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

The Cosine 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 Cosine Law:


c2=a2+b2−2abcos⁡C  or  cos⁡C=a2+b2−c22ab\Large{\boxed{c^2=a^2+b^2-2ab\cos C}}~~\text{or}~~\Large\boxed{\cos C=\dfrac{a^2+b^2-c^2}{2ab}}c2=a2+b2−2abcosC​  or  cosC=2aba2+b2−c2​​ 

Wize Tip
We can use the Cosine Law if
 we know the values of all 3 sides ➡ use cos⁡C=a2+b2−c22ab\cos C=\dfrac{a^2+b^2-c^2}{2ab}cosC=2aba2+b2−c2​ to find a missing angle
we know 2 sides and the angle in between them (the contained angle) ➡ use c2=a2+b2−2abcos⁡Cc^2=a^2+b^2-2ab\cos Cc2=a2+b2−2abcosC

*Note:
The cosine 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: Cosine Law
Consider the triangle below.
Select ALL statements that are true.

A) If we know ∠Z\angle Z∠Z, xxx, and yyy, then we can solve the triangle using Cosine Law

B) If we know ∠Z\angle Z∠Z, xxx and zzz, then we can solve the triangle using Cosine Law

C) If we know ∠X\angle X∠X, ∠Y\angle Y∠Y, and zzz, then we can solve the triangle using Cosine Law

D) If we know ∠X\angle X∠X, ∠Y\angle Y∠Y, and xxx, then we can solve the triangle using Cosine Law

E) If we know xxx, yyy, and zzz, then we can solve the triangle using Cosine Law

F) If we know ∠X\angle X∠X, yyy, and zzz, then we can solve the triangle using Cosine Law

I don't know

Practice: Simplifying Expressions
Solve for xxx in the following equations.

a) x2=a2+b2−2abcos⁡Cx^2=a^2+b^2-2ab\cos Cx2=a2+b2−2abcosC, where a=3a=3a=3, b=4b=4b=4, and ∠C=60°\angle C=60\degree∠C=60°

b) cos⁡x=a2+b2−c22ab\cos x=\dfrac{a^2+b^2-c^2}{2ab}cosx=2aba2+b2−c2​, where a=3a=3a=3, b=4b=4b=4, and c=2c=2c=2

a) Enter the exact value of

x

, do not round your answer.

b) Round your answer to 2 decimal places

I don't know

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Example: Cosine Law
Solve for zzz in the following triangle.

Since we know two of the side lengths and the contained angle, we can use the Cosine Law.
z2=x2+y2−2xycos⁡∠Zz2=72+62−2(7)(6)cos⁡50°z2=49+36−84cos⁡50°z2≈85−84(0.6428)z2≈85−54.00z2≈31z≈31z≈5.57\begin{array}{rcl}
z^2&=&x^2+y^2-2xy\cos\angle Z\\
z^2&=&7^2+6^2-2(7)(6)\cos50\degree\\
z^2&=&49+36-84\cos50\degree\\
z^2&\approx&85-84(0.6428)\\
z^2&\approx&85-54.00\\
z^2&\approx&31\\
z&\approx&\sqrt{31}\\
z&\approx&5.57
\end{array}z2z2z2z2z2z2zz​===≈≈≈≈≈​x2+y2−2xycos∠Z72+62−2(7)(6)cos50°49+36−84cos50°85−84(0.6428)85−54.003131​5.57​
So, z≈5.57\boxed{z\approx5.57}z≈5.57​.

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Example: Cosine Law
Solve for ∠Z\angle Z∠Z in the following triangle.

Since we have all 3 sides, we can use the cosine law to find a missing angle.
cos⁡∠Z=x2+y2−z22xycos⁡∠Z=92+82−522(9)(8)cos⁡∠Z=81+64−25144cos⁡∠Z=120144∠Z=cos⁡−1(120144)∠Z≈33.56°\begin{array}{rcl}
\cos\angle Z&=&\dfrac{x^2+y^2-z^2}{2xy}\\\\
\cos\angle Z&=&\dfrac{9^2+8^2-5^2}{2(9)(8)}\\\\
\cos\angle Z&=&\dfrac{81+64-25}{144}\\\\
\cos\angle Z&=&\dfrac{120}{144}\\\\
\angle Z&=&\cos^{-1}\left(\dfrac{120}{144}\right)\\\\
\angle Z&\approx&33.56\degree
\end{array}cos∠Zcos∠Zcos∠Zcos∠Z∠Z∠Z​=====≈​2xyx2+y2−z2​2(9)(8)92+82−52​14481+64−25​144120​cos−1(144120​)33.56°​

Practice: Cosine Law
Solve the following triangle. (find all missing sides lengths and angles)

\angle X\approx

(Round your answer to 2 decimal places, do not include units)

\angle Y\approx

(Round your answer to 2 decimal places, do not include units)

\angle Z\approx

(Round your answer to 2 decimal places, do not include units)

I don't know

Practice: Cosine Law
△PQR\triangle PQR△PQR has angle ∠Q=60°\angle Q=60\degree∠Q=60°, QP=10QP=10QP=10, and QR=7QR=7QR=7. Solve the triangle (find all missing angles and side lengths)

\Large{\color{black}{PR\approx}}

(Round your anwer to 2 decimal places, do not include units)

\Large{\color{black}{\angle P\approx}}

(Round your anwer to 2 decimal places, do not include units)

\Large{\color{black}{\angle R\approx}}

(Round your anwer to 2 decimal places, do not include units)

I don't know