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All courses Geometry · G-SRT.10 55 of 57
Prove the Laws of Sines and Cosines and use them to solve problems.

Use the Law of Sines to find a missing side from two angles and one opposite side

Problem
In the triangle shown, \(A~=~30°\), \(B~=~45°\), and \(a~=~10\). Use the Law of Sines to find \(b\), giving an exact value and a decimal approximation.
A scale-faithful triangle marking the two given angles, known opposite side, and requested side without a solved value. Open full size
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Hint

Start by matching each side with its opposite angle in the Law of Sines: \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Then plug in \(a=10\), \(A=30^\circ\), and \(B=45^\circ\).

Use the Law of Sines: \(\frac{a}{\sin A}=\frac{b}{\sin B}\). Here, side \(a\) goes with angle \(A=30^\circ\), and side \(b\) goes with angle \(B=45^\circ\).

Solution walkthrough

01

Match opposite pairs

\[a=10~\text{opposite}~A=30^{\circ};~b~\text{opposite}~B=45^{\circ}\]

The Law of Sines must pair each side with its opposite angle.

02

Write and solve the proportion

\[b/\sin(45^{\circ})=10/\sin(30^{\circ})~->~b=10\sin(45^{\circ})/\sin(30^{\circ})\]

Cross-multiplication isolates b without rounding.

03

Use exact special-angle values

\[b=10(\sqrt{2}/2)/(1/2)=10\sqrt{2}\]

Substitute sin(45 degrees)=sqrt(2)/2 and sin(30 degrees)=1/2, preserving the exact radical.

04

Approximate and conclude

\[b=10\sin(45^{\circ})/\sin(30^{\circ})=10~\text{square}~\text{root}~\text{of}~2,~\text{about}~14.14.\]

Since 45 degrees is larger than 30 degrees, its opposite side b being longer than a=10 checks the result.

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Another way

  1. From the sine ratio, b/10=sin45/sin30=sqrt(2), so b=10sqrt(2) directly.

!

Common mistake

Do not pair b with 30 degrees. Side b is opposite angle B=45 degrees, while side a=10 is opposite A=30 degrees.

The same labeled triangle with the missing side solved as b=10√2≈14.14.
The labeled triangle gives b=10√2≈14.14 by the Law of Sines.

Solution walkthrough video