Answer:
[tex]A = 43.0^o[/tex]
[tex]B = 55.0^o[/tex]
[tex]BC = 20.0[/tex]
Step-by-step explanation:
I will solve this question with the attached triangle
From the attachment, we have:
[tex]\angle C =82^o[/tex]
[tex]AB = 29[/tex]
[tex]AC = 24[/tex]
Required
Solve the triangle
First, we calculate [tex]\angle B[/tex] using sine law:
[tex]\frac{AC}{\sin(B)} = \frac{AB}{\sin(C)}[/tex]
This gives:
[tex]\frac{24}{\sin(B)} = \frac{29}{\sin(82^o)}[/tex]
[tex]\frac{24}{\sin(B)} = \frac{29}{0.9903}[/tex]
Cross multiply
[tex]\sin(B) * 29 = 24 * 0.9903[/tex]
[tex]\sin(B) * 29 = 23.7672[/tex]
Divide both sides by 29
[tex]\sin(B) = 0.8196[/tex]
Take arcsin of both sides
[tex]B = \sin^{-1}(0.8196)[/tex]
[tex]B = 55.0^o[/tex]
Next, calculate [tex]\angle A[/tex] using:
[tex]A + B + C = 180^o[/tex] --- angles in a triangle
[tex]A + 55^o + 82^o = 180^o[/tex]
Collect like terms
[tex]A = 180^o-55.0^o - 82^o[/tex]
[tex]A = 43.0^o[/tex]
Next, calculate BC using sine laws
[tex]\frac{BC}{\sin(A)} = \frac{AB}{\sin(C)}[/tex]
This gives:
[tex]\frac{BC}{\sin(43.0)} = \frac{29}{\sin(82)}[/tex]
[tex]\frac{BC}{0.6820} = \frac{29}{0.9903}[/tex]
Make BC the subject
[tex]BC = \frac{29*0.6820}{0.9903}[/tex]
[tex]BC = 20.0[/tex]
