Law of Cosines - Practice Questions & MCQ

Cosine Rule
For a triangle with angles A, B, and C, and opposite corresponding sides a, b, and c, respectively, the Law of Cosines is given as three equations.

$
\cos A=\frac{b^2+c^2-a^2}{2 b c} \quad \cos B=\frac{a^2+c^2-b^2}{2 a c} \quad \cos C=\frac{a^2+b^2-c^2}{2 a b}
$
Proof:

Drop a perpendicular from C to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that.

$
\cos \theta=\frac{x(\text { adjacent })}{b(\text { hypotenuse })} \text { and } \sin \theta=\frac{y(\text { opposite })}{b(\text { hypotenuse })}
$
In terms of $\theta, x=b \cos \theta$ and $y=b \sin \theta$. The $(x, y)$ point located at $C$ has coordinates $(b \cos \theta, b \sin \theta)$.
Using the side $(\mathrm{x}-\mathrm{c})$ as one leg of a right triangle and y as the second leg, we can find the length of hypotenuse a using the Pythagoras Theorem. Thus,

$
\begin{aligned}
a^2= & (x-c)^2+y^2 \\
= & (b \cos \theta-c)^2+(b \sin \theta)^2 \\
& \quad[\text { Substitute }(b \cos \theta) \text { for } x \text { and }(b \sin \theta) \text { for } y] \\
= & \left(b^2 \cos ^2 \theta-2 b c \cos \theta+c^2\right)+b^2 \sin ^2 \theta
\end{aligned}
$

[Expand the perfect square.]

$
=b^2 \cos ^2 \theta+b^2 \sin ^2 \theta+c^2-2 b c \cos \theta
$

[ Group terms noting that $\cos ^2 \theta+\sin ^2 \theta=1$ ]

$
=b^2\left(\cos ^2 \theta+\sin ^2 \theta\right)+c^2-2 b c \cos \theta
$

[ Factor out $b^2$ ]

$
a^2=b^2+c^2-2 b c \cos \theta
$
Now as $\theta$ equals angle A

$
\begin{aligned}
& a^2=b^2+c^2-2 b c \cdot \cos A \\
& \cos A=\frac{b^2+c^2-a^2}{2 b c}
\end{aligned}
$

Similarly, we can derive formulae for cosB and cosC