Cosine Theorem

In an acute triangle $ABC$, it is known that $BC=a, AC=b, \angle ACB=\alpha$. Find the height $CD$ and the angle $\angle ABC$.
Apply the Law of Sines and the Law of Cosines.

## Solution

By the Law of Cosines

$$
A B=\sqrt{B C^{2}+A C^{2}-2 B C \cdot A C \cos \angle A C B}=\sqrt{a^{2}+b^{2}-2 a b \cos \alpha}
$$

We can write the area of triangle $A B C$ in two ways:

$$
\frac{1}{2} A B \cdot C D=\frac{1}{2} B C \cdot A C \sin \angle A C B
$$

From this, we find that

$$
C D=\frac{B C \cdot A C \sin \angle A C B}{A B}=\frac{a b \sin \alpha}{\sqrt{a^{2}+b^{2}-2 a b \cos \alpha}}
$$

By the Law of Sines

$$
\frac{A C}{\sin \angle A B C}=\frac{A B}{\sin \angle A C B}
$$

from which

$$
\sin \angle A B C=\frac{A C \sin \angle A C B}{A B}=\frac{b \sin \alpha}{\sqrt{a^{2}+b^{2}-2 a b \cos \alpha}}
$$

## Answer

$C D=\frac{a b \sin \alpha}{\sqrt{a^{2}+b^{2}-2 a b \cos \alpha}}$

$$
\angle A B C=\arcsin \frac{b \sin \alpha}{\sqrt{a^{2}+b^{2}-2 a b \cos \alpha}}
$$

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