Given:
The angels and sides of the triangle are
A = 50°. B = 45°, and a=3
Aim:
We need to find the angle C and sides c and b.
Explanation:
Use sine law.
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}[/tex]
[tex]\text{ Consider }\frac{\sin A}{a}=\frac{\sin B}{b}\text{ to find side b.}[/tex]
Substitute A = 50°. B = 45°, and a=3 in the equation.
[tex]\frac{\sin 50^o}{3}=\frac{\sin 45^o}{b}[/tex]
[tex]b=\frac{\sin 45^o}{\sin 50^o}\times3[/tex][tex]b=2.77[/tex]
Use the triangle sum property to find the angle C.
[tex]A+B+C=180^o[/tex]
Substitute A = 50°. and B = 45° in the equation.
[tex]50^o+45^o+C=180^o[/tex]
[tex]95^o+C=180^o[/tex]
[tex]C=180^o-95^o[/tex]
[tex]C=85^o[/tex]
[tex]\text{ Consider }\frac{\sin A}{a}=\frac{\sin C}{c}\text{ to find side c.}[/tex]
Substitute A = 50°. C= 85°, and a=3 in the equation.
[tex]\frac{\sin50^o}{3}=\frac{\sin 85^o}{c}[/tex]
[tex]c=\frac{\sin 85^o}{\sin 50^o}\times3[/tex][tex]c=3.90[/tex]
Final answer:
[tex]C=85^o[/tex][tex]b=2.77[/tex][tex]c=3.90[/tex]
