*Monday, June 8th, 2020*

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: **5sinx + 12cosx + s is always greater than or equal to 0. What could be the possible smallest value of s?

a.) 13

b.) 10

c.) 6

d.) 2

**Solution:** 5sinx + 12cosx â‰¥ -s

â‡’ 13(5/13sinx+12/13cosx) â‰¥ -s Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Since 5,12 and 13 are Pythagorean triplets,)

â‡’ 5/13 = cosA

â‡’ 12/13 = sin A

â‡’ 13(sinx cosA + sinA cosx) â‰¥ -s

â‡’ 13(sin(x + A)) â‰¥ -s (Since sinAcosB + cosAsinB = sin(A+B))

Now, 13sin (x + A) â‰¥ -s

We know, -1 â‰¤ sin (angle) â‰¤ 1

â‡’ 13sin (x + A) â‰¥ -13

â‡’ S_{min} = 13

Answer Choice (A) . Correct answer is 13.

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** Example 2:** If cos B + cos

a.) 4

b.) 3

c.) 6

d.) 1

**Solution:** It is given that

Cos B + cos^{2}B = 1

=> cos B = 1 – cos^{2}B

=> cos B = sin^{2}B

Squaring both sides in above equation, we get

â‡’ Cos^{2}B = Sin^{4}B

â‡’ 1 â€“ Sin^{2}B = Sin^{4}Â B

â‡’ 1 = Sin^{4}B + Sin^{2}Â B

Now, we need higher terms like sin^{12}B and sin^{8}B, thus we have to take cube of the above expression (since 4^{3} = 12)

â‡’ 1^{3}= (Sin^{4}B + Sin^{2}B)^{3}

â‡’ 1 = Sin^{12}B + Sin^{6}B + 3Sin^{8}Â B + 3Sin^{10}Â B

â‡’ Sin^{12}B + Sin^{6}B + 3Sin^{8}Â B + 3Sin^{10}Â B â€“ 1 = 0

On comparing with the above equation,Â a1 sin^{12}Â B + a2 sin^{10}Â B + a3 sin^{8}Â B + a4 sin^{6}Â B – 1 = 0.

a1 = 1, a2 = 3, a3 = 3, a4 = 1

âˆ´Â (a1+a2)/(a3+a4)Â = 1

Answer choice (D)

__Example 3__**:** It is given thatÂ tanA+sinA=p, tanAâˆ’sinA=q, can you find the value of p^{2}Â – q^{2}Â ?

a.) 4âˆšpq

b.) 2âˆšpq

c.) pâ€“ q

d.) 2pq

** Solution:**Â Adding the two equations given above,

Tan A= (p+q)/2

Subtracting second equation from first one, we get

Sin A= (pâˆ’q)/2

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

cosec^{2}Aâ€“cot^{2}A=1Â Â Â Â Â Â (Since cosec and cot are inverse of sin and tan respectively)

â‡’(2/(pâˆ’q))^{2}âˆ’(2/(p+q))^{2}=1

â‡’(4((p+q)^{2}âˆ’(pâˆ’q)^{2})) / (p^{2}âˆ’q^{2})^{2}=1

â‡’(p^{2}âˆ’q^{2})^{2}=4(4pq)

â‡’p^{2} – q^{2} = 4âˆšpq

Answer choice (A)

** Example 4:** John is standing with a flag at the top of a 50 m high school building. From a point on the ground, the angle of elevation of the top of the John is 45Â° and from the same point, the angle of elevation of the top of the school building is 30Â°. Find John’s height?

a.) 36.6m

b.) 15 m

c.) 20 m

d.) 45 m

__Solution:__

Let BC be the height of the school building and DC be the height of John.

In rt. âˆ† ABC

AB = BC cot 30Â°

AB = 50v3 mÂ â€¦â€¦â€¦â€¦â€¦ (1)

In rt. âˆ† ABD

AB = BD cot 45Â°

AB = (BC + CD) cot 45Â°

AB = (50 + CD) * 1

Equating (i) and (ii)

(50 + CD)Â = 50âˆš3

CD = 50âˆš3 â€“ 50

= 50(1.732 â€“ 1) = 50 x 0.732 = 36.6 m

Answer choice (A)

** Example 5: **There is a right-angled triangle. The base is ‘b’, height is ‘p’ and hypotenuse is ‘h’. It is known that p and b are positive integers. h

a.) 52

b.) 61

c.) 25

d.) 24

__Solution:__

We know that,

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

We know that

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

b) 6^{2} + 5^{2} = 61

c) 3^{2} + 4^{2} = 25

d) Canâ€™t be expressed as a sum of two perfect squares

Answer choice (D)

**Example 6: **What is the value of (1-CosA) / (1+CosA) given that tan A = Â¾?

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 7:** IfÂ (2Sinx) (1+cosx+Sinx) =t, (1â€“Cosx+Sinx)/(1+Sinx)Â can be written as?

a.) 1/t

b.) t

c.) âˆšt

d.) tSinx

__Solution:Â Â Â __

=Â (2Sinx)/(1+cosx+Sinx)âˆ— [(1+Sinxâˆ’Cosx)(1+Sinxâˆ’Cosx) ]

= 2 Sin x *Â (1+Sinxâ€“Cosx) / ((1+Sinx)^{2}â€“Cos^{2}x)

= 2 Sin x *Â (1+Sinxâ€“Cosx) / (1+Sin^{2}x+2Sinxâ€“Cos^{2}x)

= 2 Sin x .Â (1+Sinxâ€“Cosx) / (Cos^{2}x+Sin^{2}x+Sin^{2}x+2Sinxâ€“Cos^{2}x)

(Since, 1 = Cos^{2}x + SinÂ ^{2}x)

= 2 Sin x *Â (1+Sinxâ€“Cosx) / (2Sin^{2}x+2Sinx)

= 2 Sin x *Â (1+Sinxâ€“Cosx) / (2Sinx(1+Sinx))

=Â (1â€“Cosx+Sinx)(1+Sinx)Â = t

Answer Choice B.

** Example 8**:Â A father and his son are waiting at a bus stop in the evening. There is a lamp post behind them. The lamp post, father and his son stand on the same straight line. The father observes that the shadows of his head and his son’s head are incident at the same point on the ground. If the height of the lamp post, the father and the son are 6 meters, 1.8 and 0.9 meters respectively and the father is standing 2.1 meters away from the post, then how far (in meters) is son standing from his father?

**Solution: **If we look at the below figure of father, son and the lamp post

Î” ABE ~ Î” FCE

=> (6/1.8) = (2.1 + x + y) / (x + y)Â —–Â Â Â Â Â Â Â equation 1

Also, we see that Î” ABE ~ Î” GDE

=> 6/ 0.9 = (2.1 + x + y) / yÂ Â Â Â —– equation 2

From equation 1 and 2, we would get x = 0.45 m.

** Example 9:**Â The shadow of a tower standing on a level ground is found to be 40 m longer when the sun’s altitude is 30 degrees than when it is 60 degrees. Find the height of the tower.

** Solution:**Â Â In the figure below AB is the tower and BC is the length of the shadow when the sun’s altitude is 60 degrees and DB is the length of the shadow when the angle of elevation is 30 degrees.

Now, let AB be h m and BC be x m. According to the question, DB is 40 m longer than BC.

So, DB = (40 + x ) m

Now, we have two right triangles ABC and ABD.

In Î” ABC, tan 60 = AB/ BC

=> âˆš3 = h / x

In Î”ABD, tan 30 = AB/ BD

=> 1/ âˆš3 = h / (x + 40)

Solving the two equations, we get

=> x = 20

=> h = 20âˆš3

** Example 10:**Â From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30 degree and 45 degree respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

** Solution:** In the below figure, A and B represent points on the bank on the opposite sides of the river. P is a point on the bridge at a height of 3 m. Length of the side AB is the width of the river.

Now, AB = AD + DB

In right Î” APD, angle A = 30Â°

So, tan 30Â° = PD / AD

Or 1/ âˆš3 = 3 / AD or AD = 3âˆš3 m

Also, In right Î” PBD, angle B = 45Â°. So, BD = PD = 3 m

Now, AB = BD + AD = 3 + 3âˆš3 m

** Conclusion:**Â Â 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.

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