Chapter 4.8, Problem 14E

To determine

To solve: The right triangle.

Answer to Problem 14E

The solution of the triangle is $\text{A = 24}{\text{.8}}^{\circ },\text{ B = 65}{\text{.2}}^{\circ },{\text{ C = 90}}^{\circ }\text{ and a = 145}\text{.5, b =314}\text{.938, c = 346}\text{.924}$.

Explanation of Solution

Given information:

  

$\text{B}={65}^{\circ }{12}^{\text{'}}={\left(65+\frac{12}{60}\right)}^{\circ }={65.2}^{\circ },\text{ a}=145.5$

Calculation: From the properties of trigonometric function,

  $\begin{array}{l}\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}},\text{ csc}\theta \text{=}\frac{\text{Hypotenuse}}{\text{Perpendicular}}\\ \cos \theta =\frac{\text{Base}}{\text{Hypotenuse}},\text{ sec}\theta \text{=}\frac{\text{Hypotenuse}}{\text{Base}}\\ \tan \theta =\frac{\text{Perpendicular}}{\text{Base}},\text{ cot}\theta =\frac{\text{Base}}{\text{Perpendicular}}\end{array}$   

From the triangle,

  $\text{B}={65}^{\circ }{12}^{\text{'}}={\left(65+\frac{12}{60}\right)}^{\circ }={65.2}^{\circ },\text{ C}={90}^{\circ }$

Therefore,

  $\begin{array}{c}\text{A+B+C}={180}^{\circ }\\ \text{A}+{65.2}^{\circ }+{90}^{\circ }={180}^{\circ }\\ \text{A}={180}^{\circ }-{90}^{\circ }-{65.2}^{\circ }\\ \text{A}={24.8}^{\circ }\end{array}$

Now, $\text{a}=\text{Perpendicular, b = Base, c = Hypotenuse}$

Since, $\text{a}=145.5$

From the properties of a right triangle,

  $\begin{array}{c}\sin {24.8}^{\circ }=\frac{\text{a}}{\text{c}}\\ 0.4194=\frac{145.5}{\text{c}}\\ \text{c}=\frac{145.5}{0.4194},\text{ doing cross-multiplication}\\ \text{c = 346}\text{.924}\end{array}$

Therefore, Perpendicular = $145.5$ and Hypotenuse = $\text{346}\text{.924}$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ 346.924=\sqrt{{145.5}^2+{\left(\text{Base}\right)}^2}\\ 120356.2618=21170.25+{\text{Base}}^2\\ \text{Base}=\sqrt{120356.2618-21170.25}\\ \text{Base}=\sqrt{99186.0118}\\ \text{Base}=314.938\end{array}$

Hence, The third side of the triangle is, base $=314.938$

Hence, The solution of the triangle is $\text{A = 24}{\text{.8}}^{\circ },\text{ B = 65}{\text{.2}}^{\circ },{\text{ C = 90}}^{\circ }\text{ and a = 145}\text{.5, b =314}\text{.938, c = 346}\text{.924}$