Chapter 8.6, Problem 53PPS

(a)

To determine

Draw an acute, scalene $\Delta ABC$ including an altitude of length h originating at vertex A .

Explanation of Solution

Calculation:

Draw scalene $\Delta ABC$ including an altitude of length h originating at vertex A .

(b)

To determine

Use trigonometry to represent h in terms of $m\angle B$ .

Answer to Problem 53PPS

$h=AB\sin \left(m\angle B\right)$

Explanation of Solution

Given :

From part (a) , h is the altitude.

Calculation:

In right $\Delta ADB$

$\begin{array}{l}\sin \left(m\angle B\right)=\frac{\text{length of side opposite to }\angle \text{B}}{\text{length of hypotenuse}}\\ \Rightarrow \sin \left(m\angle B\right)=\frac{h}{AB}\\ \Rightarrow h=AB\sin \left(m\angle B\right)\mathrm{....................}[\text{multiply each side by AB]}\end{array}$

(c)

To determine

Write an equation to find the area of $\Delta ABC$ using trigonometry.

Answer to Problem 53PPS

$Area=\frac{1}{2}BC\cdot AB\sin \left(m\angle B\right)$

Explanation of Solution

Given :

From part (a) and (b) , h is the altitude and $h=AB\sin \left(m\angle B\right)$

Calculation:

Area of $\Delta ABC$ is:

$\begin{array}{l}Area=\frac{1}{2}(base)(altitude)\\ \Rightarrow Area=\frac{1}{2}BC(h)\\ \Rightarrow Area=\frac{1}{2}BC\cdot AB\sin \left(m\angle B\right)\end{array}$

(d)

To determine

Find the area of $\Delta ABC$ .

Answer to Problem 53PPS

$Area\approx 57.2{\text{ units}}^2$

Explanation of Solution

Given :

$m\angle B$ is 47, $AB=11.1,BC=14.1$ , and $CA=10.4$

From part (c) , $Area=\frac{1}{2}BC\cdot AB\sin \left(m\angle B\right)$

Calculation:

Area of $\Delta ABC$ is:

$\begin{array}{l}Area=\frac{1}{2}BC\cdot AB\sin \left(m\angle B\right)\\ \Rightarrow Area=\frac{1}{2}(14.1)\cdot (11.1)\sin \left(47°\right)\\ \Rightarrow Area\approx 57.2{\text{ units}}^2\end{array}$

(e)

To determine

Write an equation to find the area of $\Delta ABC$ using trigonometry in terms of a different angle.

Answer to Problem 53PPS

$Area=\frac{1}{2}BC\cdot AC\sin \left(m\angle C\right)$

Explanation of Solution

Given :

From part (a) , h is the altitude.

Calculation:

In right $\Delta ADC$

$\begin{array}{l}\sin \left(m\angle C\right)=\frac{\text{length of side opposite to }\angle C}{\text{length of hypotenuse}}\\ \Rightarrow \sin \left(m\angle C\right)=\frac{h}{AC}\\ \Rightarrow h=AC\sin \left(m\angle C\right)\mathrm{....................}[\text{multiply each side by AC]}\end{array}$

So, Area of $\Delta ABC$ is :

$\begin{array}{l}Area=\frac{1}{2}(base)(altitude)\\ \Rightarrow Area=\frac{1}{2}BC(h)\\ \Rightarrow Area=\frac{1}{2}BC\cdot AC\sin \left(m\angle C\right)\end{array}$