Topic
For triangle ABC have \(AB = 5,AC = 8,BC = 9\). Calculate (round up to tenths)
a) Measures of angles A, REMOVE, OLD
b) Area of triangle ABC
Solution method – See details
Method
Step 1: Use the cosine theorem and the sum of the angles in a triangle to calculate the measure of the angles A, REMOVE, OLD
Step 2: Use the area formula \(S = \frac{1}{2}AB.AC\sin A\) to calculate the area ∆ABC
Detailed explanation
a) Apply the law of cosines toABC We have:
\(\left\{ \begin{array}{l}B{C^2} = A{B^2} + A{C^2} – 2.AB.AC.\cos A\\A{C^ 2} = A{B^2} + B{C^2} – 2.AB.BC.\cos B\end{array} \right.\)
\( \Rightarrow \left\{ \begin{array}{l}\cos A = \frac{{A{B^2} + A{C^2} – B{C^2}}}{{2. AB.AC}} = \frac{{{5^2} + {8^2} – {9^2}}}{{2.5.8}} = \frac{1}{{10}}\\\ cos B = \frac{{A{B^2} + B{C^2} – A{C^2}}}{{2.AB.BC}} = \frac{{{5^2} + { 9^2} – {8^2}}}{{2.5.9}} = \frac{7}{{15}}\end{array} \right.\)
\( \Rightarrow \left\{ \begin{array}{l}\widehat A \approx 84,{3^0}\\\widehat B \approx 62,{2^0}\end{array} \right. \)
We have: \(\widehat C = {180^0} – (\widehat A + \widehat B) = 33,{5^0}\)
b) \({S_{ABC}} = \frac{1}{2}AB.AC\sin A = \frac{1}{2}.5.8.\sin 84,{3^0} \approx 19, 9\)
