π or –2π or –π or 0 or π or 2π or 3π or 4π or 5π.

For all other values of x, Sin2014 x will be strictly lesser than Sin2x.

For all other values of x, Cos2014 x will be strictly lesser than Cos2x.

We know that Sin2x + Cos­2x is equal to 1. Hence, Sin2014x + Cos2014x will never be equal to 1 for all other values of x. Thus there are 21 values.

Answer choice (C)

Example 3: If cos A + cos2 A = 1 and a sin12 A + b sin10 A + c sin8 A + d sin6 A – 1 = 0. Find the value of a+b/c+d

a.) 4

b.) 3

c.) 6

d.) 1

Solution: Given,

Cos A = 1- Cos2A

⇒ Cos A = Sin2A

⇒ Cos2A = Sin4A

⇒ 1 – Sin2A = Sin4 A

⇒ 1 = Sin4A + Sin2 A

⇒ 13= (Sin4A + Sin2A)3

⇒ 1 = Sin12A + Sin6A + 3Sin8 A + 3Sin10 A

⇒ Sin12A + Sin6A + 3Sin8 A + 3Sin10 A – 1 = 0

On comparing,
a = 1, b = 3, c = 3, d = 1
∴ a+b/c+d = 3
Answer choice (B)

 

Example 4: If tanϕ+sinϕ=m, tanϕ−sinϕ=n, Find the value of m2 – n2

a.) 4√mn

b.) 2√mn

c.) m – n

d.) 2mn

Solution:  Adding the two equations given above,

Tan ϕ= (m+n)/2

Subtracting the same,

Sin ϕ= (m−n)/2

Since, there are no available direct formula for relation between sin ϕ , tan ϕ but we know that

cosec2ϕ–cos2ϕ=1

⇒ (2/(m−n))2−(2(m+n))2=1

⇒ (4((m+n)2−(m−n)2)) / (m2−n2)2=1

⇒ (m2−n2)2=4(4mn)

⇒ m2 – n2 = 4√mn

Answer choice (A)

Example 5: A student is standing with a banner at the top of a 100 m high college building. From a point on the ground, the angle of elevation of the top of the student is 60° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the student. 

a.) 73.2 m

b.) 35 m

c.) 50 m

d.) 75 m

Let BC be the height of the tower and DC be the height of the student.

In rt. ∆ ABC

AB = BC cot 45°

AB = 100 m …………… (1)

In rt. ∆ ABD

AB = BD cot 60°

AB = (BC + CD) cot 60°

AB = (10 + CD) * 1√3

Equating (i) and (ii)

(10 + CD) * 1√3 = 100

(10 + CD)= 100√3

CD = 100√3 – 100

= 10(1.732 – 1) = 100 x 0.732 = 73.2 m

Answer choice (A)

Example 6: A right angled triangle has a height 'p', base 'b' and hypotenuse 'h'. Which of the following value can h2 not take, given that p and b are positive integers? 

a.) 52

b.) 74

c.) 13

d.) 23

Solution:

We know that,
h2 = p2 + b2 Given, p and b are positive integer, so h2 will be sum of two perfect square.
We see

a) 62 + 42 = 52

b) 72 + 52 = 74

c) 32 + 22 = 13

d) Can't be expressed as a sum of two perfect squares

Answer choice (D)

Correct Answer: 23

 

Example 7: A tall tree AB and a building CD are standing opposite to each other. A portion of the tree breaks off and falls on top of the building making an angle of 30°. After a while it falls again to the ground in front of the building, 4 m away from foot of the tree, making an angle of 45°. The height of the building is 6 m. find the total height of the tree in meters before it broke. 

a.) 27√3 + 39

b.) 12√3 + 10

c.) 15√3 + 21

d.) Insufficient Data

Let the broken portion of tree AA' be x. Hence A'C = A'G = x

From the figure, total height of the tree = x + y + 6

Consider triangle A'BG, tan 45° = A’B / BG

A'B = BG

Or BG = y + 6

Consider triangle A'C'C, tan 30° = A'C'/C'C

1/√3 = y/ (BG+4)

⇒ 1/√3 = y(y+6+4 )

⇒ y/√3 = y + 10

⇒ y = 10/(√3−1)

= 10/ (√3−1) * [ (√3+1) / (√3+1 )]

= 10(√3+1) / 2

Therefore, y = 5(√3 + 1)

Take sin 30° to find x

sin 30° = A’C’ / A’C

1/2 = y/x

or x = 2y

 

Height of the tree = x + y + 6

= 2 * 5(√3 + 1) + 5(√3 + 1) + 6

= 10√3 + 10 + 5√3 + 5 + 6

= 15√3 + 21 meters

Answer choice (C)

Correct Answer: 15√3 + 21

Example 8:  Q.16: A flag is hoisted on top of a building of height 7√3 m. A man of height √3 m,standing on the ground, sees the top and bottom of the flag pole at 2 different angles of elevation that are found to be complementary. If the man is standing √135 m away from the building, find the height of the flag pole. 

a.) 5√3 m

b.) 3√3 m

c.) 2 / √3 m

d.) 6 / √3 m

Let height of the flag pole PQ be x
Height of building QR = 7√3 m
Height of man AB = √3
Therefore QQ' = 7√3 – √3 = 6√3

From diagram, tan y° = QQ'/AQ'
Or tan y° = 6√3/√135 ——- (1)

Also, tan (90 – y)° = PQ'/AQ'
Or tan (90 – y)° = (x+QQ′)√135
But tan (90 – y)° = cot y°
Therefore cot y = (x+6√3) / √135
or tan y = √135 / (x+6√3) ——–(2)

Equating (1) and (2), we get
6√3/√135 = √135/ (x+6√3 )
6√3x + 108 = 135
6√3x = 27
x = 9 / 2√3 or x = 1.5√3
Answer choice (A)
Correct Answer: 1.5√3 m

Example 9: What is the value of (1-CosA) / (1+CosA) given that tan A = ¾?

a.) 1/4

b.) 9/16

c.) 1/9

d.) 1/3

Solution: We know that tan A = opposite side / adjacent side = 3/4

Therefore, opposite side = 3, adjacent side = 4 and the hypotenuse = 5

Hence, cos A = 4/5

Therefore, (1-Cos A) / (1 + Cos A) = (1- (4/5)) / (1 + (4/5)) = 1/9

Answer Choice 3

 

Example 10: If (2Sinx)(1+cosx+Sinx)=t,          (1–Cosx+Sinx)/(1+Sinx)  can be written as:

a.) 1/t

b.) t

c.) √t

d.) tSinx

Solution:   

= (2Sinx)/(1+cosx+Sinx)∗ [(1+Sinx−Cosx)(1+Sinx−Cosx) ]

= 2 Sin x * (1+Sinx–Cosx) / ((1+Sinx)2–Cos2x)

= 2 Sin x * (1+Sinx–Cosx) / (1+Sin2x+2Sinx–Cos2x)

= 2 Sin x . (1+Sinx–Cosx) / (Cos2x+Sin2x+Sin2x+2Sinx–Cos2x)

(Since, 1 = Cos2x + Sin 2x)

= 2 Sin x * (1+Sinx–Cosx) / (2Sin2x+2Sinx)

= 2 Sin x * (1+Sinx–Cosx) / (2Sinx(1+Sinx))

= (1–Cosx+Sinx)(1+Sinx) = t

Answer Choice B.

Conclusion:   As we have seen, questions from trigonometry are based on mostly heights and distances and trigonometric identities. The questions above on height and distances are good representations of types of problems one can expect in the exams also. For solving Identities related questions, if we remember some of the main formulae, value of the angles for ratios and Pythagorean triplets we should be good to go.

You can also see: Should You Quit Your Job To Prepare For CAT?

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Basic Trigonometry Concepts 2 – Questions and Solved Examples for CAT

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