Question

Find the missing side length and angles of ABC given that m A = 41°, b = 4, and c = 10.

Answer 1

To solve this question, we would use cosine rule which is given as

`a^2=b^2+c^2-2bc\cos A`

Our values have been defined for us and we will proceed to evaluate

`\begin{gathered} a^2=4^2+10^2-2(4)(10)\cos 41 \\ a^2=16+100-80\cos 41 \\ a^2=116-60.376 \\ a^2=55.624 \\ \text{take the square root of both sides} \\ a=\sqrt[]{55.624} \\ a=7.458\approx7.5 \end{gathered}`

From the calculations above, the value of the missing side a is 7.5 units

To find angle B,

we can use sine rule

`\begin{gathered} (a)/(\sin A)=(b)/(\sin B) \\ (7.5)/(\sin 41)=(4)/(\sin B) \\ \sin B=(4*\sin 41)/(7.5) \\ \sin B=0.3498 \\ B=\sin ^(-1)0.3498 \\ B=20.5^0 \end{gathered}`

We can still approach C with sine rule or sum of angle in a triangle

`\begin{gathered} A+B+C=180 \\ 41+20.5+C=180 \\ c=118.5^0 \end{gathered}`

From the calculations above, the value of a = 7.5 , B = 22⁰ and C = 118.5⁰ respectively which is option B