Question

In ΔQRS, q = 5.5 cm, r = 7.9 cm and ∠S=154°. Find ∠R, to the nearest 10th of a degree.

Answer 1

Check the picture below.

let's firstly find side "s"

`\textit{Law of Cosines}\\\\ c^2 = a^2+b^2-(2ab)\cos(C)\implies c = √(a^2+b^2-(2ab)\cos(C)) \\\\[-0.35em] ~\dotfill\\\\ s = √(5.5^2+7.9^2~-~2(5.5)(7.9)\cos(154^o)) \implies s = √( 92.66 - 86.90 \cos(154^o) ) \\\\\\ s \approx √( 92.66 - (-78.105203) ) \implies s \approx √( 170.765203 ) \implies s \approx 13.07`

now, let's use the Law of Sines to get ∡R

`\textit{Law of sines} \\\\ \cfrac{\sin(\measuredangle A)}{a}=\cfrac{\sin(\measuredangle B)}{b}=\cfrac{\sin(\measuredangle C)}{c} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{\sin(\measuredangle S)}{s}=\cfrac{\sin(\measuredangle R)}{r}\implies \cfrac{r\sin(\measuredangle S)}{s}=\sin(\measuredangle R) \\\\\\ \sin^(-1)\left[ \cfrac{r\sin(\measuredangle S)}{s} \right]=\measuredangle R\implies \sin^(-1)\left[ \cfrac{5.5\sin(154^o)}{13.07} \right]\approx\measuredangle R\implies 10.6^o\approx R`

Make sure your calculator is in Degree mode.