Chapter 5.6, Problem 9E

Solution

To find: The measure of angles $A$ , $B$ , and $C$ in the triangle.

The triangle is not possible.

Given information: In the triangle, the measure of sides are $a=1$ , $b=5$ , and $c=4$ .

Formula used:

The law of cosines to find the angles and sides are given by,

  $a^2=b^2+c^2-2bc\cos A$  ...... (1)

  $b^2=c^2+a^2-2ca\cos B$  ...... (2)

  $c^2=a^2+b^2-2ab\cos C$  ...... (3)

Calculation:

Substitute $1$ for $a$ , $5$ for $b$ , $4$ for $c$ in equation (1).

  $\begin{array}{c}{\left(1\right)}^2={\left(5\right)}^2+{\left(4\right)}^2-2\left(5\right)\left(4\right)\cos A^{\circ }\\ 1=25+16-40\cos A\\ \cos A=\frac{40}{40}\\ A=0^{\circ }\end{array}$

Substitute $1$ for $a$ , $5$ for $b$ , $4$ for $c$ in equation (2).

  $\begin{array}{c}{\left(5\right)}^2={\left(4\right)}^2+{\left(1\right)}^2-2\left(4\right)\left(1\right)\cos B\\ 25=16+1-8\cos B\\ \cos B=\frac{8}{-8}\\ B={180}^{\circ }\end{array}$

Substitute $1$ for $a$ , $5$ for $b$ , $4$ for $c$ in equation (3).

  $\begin{array}{c}{\left(4\right)}^2={\left(1\right)}^2+{\left(5\right)}^2-2\left(1\right)\left(5\right)\cos C\\ 16=1+25-10\cos C\\ \cos C=\frac{-10}{-10}=1\\ C=0^{\circ }\end{array}$

From the above, it is clear that the angle of a triangle can’t be zero.

Therefore, the triangle is not possible.