Question

InAABC,a = 13, and bFind c to the nearesttenth.AmZA = 19°,b= 14.Baс

Answer 1

Given:

$m\angle A=19\degree,a=13\text{ and b=14.}$

Required:

We need to find the value of c.

Explanation:

Consider the sine law.

$\frac{sinA}{a}=\frac{sinB}{b}$$Substitute\text{ }m\angle A=19\degree,a=13\text{ and b=14 in the formula,}$$\frac{sin19\degree}{13}=\frac{sinB}{14}$$\frac{sin19\degree}{13}\times14=\frac{sinB}{14}\times14$$\frac{sin19\degree}{13}\times14=sinB$$sinB=\frac{sin19\degree}{13}\times14$$sin^{-1}sinB=sin^{-1}(\frac{sin19\degree}{13}\times14)$$B=sin^{-1}(\frac{sin19\degree}{13}\times14)$$m\angle B=20.5\degree$

Use the triangle sum properly.

$m\angle A+m\angle B+m\angle C=180\degree$$Substitute\text{ }m\angle A=19\degree,and\text{ }m\angle B=20.5\text{ in the formula.}$$19\degree+20.5\degree+m\angle C=180\degree$$39.5\degree+m\angle C=180\degree$$39.5\degree+m\angle C-39.5\degree=180\degree-39.5\degree$$m\angle C=140.5\degree$

Consider the sine law.

$\frac{sinA}{a}=\frac{sinC}{c}$$Substitute\text{ }m\angle A=19\degree,a=13\text{ and }m\angle C=140.5\text{ in the formula,}$$\frac{sin19\degree}{13}=\frac{sin140.5\degree}{c}$$\frac{13}{sin19\degree}=\frac{c}{sin140.5\degree}$$\frac{13}{sin19\degree}\times sin140.5\degree=\frac{c}{sin140.5\degree}\times sin140.5\degree$$\frac{13}{sin19\degree}\times sin140.5\degree=c$$c=25.3987$$c=25.4$

Final answer:

$c=25.4units$