Chapter 8.3, Problem 59AYU

To determine

What do you do first if you are asked to solve a triangle and are given two sides and the included angle?

Answer to Problem 59AYU

  $\boxed{a=\sqrt{\left(b^2+c^2\right)-2bc\times \cos A}}$ , $\boxed{B\approx {\cos }^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)}$ and $\boxed{C\approx {\cos }^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)}$ proved.

Explanation of Solution

Given information:

What do you do first if you are asked to solve a triangle and are given two sides and the included angle?

Calculation:

Let us consider the triangle having sides a,b,c and corresponding opposite angle A,B,C. Suppose two sides b, c and included angle A is given i.e., it is anSSA problem.

Apply the Law of Cosines, we get

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

  $a^2=\left(b^2+c^2\right)-2bc\times \cos A$

  $a=\sqrt{\left(b^2+c^2\right)-2bc\times \cos A}\mathrm{......}\left(1\right)$

At this stage all the three sides and one angle are known.

Apply the Law of Cosines again, we get

  $\cos B=\frac{a^2+c^2-b^2}{2ac}$

  $B\approx {\cos }^{-1}\left(\frac{a^2+c^2-b^2}{2ac}\right)\mathrm{.....}\left(2\right)$

At this stage all the three sides and two angles are known.

Apply the Law of Cosines again, we get

  $\cos C=\frac{a^2+b^2-c^2}{2ab}$

  $C\approx {\cos }^{-1}\left(\frac{a^2+b^2-c^2}{2ab}\right)\mathrm{.....}\left(3\right)$

At this stage all the three sides and all three angles are known.