Chapter 15, Problem 1PP

Basic Trigonometry

Refer to Figure 14-22 to find the missing values in the following chart.

FIGURE 14-22 Finding the values of a right triangle.

To determine

The unknowns in the table.

Answer to Problem 1PP

$\angle x$	$\angle y$	Side A	Side B	Side C
$40°$	$50°$	5.1423	6.1284	8
$57°$	$33°$	72	46.7573	85.8501
$37.0976°$	$52.9024°$	38	50.2493	63
$52°$	$38°$	17.9190	14	22.7397
$48°$	$42°$	173.2555	156	233.1383

Explanation of Solution

ROW 1

$\begin{array}{l}\angle x+\angle y+90°=180°\\ 40°+\angle y+90°=180°\\ \angle y=180°-130°\\ \angle y=50°\end{array}$

$\begin{array}{l}\sin x=\frac{A}{C}\\ \Rightarrow A=C\sin x\\ \Rightarrow A=8\sin 40°\\ \Rightarrow A=5.1423\end{array}$

$\begin{array}{l}\cos x=\frac{B}{C}\\ \Rightarrow B=C\cos x\\ \Rightarrow B=8\cos 40°\\ \Rightarrow B=6.1284\end{array}$

ROW 2

$\begin{array}{l}\angle x+\angle y+90°=180°\\ \angle x+33°+90°=180°\\ \angle x=180°-123°\\ \angle x=57°\end{array}$

$\begin{array}{l}\sin x=\frac{A}{C}\\ \Rightarrow C=\frac{A}{\sin x}\\ \Rightarrow C=\frac{72}{\sin 57°}\\ \\ \Rightarrow C=85.8501\\ \end{array}$

$\begin{array}{l}\tan x=\frac{A}{B}\\ \Rightarrow B=\frac{A}{\tan x}\\ \Rightarrow B=\frac{72}{\tan 57°}\\ \\ \Rightarrow B=46.7573\\ \end{array}$

ROW 3

$\begin{array}{l}{A}^2+B^2=C^2\\ \Rightarrow B^2=C^2-A^2\\ \Rightarrow B=\sqrt{C^2-A^2}\\ \Rightarrow B=\sqrt{{63}^2-{38}^2}\\ \Rightarrow B=\sqrt{2525}\\ \Rightarrow B=50.2493\end{array}$

$\begin{array}{l}\sin x=\frac{A}{C}\\ \Rightarrow x={\sin }^{-1}\left(\frac{38}{63}\right)\\ \Rightarrow x={\sin }^{-1}\left(0.6031\right)\\ \Rightarrow x=37.0976°\end{array}$

$\begin{array}{l}\angle x+\angle y+90°=180°\\ 37.0976°+\angle y+90°=180°\\ \angle y=180°-90°-37.0976°\\ \angle y=52.9024°\end{array}$

ROW 4

$\begin{array}{l}\angle x+\angle y+90°=180°\\ 52°+\angle y+90°=180°\\ \angle y=180°-142°\\ \angle y=38°\end{array}$

$\begin{array}{l}\cos x=\frac{B}{C}\\ \Rightarrow C=\frac{B}{\cos x}\\ \Rightarrow C=\frac{14}{\cos 52°}\\ \\ \Rightarrow C=22.7397\end{array}$

$\begin{array}{l}{A}^2+B^2=C^2\\ \Rightarrow {\text{ A}}^2=C^2-B^2\\ \Rightarrow \text{ A}=\sqrt{C^2-B^2}\\ \Rightarrow \text{ A}=\sqrt{{22.7397}^2-{14}^2}\\ \Rightarrow \text{ A}=\sqrt{321.0939}\\ \Rightarrow \text{ A}=17.9190\end{array}$

ROW 5

$\begin{array}{l}\angle x+\angle y+90°=180°\\ \angle x+42°+90°=180°\\ \angle x=180°-132°\\ \angle x=48°\end{array}$

$\begin{array}{l}\cos x=\frac{B}{C}\\ \Rightarrow C=\frac{B}{\cos x}\\ \Rightarrow C=\frac{156}{\cos 48°}\\ \\ \Rightarrow C=233.1383\end{array}$

$\begin{array}{l}{A}^2+B^2=C^2\\ \Rightarrow {\text{ A}}^2=C^2-B^2\\ \Rightarrow \text{ A}=\sqrt{C^2-B^2}\\ \Rightarrow \text{ A}=\sqrt{{233.1383}^2-{156}^2}\\ \Rightarrow \text{ A}=\sqrt{30017.4864}\\ \Rightarrow \text{ A}=173.2555\end{array}$