Question

solve for the remaining angles and side of the one triangle that can be created. round to the nearest hundredth: b = 105°, c = 4, b = 4.5

Explanation:

Step1: Use the Law of Sines

By the Law of Sines, $\frac{\sin B}{b}=\frac{\sin C}{c}$. Substitute $B = 105^{\circ}$, $c = 4$, and $b = 4.5$ into the formula: $\sin C=\frac{c\sin B}{b}=\frac{4\times\sin(105^{\circ})}{4.5}$.
Since $\sin(105^{\circ})=\sin(60^{\circ}+ 45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}+\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}\approx0.9659$. Then $\sin C=\frac{4\times0.9659}{4.5}\approx0.8586$. So $C=\sin^{- 1}(0.8586)\approx59.11^{\circ}$.

Step2: Find angle A

Since the sum of angles in a triangle is $180^{\circ}$, $A = 180^{\circ}-B - C$. Substitute $B = 105^{\circ}$ and $C\approx59.11^{\circ}$ into the formula. So $A=180^{\circ}-105^{\circ}-59.11^{\circ}=15.89^{\circ}$.

Step3: Use the Law of Sines again to find side a

Using the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}$. We know $A = 15.89^{\circ}$, $B = 105^{\circ}$, and $b = 4.5$. Then $a=\frac{b\sin A}{\sin B}=\frac{4.5\times\sin(15.89^{\circ})}{\sin(105^{\circ})}$. Since $\sin(15.89^{\circ})\approx0.2747$ and $\sin(105^{\circ})\approx0.9659$, $a=\frac{4.5\times0.2747}{0.9659}\approx1.27$.

Answer:

$A = 15.89^{\circ}$, $C = 59.11^{\circ}$, $a\approx1.27$