Chapter 4.8, Problem 11E

To determine

To solve: The right triangle.

Answer to Problem 11E

The solution of the triangle is $\text{A = 17}{\text{.235}}^{\circ },\text{ B = 72}{\text{.765}}^{\circ },{\text{ C = 90}}^{\circ }\text{ and a = 16, b =51}\text{.575, c = 54}$.

Explanation of Solution

Given information:

  

$\text{a}=16,\text{ c}=54$

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,

  $\text{a}=16,\text{ c}=54$ and let, $\text{A = }\theta$

Therefore,

  $\begin{array}{c}\sin \theta =\frac{16}{54}\\ \theta =\arcsin \left(\frac{16}{54}\right)\\ ={17.235}^{\circ }\end{array}$

Since, $\text{A = 17}{\text{.325}}^{\circ }{\text{ and C = 90}}^{\circ }$

Therefore,

  $\begin{array}{c}\text{A+B+C}={180}^{\circ }\\ {17.235}^{\circ }+\text{B}+{90}^{\circ }={180}^{\circ }\\ \text{B}={180}^{\circ }-{90}^{\circ }-{17.235}^{\circ }\\ \text{B}={72.765}^{\circ }\end{array}$

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

Since, $\text{a}=16,\text{ c}=54$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ 54=\sqrt{{16}^2+{\left(Base\right)}^2}\\ \text{2916}=256+{\text{Base}}^2\\ \text{Base}=\sqrt{2916-256}\\ \text{Base}=51.575\end{array}$

Hence, The third side of the triangle is, hypotenuse $=51.575$

Hence, The solution of the triangle is $\text{A = 17}{\text{.235}}^{\circ },\text{ B = 72}{\text{.765}}^{\circ },{\text{ C = 90}}^{\circ }\text{ and a = 16, b =51}\text{.575, c = 54}$