Chapter 4.8, Problem 6E

To determine

Tofind:the unknown side and angle of the right- angle triangle shown below.

Answer to Problem 6E

The required value are $a=8.82$ , $b=12.14$ and ${\text{A=36}}^0$ .

Explanation of Solution

Given:

  ${\text{B=54}}^0,c=15$

Concept used:

Sum of the angle of the triangle is ${180}^0$ .

Trigonometric formula:

  $\text{sinθ=}\frac{\text{perpendicular}}{\text{hypotenuse}}$

  $\text{cosθ=}\frac{\text{base}}{\text{hypotenuse}}$

Calculation:

Assume that ${\text{B=54}}^{\text{0}}$ , $\text{c=15}$

Since the sum of the three angles of triangle is ${\text{180}}^0$ and ${\text{C=90}}^{\text{0}}$ . Therefore

  ${\text{A+B=90}}^{\text{0}}$

Put ${\text{B=54}}^{\text{0}}$ in the above question

  ${\text{A+B=90}}^{\text{0}}$

  ${\text{A+54}}^0{\text{=90}}^{\text{0}}$

  ${\text{A=90}}^{\text{0}}{\text{-54}}^0$

  ${\text{A=36}}^0$ .

Since the sine of the angle is given by the ratio od the length of the perpendicular to the hypotenuse, therefore

  $\text{sinB=}\frac{\text{b}}{\text{c}}$

Put the values

  ${\text{sin54}}^0\text{=}\frac{\text{b}}{15}$

  $b=15\times \sin {54}^0$

  $b=15\times 0.8090$

  $b=12.14$

Again, since the cosine of the angle is given by the ratio of the length of the base to the hypotenuse, therefore

  $\text{cosB=}\frac{\text{a}}{\text{c}}$

  ${\text{cos45}}^0\text{=}\frac{\text{a}}{15}$

  $a=15\times \cos {54}^0$

  $a=15\times 0.5877$

  $a=8.82$

Hence, the required value are $a=8.82$ , $b=12.14$ and ${\text{A=36}}^0$ .