*Friday, July 13th, 2018*

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

⇒ r_{min} = 5

Answer Choice (A)

Correct Answer: 5

__Example 2:__** Sin ^{2014}x + Cos^{2014}x = 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 Sin

Now, If Sin x or Cos x is equal to –1 or 1, then Sin^{2014}x + Cos^{2014}x 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, Sin^{2014 }x will be strictly lesser than Sin^{2}x.

For all other values of x, Cos^{2014} x will be strictly lesser than Cos^{2}x.

We know that Sin^{2}x + Cos^{2}x is equal to 1. Hence, Sin^{2014}x + Cos^{2014}x 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 + cos

a.) 4

b.) 3

c.) 6

d.) 1

**Solution:** Given,

Cos A = 1- Cos^{2}A

⇒ Cos A = Sin^{2}A

⇒ Cos^{2}A = Sin^{4}A

⇒ 1 – Sin^{2}A = Sin^{4} A

⇒ 1 = Sin^{4}A + Sin^{2} A

⇒ 1^{3}= (Sin^{4}A + Sin^{2}A)^{3}

⇒ 1 = Sin^{12}A + Sin^{6}A + 3Sin^{8} A + 3Sin^{10} A

⇒ Sin^{12}A + Sin^{6}A + 3Sin^{8} A + 3Sin^{10} 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 m ^{2} – n^{2}**

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

cosec^{2}ϕ–cos^{2}ϕ=1

⇒ (2/(m−n))^{2}−(2(m+n))^{2}=1

⇒ (4((m+n)^{2}−(m−n)^{2})) / (m^{2}−n^{2})^{2}=1

⇒ (m^{2}−n^{2})^{2}=4(4mn)

⇒ m^{2} – n^{2} = 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 h ^{2} not take, given that p and b are positive integers? **

a.) 52

b.) 74

c.) 13

d.) 23

__Solution:__

We know that,

h^{2} = p^{2} + b^{2} Given, p and b are positive integer, so h^{2} will be sum of two perfect square.

We see

a) 6^{2} + 4^{2} = 52

b) 7^{2} + 5^{2} = 74

c) 3^{2} + 2^{2} = 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}–Cos^{2}x)

= 2 Sin x * (1+Sinx–Cosx) / (1+Sin^{2}x+2Sinx–Cos^{2}x)

= 2 Sin x . (1+Sinx–Cosx) / (Cos^{2}x+Sin^{2}x+Sin^{2}x+2Sinx–Cos^{2}x)

(Since, 1 = Cos^{2}x + Sin ^{2}x)

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