Chapter 9.10, Problem 20PSD

To determine

To prove:The law of cosines $c^2=a^2+b^2+2ab\cos (\angle C)$

Explanation of Solution

Given information:The following information has been given

ABC is an acute triangle with sides as a, b and c

Formula used:basic trigonometric and Pythagorean identities are used

Proof:Let us assume that in the acute triangle ABC, an altitude is drawn from vertex B to side AC. Hence, we can say that the following are true

  $\cos C=\frac{CD}{a}\to CD=a\cos C$

Thus, we know that

  $DA=b-a\cos C$

  $\sin C=\frac{BD}{a}\to BD=a\sin C$

Now, applying Pythagoras theorem to triangle ADB, we get

  $B{D}^2+D{A}^2=c^2$

Now, we substitute and simplify the values of BD and DA, and we get

  $c^2=a^2+b^2+2ab\cos (\angle C)$

Hence, given identity is proved