Chapter 14.4, Problem 22PPSE

Solution

a.

To identify: Whether the triangle has a greater area when $m\angle C=1^{\circ }\text{ or when }m\angle C={50}^{\circ }$ in $\Delta ABC,a=10\text{ and }b=15$ .

The area is greater when $m\angle C={50}^{\circ }$ .

Given information:

In $\Delta ABC$ , the given values are $a=10\text{ and }b=15$ .

Explanation:

The area of a triangle is given by: $A=\frac{1}{2}ab\sin C$

Substitute the values of $a=10,b=15\text{ and }C=1^{\circ }$ ,

  $\begin{array}{c}A=\frac{1}{2}\left(10\right)\left(15\right)\sin 1^{\circ }\\ =75\sin 1^{\circ }\\ =1.31\end{array}$

Now substitute the values of $a=10,b=15\text{ and }C={50}^{\circ }$ ,

  $\begin{array}{c}A=\frac{1}{2}\left(10\right)\left(15\right)\sin {50}^{\circ }\\ =75\sin {50}^{\circ }\\ =57.5\end{array}$

Therefore, the area is greater when $m\angle C={50}^{\circ }$ .

b.

To identify: Whether the triangle has a greater area when $m\angle C={50}^{\circ }\text{ or when }m\angle C={179}^{\circ }$ in $\Delta ABC,a=10\text{ and }b=15$ .

The area is greater when $m\angle C={50}^{\circ }$ .

Given information:

The value of $m\angle A$ from part (a) is ${\cos }^{-1}\left(\frac{1200}{d}\right)$ .

Explanation:

The area of a triangle is given by: $A=\frac{1}{2}ab\sin C$

Substitute the values of $a=10,b=15\text{ and }C={50}^{\circ }$ ,

  $\begin{array}{c}A=\frac{1}{2}\left(10\right)\left(15\right)\sin {50}^{\circ }\\ =75\sin {50}^{\circ }\\ =57.5\end{array}$

Now substitute the values of $a=10,b=15\text{ and }C={179}^{\circ }$ ,

  $\begin{array}{c}A=\frac{1}{2}\left(10\right)\left(15\right)\sin {179}^{\circ }\\ =75\sin {179}^{\circ }\\ =1.31\end{array}$

Therefore, the area is greater when $m\angle C={50}^{\circ }$ .

c.

To identify: For what measure of $\angle C$ , $\Delta ABC$ has a greater area.

For the measure of $\angle C={50}^{\circ }$ the triangle will have the greatest area.

Given information:

A triangle $\Delta ABC$ is given.

Explanation:

The angle C should be ${50}^{\circ }$ near to ${30}^{\circ }$ . The reason being at this degree the area will be greater compared to any other value.

And the values of $\sin 1\text{ and }\sin 179$ are nearly equal to zero.

Therefore, in triangle $ABC$ its greatest area will be when measure of $\angle C={50}^{\circ }$ .