Problems Based on Sine and Cosine Rules

Trigonometry is the term that first originated from Greek words; “Trigonon” and “Metron.” “Trigonon” transcribes the triangle “metron” transcribes to measure. So, trigonometry is about triangles, and in higher classes, it is also used to calculate the amplitude and sound waves. The subject will be familiarized to you in Class 9th, and then you will learn its complex functions in class 12th. You will learn these concepts of sine and cosine if you opt for science in your class 11th. Keep on reading to know more about Problems Based on Sine and Cosine Rules.

Trigonometric Ratios

After understanding the history of trigonometry, we will recollect the six trigonometric ratios that are used in mathematics; let’s take a look at sine and cosine ratios and revise:

1. Ø is usually considered as the angle that is used to solve any type of trigonometric equation in Problems Based on Sine and Cosine Rules.

1. A right-angled triangle’s three sides are called the hypotenuse, opposite and adjacent.
2. The hypotenuse is the side opposite to the right angle in the triangle; the opposite is the side reversed of ø angle, and lastly is the adjacent side.

The basic 6 trigonometric ratios are known as:

• Sin ø
• Cosec ø
• Sec ø
• Cos ø
• Cot ø
• Tan ø

 We can take a look about this in the list below:

• Sin ø = opposite/hypotenuse
• Cos ø = adjacent/hypotenuse
• Tan ø = opposite/adjacent

The law of Sine (Sine Rule)

The sine rule is the ratio of a triangle, and their sine angles are equal to each other. It is also called sine law, sine rule and sine formula. The law of sine helps to find the unknown angle of the oblique angle. The oblique triangle is a triangle that is not a right-angle triangle. The law will work if there are at least two angles and its side measurements at a time.

The formula of the law of sine used in Problems Based on Sine and Cosine Rules:

a / Sin A= b/ Sin B= c / Sin C

a: b: c = Sin A: Sin B: Sin C

a / b = Sin A / Sin B

b / c = Sin B / Sin C

The law of Cosine (Cosine Rule)

The law of Cosine says any square length of a side of the triangle is equal to the sum of the squares of the length minus twice the product of the other two sides multiplied by the cosine of the angle between them. It is also known as the law of cosines or Cosine Formula.

Formula of law of cosine used in Problems Based on Sine and Cosine Rules:

• a = \sqrt{b^2 + c^2 – 2~b~c~ cos x}
• b = \sqrt{a^2 + c^2 – 2~a~c~ cos y}
• c = \sqrt{a^2 + b^2 – 2~a~b~cos z}

Examples of sine and cosine rule

Example 1

It is given to us that sine (A) = ⅔. You have to calculate angle ∠ B.

Sine (A)/a = Sine (B)/b

Then substituting,

(2/3)/2 = sine (B)/3

3(2/3) = 2 sine B

2 = 2 sine B

Dividing both sides by 2

1 = sine B

Finding the sine inverse of 1 with a scientific calculator.

Sine-1 1 = B

Therefore, ∠B = 90˚

Example 2

​​A triangle, BAC is considered, where the measurement between B and A is 8mm and the measurement between A and C is 12mm. The value of angle CBA is 50°, whereas the value of angle BAC is 100° and that of angle BCA is 30°. Find out the length of length BC of the triangle ABC

a/sine (A) = b/sine (B)

We substitute.

a/sine 100 ˚ = 12/sine 50 ˚

 Now Cross multiply.

12 sine 100 ˚= a sine 50 ˚

Dividing both sides by sine 50 ˚

a = (12 sine 100 ˚)/sine 50 ˚

By making use of a calculator, the answer is;

a = 15.427

Example 3

In the triangle BAC, side BA is 16cm. The angles are as follows:

• Angle BAC = 110°
• Angle ACB = 30°
• Angle CBA= 40

Calculate the missing lengths of the given triangle

a/sine (A) = b/sine (B) = c/ sine (C)

By substituting, we get,

a/sine 110 ˚ = 16/sine 30 ˚

By Cross multiply

a = (16 sine 110 ˚)/sine 30 ˚

a = 30.1

Solve for b.

b/sine 40 ˚ = 16/sine 30 ˚

b = (16 sine 40 ˚)/sine 30 ˚

= 20.6

Example 4

​​PQR is a triangle where side PQ is 4m, side QR is 7m and side RP is 9m. The angle PQR is 76°. 

Calculate the angles of the triangle

Apply the sine rule in the form;

sine(Q)/q = Sine (P)/p = Sine R/r

(Sine 76 ˚)/9 = sine (P)/7

Solve for angle P

By Cross multiply.

7 Sine 76 ˚ = 9 sine P

Dividing both sides by 9

Sine P = 7/9 sine 76 ˚

Sine P = 0.7547

Find the sine inverse of 0.7547.

Sine -1 0.7547 = P

P = 48.99 ˚

Solve for angle R

Sine R/4 = Sine 76 ˚/9

By Cross multiply.

9 Sine R = 4 sine 76 ˚

Dividing both sides by 9

Sine R = 4/9 sine 76 ˚

Sine R = 0.43124.

Sine -1 0.43124 = R

R = 25.54 ˚

Examples of cosine rule

Example 1

In triangle BAC, the following data is given

• Side BA= 3cm 
• Side CB= 4cm
• Angle CBA= 50°

Calculate the length of AC in the triangle

cosine rule in the form of;

⇒ (b) 2 = [a2 + c2 – 2ac] cos (B)

By substituting, we have,

b2 = 42 + 32 – 2 x 3 x 4 cos (50)

b2 = 16 + 9 – 24cos50

= 25 – 24cos 50

b2 = 9.575

Determine the square root of both sides to get,

b = √9.575 = 3.094.

Therefore, the length of AC = 3.094 cm

Example 2

Let’s consider a triangle, BAC where side BA is ‘c’, side AC is ‘b’ and side CB is ‘a’. 

Calculate all value of the three angles of the triangle

Since we have the length of the triangle we measure of A,B,C

⇒ Cos (A) = [b2 + c2 – a2]/2bc

⇒ Cos (B) = [a2 + c2– b2]/2ac

⇒ Cos (C) = [a2 + b2– c2]/2ab

Solve for angle A:

Cos A = (72 + 52 – 102)/2 x 7 x 5

Cos A = (49 + 25 – 100)/70

Cos A = -26/70

Cos A = – 0.3714.

Now, determine the cos inverse of – 0.3714.

A = Cos -1 – 0.3714.

A = 111.8°

Solve for angle B:

By substituting,

cos B = (102 + 52– 72)/2 x 10 x 7

Simplify.

Cos B = (100 + 25 – 49)/140

Cos B = 76/140

Determine the cos inverse of 76/140

B = 57.12°

Solve for angle C:

By substituting,

cos C = (102 + 72– 52)/2 x 10 x 7

Cos C = (100 + 49 – 25)/140

Cos C = 124/140

Determine the cos inverse of 124/140.

C = 27.7°

Conclusion

Trigonometry helps us to find the relationship between the sides and angles of a triangle. The term is derived from the Latin word. It is used widely in different fields like astronomy, geography, navigation, engineering, etc. The values of sine and cosine, etc., are mostly used in all the above-discussed fields.