Answer:
[tex]a =11.0[/tex]
[tex]\angle A =21.5^o[/tex]
[tex]c = 28.9[/tex]
Step-by-step explanation:
Given
[tex]\angle B = 53.6^o[/tex]
[tex]\angle C = 104.9^o[/tex]
[tex]b=24.1[/tex]
Required
Solve the triangle
We have:
[tex]\angle A + \angle B + \angle C =180^o[/tex] --- angles in a triangle
Substitute known values
[tex]\angle A + 53.6^o + 104.9^o =180^o[/tex]
So, we have:
[tex]\angle A =180^o-53.6^o - 104.9^o[/tex]
[tex]\angle A =21.5^o[/tex]
To solve for the sides, we make use of sine rule:
[tex]\frac{a}{\sin A} =\frac{b}{\sin B} = \frac{c}{\sin C}[/tex]
So, we have:
[tex]\frac{a}{\sin (21.5)} =\frac{24.1}{\sin 53.6} = \frac{c}{\sin 104.9}[/tex]
Solving for (a), we have:
[tex]\frac{a}{\sin (21.5)} =\frac{24.1}{\sin 53.6}[/tex]
Make (a) the subject
[tex]a =\frac{24.1}{\sin 53.6} * \sin (21.5)[/tex]
[tex]a =\frac{24.1}{0.8049} * 0.3665[/tex]
[tex]a =11.0[/tex]
To solve for (c), we have:
[tex]\frac{24.1}{\sin 53.6} = \frac{c}{\sin 104.9}[/tex]
Make (c) the subject
[tex]c = \frac{24.1}{\sin 53.6} * \sin 104.9[/tex]
[tex]c = \frac{24.1}{0.8049} * 0.9664[/tex]
[tex]c = 28.9[/tex]
