How To Solve Mathematical Problems Involving Complementary Angles

This is a subtopic of trigonometric ratio. This topic is sometimes confusing to some students because of the relationship between the sin ө and cos ө involved, do not panic.

I will make it very easy for you. I will break it down to your understanding but first of all, let’s establish the formulas we are going to use

Drawing

In ΔABC, angle B = 90^o

Given that C = θ then,

A = (90 –  θ)^o

Sin θ = c/b

Cos (9^o – θ) = c/b

Therefore, sin θ = cos (90 – θ)

Similarly,

Cos θ = 9/b

Sin (90 – θ) = 9/b

Therefore, cos θ = sin (90 – θ) – (2)

Angles (90^o– θ) and θ one complementary since

90^o – θ + θ = 90^o

Note These formula

• Sin θ = cos (90 – θ)
• Cos θ = sin (90 – θ)

Write it down for reference purposes. And also know it so having establish the two formula we are going to use for solving complementary angle, let’s solve some problems

Example 1:

If sin θ = cos 26^o, find the value of θ

Solution

From the formula we have established

Sin θ = cos (90 – θ)

Relating with the problem given

Cos (90 – θ) = Cos 26^o   

Carear the cos

90 – θ = 26

90 -26= θ

θ = 64^o

Example 2:

Solve the following

• Sin 3θ = cos 2θ
• Cos(θ+10)^o=sin θ

Solution

From the formula established

sin θ = cos (90 – θ)

then sin 3θ = cos (90 – θ)

therefore,

cos (90 – 3θ) = cos 2θ

90 – 3θ = 2θ

• Collect like terms

50/5 = 90/5

θ = 90/5 = 18^o

• for cos (θ + 10)^o = sin θ

using sin θ = cos (90- θ)

therefore

cos (θ+10) = cos(90- θ)

θ+10 = 90 – θ

collect like terms

θ+ θ = 90 – 10

2θ/2 = 80^o/2

Therefore

θ = 40^o

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Example 3:

Solve the following equations

• sin 3θ = cos (θ – 60^o)
• sin (3θ+30) = cos 4θ
• cos 2θ = sin (θ + 10)^o

solution

• using the formula

sin if sin θ = cos (90^o – θ)

sin 3θ = cos (90^o – 3θ)

therefore,

cos(90-3θ) = cos θ – 60^o

90-30=0-60^o

Collect like terms

90 + 60^o = θ+3θ

40/4 = 150^o/4

θ = 150/4

Therefore, θ = 37.5^o

• sin (3θ+30) = cos 40

using

if cos θ = sin (90 – θ)

then cos 40 = sin (90-4θ)

Therefore

sin (3θ+30) = sin(90 – 40)

3θ + 30^o = 90 – 4θ

Collect like terms

3θ+4θ = 90^o – 30^o

70/7 = 60^o/7

θ = 60/7

θ = 8.57^o

• cos 2θ = sin (θ + 10)^o

cos θ = sin 90 – θ

cos 20 = sin (90 – 20)

therefore, sin 90 – 20 = sin (θ + 10)

90^o – 20 = θ + 10

θ + 20 = 90 – 10

3θ/3θ = 80/3

θ = 80/3 

θ = 26.7

Example 4:

Solve the equation

• cos 2θ = sin 3θ
• cos x = sin (x+22^o)

solution

using the formula

cos θ = sin (90- θ)

cos 2θ = sin (90 – 2θ)

therefore,

sin (90-2θ) = sin 3θ

90 – 2θ  3θ

3θ + 2θ  = 90^o

θ = 90/5

θ = 18^o

• cos x = sin (x + 22)

again cos x = sin (90 – x)

90 –x = x + 22

x + x = 90 – 22

2x/2 = 68/2

x = 68/2

x = 34^o

In conclusion, problems involving complementary angles are not difficult to solve. The rules remain the same i.e your ability to relate the formulas given.

If you are able to play along with the formulas and use them effectively when you encounter problems on this topic, I strongly believe you will be able to solve any problem you are faced with.