Consider a triangle ABC like the one below. Suppose that A=68°, C = 55°, and b= 42. (The figure is not drawn to scale.) Solve the triangle. Round your answers to the nearest tenth. If there is more than one solution, use the button labeled "or". E E B 12 c T B = 57°, a = = 0,c=0 No solution X Х 6 ?

Respuesta :

At first, let us find the measure of angle B

The sum of angles of a triangle is 180 degrees

[tex]\begin{gathered} m\angle A+m\angle B+m\angle C=180 \\ 68+55+m\angle B=180 \\ 123+m\angle B=180 \end{gathered}[/tex]

Subtract 123 from both sides to fin m < B

[tex]\begin{gathered} 123-123+m\angle B=180-123 \\ m\angle B=57 \end{gathered}[/tex]

Now, let us use the sin rule

[tex]\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}[/tex]

Since b = 42, then

[tex]\frac{a}{\sin68}=\frac{42}{\sin 57}[/tex]

By using cross multiplication

[tex]a\sin 57=42\sin 68[/tex]

Divide both sides by sin 57

[tex]\begin{gathered} a=\frac{42\sin 68}{\sin 57} \\ a=46.43267974 \\ a=46.4 \end{gathered}[/tex]

Do the same to find c

[tex]\frac{42}{\sin57}=\frac{c}{\sin 55}[/tex]

By using cross multiplication

[tex]c\sin 57=42\sin 55[/tex]

Divide both sides by sin 57

[tex]\begin{gathered} c=\frac{42\sin 55}{\sin 57} \\ c=41.02252681 \\ c=41.0 \end{gathered}[/tex]