Given:
• A = 100 degrees
,
• a = 3.5
,
• b = 3
Let's solve for the remaining angles and side of the triangle.
Here, we are given one angle and two sides.
To solve, apply the Law of Sines:
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}[/tex]
• To solve for measure of angle B, we have:
[tex]\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \\ \frac{\sin100}{3.5}=\frac{\sin B}{3} \\ \\ \sin B=\frac{3\sin 100}{3.5} \\ \\ \sin B=\frac{2.954}{3.5} \\ \\ \sin B=0.844 \end{gathered}[/tex]
Take the sine inverse of both sides:
[tex]\begin{gathered} B=\sin ^{-1}(0.844) \\ \\ B=57.58^0 \end{gathered}[/tex]
Therefore, the measue of angle B is = 57.58 degrees.
• To solve for angle C, apply the Triangle Angle Sum Theorem.
m∠A + m∠B + m∠C = 180
m∠C = 180 - m∠A - m∠B
m∠C = 180 - 100 - 57.68
m∠C = 22.32
The measure of angle C is 22.32 degrees.
• To find the length of c, apply the Law of Sines:
[tex]\begin{gathered} \frac{\sin A}{a}=\frac{\sin C}{c} \\ \\ \frac{\sin100}{3.5}=\frac{\sin 22.32}{c} \\ \\ c=\frac{3.5\sin 22.32}{\sin 100}\tan ^{-1}\tan ^{-1} \\ \\ c=\frac{1.329}{0.9848} \\ \\ c=1.35 \end{gathered}[/tex]
The length of side c is 1.35 units.
ANSWER:
• B = 57.58,°
,
• C = 22.32,°
,
• c = 1.35
