COMPLEMENTARY ANGLES

Two angles are complementary if their sum is 90°. Consider the right – angled triangle below,

Since one angle is 90°, the other two will be complementary. i.e. if one is θ the other will be 90 – θ.

** if one is 40° the other will be 50°

** if one is 85° the other will be 05°

** if one is 27° the other will be ……..

From the triangle above, sin θ = a/c . Also from the other angle (90 – θ), cos (90 – θ) = a/c. Since the two are equal, we say sin θ = cos (90 – θ)

Also, cos θ =  and sin (90 – θ) = , i.e. cos θ = sin (90 – θ)

 

      Example

1.   Given that cos 2θ = sin θ, find the value of θ.

 

Solution 

cos 2θ = sin θ

but Sin θ = cos (90 – θ). Substitute cos (90 – θ) for Sin θ.

∴ cos 2θ = cos (90 – θ)

Cos will cancel cos

 

We now have 2θ = 90 – θ

Collect like terms

2θ + θ = 90

3θ = 90

Divide both sides by 3,

θ = 30°.

                         OR

cos 2θ = sin θ

sin (90 - 2θ) = sin θ

90 - 2θ = θ

90 = 2θ + θ

90 = 3θ

30 = θ

θ = 30

2.  Find the value of K if cos K = sin 7K.

Solution

Cos K = sin 7K

Sin (90 – K) = sin 7K

90 – K = 7K

90 = 7K + K

90 = 8K

Divide both sides by 8

90/8 = k

11.25 = K

∴ K = 11.25°.

 

3.   Find the value of P if cot P = tan 14P

 

Solution 

Cot P = tan 14P

Substitute tan (90 – P) for Cot P

Tan (90 – P) = tan 14P

90 – P = 14P

90 = 14P + P

90 = 15P

Divide through by 15

90/15 = P

6 = P

∴ P = 6°.

 

4. Solve for y in the equation below sin3y = cos 2y

Solution

sin3y = cos2y

sin3y = sin (90 - 2y)

sin cancel sin

3y = 90 - 2y

3y + 2y = 90 

5y = 90

y = 18°

 

 

Exercise E 

Find the value of the angles below given that:

1. Cos Q = sin 5Q

2. Tan 7P = cot P

3. Cos 3X = sin X

4. Sin 7K = cos K

5. Tan B = cot 5B

6. Sin 10Q = cos 5Q

7. cos 7t = sin 3t

8. sin 2p = cos 5p

9. cos 6.5x = sin 2.3x

10. cot 8a = tan 5a

 

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