The cosine law

Solve the unknown dimensions for the following triangle: Triangle ABC: Angle A=43 degrees, b=7.0 cm, c=6.0 cm

Question 1. Angle B with units written as degrees
Question 2. Angle C with units written as degrees
Question 3. a, rounded to the nearest tenth of a cm with the unit.

Final Answer:

a =  4.8545 cm
B =  79.5497 °
C =  57.4503 °

Step-by-step explanation:

$$\begin{gathered} A=43{}^{\circ } \\ b=7.0\text{cm} \\ c=6.0\text{cm} \\ {a}^2=b^2+c^2-2bc\cos A \\ a=\sqrt{b^2+c^2-2\cdot b\cdot c\cdot \cos A}=\sqrt{b^2+c^2-2\cdot b\cdot c\cdot \cos 43°}=\sqrt{7^2+6^2-2\cdot 7\cdot 6\cdot \cos 43°}=\sqrt{7^2+6^2-2\cdot 7\cdot 6\cdot 0.731354}=4.855=4.8545\text{cm} \end{gathered}$$

$$\begin{gathered} \sin B:\sin A=b:a=7:4.8545\doteq 1.442=82653:57320 \\ {b}_0=\sin A\cdot \frac{b}{a}=\sin 43°\cdot \frac{b}{a}=\sin 43°\cdot \frac{7}{4.8545}=0.681998\cdot \frac{7}{4.8545}=0.98341 \\ B=\frac{180°}{\pi }\cdot \arcsin (b_0)=\frac{180°}{\pi }\cdot \arcsin 0.9834=79.5497^{\circ }=79°32^{\prime }59" \end{gathered}$$

$$\begin{gathered} \sin C:\sin A=c:a=6:4.8545\doteq 1.236=68968:55801 \\ {c}_0=\sin A\cdot \frac{c}{a}=\sin 43°\cdot \frac{c}{a}=\sin 43°\cdot \frac{6}{4.8545}=0.681998\cdot \frac{6}{4.8545}=0.84293 \\ C=\frac{180°}{\pi }\cdot \arcsin (c_0)=\frac{180°}{\pi }\cdot \arcsin 0.8429\doteq 57.4503^{\circ } \\ \text{ Verifying Solution: } \\ s=A+B+C=43+79.5497+57.4503=180{}^{\circ } \end{gathered}$$

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