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Maths II

Sine and cosine rule and area of a triangle

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The trigonometric ratios sin, cos and tan are used to solve for angles and sides in right angled triangles. We are now going to extend trigonometry beyond right angled triangles and use it to solve problems involving any triangle.

The sine rule

{a \over sinA}={b \over sinB}={c \over sinC}

We can use the sine rule when we're given -

  • two sides and an angle opposite to one of the two sides
  • one side and any two angles

In the first case, the sine rule could be used to find another angle and, in the second case, it could be used to find another side.

Example

Find the size of angle R.

A triangle. Angle P is 75 degrees. the edge between angles P and Q is 4 centimetres. The edge between angles Q and R is 9 centimetres.

Substituting in: {4 \over sinR}={9 \over sin75^{\circ}}

Cross multiplying: 9 sin R = 4 sin {75^o}

Dividing by 9: {sin R}={{sin75^{\circ}} \over 9}={0.4293...}

Taking inverse sine: R = {25.4^o}

Now try this one.

Question

Find the length of YZ.

A scalene triangle. Angle Y is 95 degrees. Angle Z is 40 degrees. The edge between Y and X is 4 centimetres.

toggle answer

Answer

We know two angles and so can calculate the third angle in triangle. ie x= 45°

Substituting in: {yz \over {sin45^{\circ}}}={4 \over {sin40^{\circ}}}

Cross multiplying: sin 40° yz = 4 sin 45°

Dividing by sin 40: zy = {4sin45^{\circ} \over {sin40^{\circ}}} = 4.40cm

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