Answer:
[tex]A = 100^o[/tex]
[tex]B = 43.0^o[/tex]
[tex]C =37.0^o[/tex]
Step-by-step explanation:
Given
[tex]A = 100^o[/tex]
[tex]b = 51[/tex]
[tex]c =45[/tex]
Required
Complete the triangle (See attachment)
First, calculate side a using cosine's equation
[tex]a^2 = b^2 + c^2 -2bc \cos A[/tex]
[tex]a^2 = 51^2 + 45^2 -2*51*45 \cos (100^o)[/tex]
[tex]a^2 = 4626 -4590 *-0.1736[/tex]
[tex]a^2 = 4626+ 796.8[/tex]
[tex]a^2 = 5422.8240[/tex]
Take square roots
[tex]a = 73.6398[/tex]
The measure of B and C can then be calculated using sine's equation
[tex]\frac{a}{\sin A} =\frac{b}{\sin B} =\frac{c}{\sin C}[/tex]
So:
[tex]\frac{73.6398}{\sin (100)} =\frac{51}{\sin B}[/tex]
[tex]\frac{73.6398}{0.9848} =\frac{51}{\sin B}[/tex]
Cross multiply
[tex]\sin B * 73.6398 = 51 * 0.9848[/tex]
[tex]\sin B * 73.6398 = 50.2248[/tex]
Solve for sin B
[tex]\sin B = \frac{50.2248}{73.6398}[/tex]
[tex]\sin B = 0.6820[/tex]
Take arc sin of both sides
[tex]B = \sin^{-1}(0.6820)[/tex]
[tex]B = 43.0^o[/tex]
To calculate C, we make use of:
[tex]A + B + C = 180^o[/tex] --- angles in a triangle
[tex]100 + 43 + C = 180[/tex]
[tex]143 + C = 180[/tex]
Collect like terms
[tex]C =180-143[/tex]
[tex]C =37.0^o[/tex]
