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Wize High School Grade 11 Math Textbook > Trigonometry with Non-Right Triangles

Sine Law

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The Sine Law

Watch Out!
If you are given a non-right angle triangle, we cannot use SOH CAH TOA to solve for missing side lengths and angles

When we are given a non-right angle triangle, the side lengths and interior angles are related using the Sine Law:


sin⁡Aa=sin⁡Bb=sin⁡Cc  or  asin⁡A=bsin⁡B=csin⁡C\Large\boxed{\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}}~~\text{or}~~\Large\boxed{\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}}

Wize Tip
  • We want to use the Sine Law when we know the values of an angle and its opposite side
  • Whatever you are trying to solve for should be in the numerator:
  • If you are solving for a missing angle, use sin⁡Aa=sin⁡Bb=sin⁡Cc\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}
  • If you are solving for a missing side, use asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}
*You can technically use either version of the Sine Law to solve the problem, but picking the correct version will make the calculations simpler.

*Note:
The sine law actually works for right-angle triangles as well, but if you have a right-angle triangle, it's easier to use SOH CAH TOA.

Practice: Sine Law

Consider the triangle below:
Select ALL of the statements that are true.

Practice: Solving for a Missing Side

Solve for xx in the following equations.

a) 6sin⁡30°=xsin⁡20°\dfrac{6}{\sin 30\degree}=\dfrac{x}{\sin 20\degree}

b) sin⁡30°5=sin⁡60°x\dfrac{\sin 30\degree}{5}=\dfrac{\sin 60\degree}{x}

Practice: Solving for a Missing Angle

Solve for θ\theta in the following equations.

a) sin⁡30°5=sin⁡θ4\dfrac{\sin 30\degree}{5}=\dfrac{\sin \theta }{4}

b) 6sin⁡30°=7sin⁡θ\dfrac{6}{\sin 30\degree}=\dfrac{7}{\sin \theta}
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Example: Sine Law - Solving for a Missing Angle

Solve for xx in the following triangle.
Since we know ∠B\angle B and bb, we can use the Sine Law.

Finding x\bco x

Since xx is a missing angle, let's use
sin⁡Aa=sin⁡Bb=sin⁡Ccsin⁡Az=sin⁡40°5=sin⁡x7\begin{array}{ccccc} \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\\\\ \dfrac{\sin A}{z}=\dfrac{\sin 40\degree}{5}=\dfrac{\sin x}{7} \end{array}

We want to find xx, so let's only focus on the last 2 fractions:
sin⁡40°5=sin⁡x7sin⁡40°5×7=sin⁡x7×7sin⁡40°5×7=sin⁡x0.8999≈sin⁡xsin⁡−1(0.8999)≈x64.14≈xx≈64.14\begin{array}{ccc} \dfrac{\sin 40\degree}{5}&=&\dfrac{\sin x}{7}\\\\ \dfrac{\sin 40\degree}{5}\colorTwo{\times 7}&=&\dfrac{\sin x}{7}\colorTwo{\times 7}\\\\ \dfrac{\sin 40\degree}{5}\times 7&=&\sin x\\\\ 0.8999&\approx&\sin x\\\\ \sin^{-1}(0.8999)&\approx&x\\\\ 64.14&\approx&x\\\\ x&\approx&64.14 \end{array}
So, x≈64.14°\boxed{x\approx 64.14\degree}.


Alternative method

We could have also used asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C} instead, but the calculations would be a bit messier.
asin⁡A=bsin⁡B=csin⁡Czsin⁡A=5sin⁡40°=7sin⁡x\begin{array}{ccccc} \dfrac{a}{\sin A}&=&\dfrac{b}{\sin B}&=&\dfrac{c}{\sin C}\\\\ \dfrac{z}{\sin A}&=&\dfrac{5}{\sin 40\degree}&=&\dfrac{7}{\sin x}\\\\ \end{array}
Let's focus on the last 2 fractions:
5sin⁡40°=7sin⁡x5sin⁡40°×sin⁡x=7sin⁡x×sin⁡x5sin⁡40°×sin⁡x=75sin⁡40°×sin⁡x5sin⁡40°=75sin⁡40°sin⁡x=7×sin⁡40°5sin⁡x≈7×0.64285sin⁡x≈0.8999x≈sin⁡−1(0.8999)x≈64.14°\begin{array}{ccc} \dfrac{5}{\sin 40\degree}&=&\dfrac{7}{\sin x}\\\\ \dfrac{5}{\sin 40\degree}\colorTwo{\times \sin x}&=&\dfrac{7}{\sin x}\colorTwo{\times \sin x}\\\\ \dfrac{5}{\sin 40\degree}\times\sin x&=&7\\\\ \dfrac{\dfrac{5}{\sin 40\degree}\times\sin x}{\scriptsize\colorTwo{\dfrac{5}{\sin 40\degree}}}&=&\dfrac{7}{\scriptsize\colorTwo{\dfrac{5}{\sin 40\degree}}}\\\\ \sin x&=&7\times\dfrac{\sin 40\degree}{5}\\\\ \sin x&\approx&7\times\dfrac{0.6428}{5}\\\\ \sin x&\approx&0.8999\\\\ x&\approx&\sin^{-1}(0.8999)\\\\ x&\approx&64.14\degree \end{array}
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Example: Sine Law - Solving for a Missing Side

If x=64.14°x=64.14\degree, solve for zz in the following triangle.
Since we know ∠B\angle B and bb, we can use the Sine Law.

Finding z\bco z

Since zz is a missing side, let's use
asin⁡A=bsin⁡B=csin⁡Czsin⁡(180°−40°−64.14°)=5sin⁡40°=7sin⁡64.14°\begin{array}{ccccc} \dfrac{a}{\sin A}&=&\dfrac{b}{\sin B}&=&\dfrac{c}{\sin C}\\\\ \dfrac{z}{\sin(180\degree-40\degree-64.14\degree)}&=&\dfrac{5}{\sin 40\degree}&=&\dfrac{7}{\sin 64.14\degree} \end{array}

We want to find zz, so let's only focus on the first 2 fractions:
zsin⁡(75.86°)=5sin⁡40°zsin⁡(75.86°)×sin⁡(75.85°)=5sin⁡40°×sin⁡(75.86°)z=5sin⁡40°×sin⁡(75.86°)z≈50.6428×0.9697z≈7.54\begin{array}{ccc} \dfrac{z}{\sin(75.86\degree)}&=&\dfrac{5}{\sin 40\degree}\\\\ \dfrac{z}{\sin(75.86\degree)}\colorTwo{\times\sin(75.85\degree)}&=&\dfrac{5}{\sin 40\degree}\colorTwo{\times\sin(75.86\degree)}\\\\ z&=&\dfrac{5}{\sin40\degree}\times\sin(75.86\degree)\\\\ z&\approx&\dfrac{5}{0.6428}\times0.9697\\\\ z&\approx&7.54 \end{array}
So, z≈7.54m\boxed{z\approx 7.54 m}

Practice: Sine Law

Solve the following triangle for all of the missing side lengths and angles.


Practice: Sine Law

Triangle ABC is an acute triangle with ∠B=30°\angle B=30\degree, ∠A=70°\angle A=70\degree, and side AC=5mAC=5m. Find the length of side ABAB.