Chapter 4.8, Problem 10E

To determine

To solve: The right triangle.

Answer to Problem 10E

The solution of the triangle is $\text{A = 26}{\text{.565}}^{\circ },\text{ B = 63}{\text{.435}}^{\circ },{\text{ C = 90}}^{\circ }\text{ and a = 6, b =12, c = 13}\text{.416}$.

Explanation of Solution

Given information:

  

$\text{a}=25,\text{ b}=37$

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,

  $\text{a}=25,\text{ b}=37$ and let, $\text{A = }\theta$

Therefore,

  $\begin{array}{c}\tan \theta =\frac{25}{37}\\ \theta =\arctan \left(\frac{25}{37}\right)\\ ={34.06}^{\circ }\end{array}$

Since, $\text{A = 34}{\text{.06}}^{\circ }{\text{ and C = 90}}^{\circ }$

Therefore,

  $\begin{array}{c}\text{A+B+C}={180}^{\circ }\\ {34.06}^{\circ }+\text{B}+{90}^{\circ }={180}^{\circ }\\ \text{B}={180}^{\circ }-{90}^{\circ }-{34.06}^{\circ }\\ \text{B}={55.94}^{\circ }\end{array}$

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

Since, $\text{a}=25,\text{ b}=37$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ \text{Hypotenuse}=\sqrt{{25}^2+{37}^2}\\ \text{Hypotenuse}=\sqrt{625+1369}\\ \text{Hypotenuse}=\sqrt{1994}\\ \text{Hypotenuse}=44.654\end{array}$

Hence, The third side of the triangle is, hypotenuse $=44.654$

Hence, The solution of the triangle is $\text{A = 34}{\text{.06}}^{\circ },\text{ B = 55}{\text{.94}}^{\circ },{\text{ C = 90}}^{\circ }\text{ and a = 25, b =37, c = 44}\text{.654}$