## Solve Lesson 14 Page 79 Math 10 – Kite >

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) $${S_{ABC}} = \frac{1}{2}AB.AC\sin A = \frac{1}{2}.5.8.\sin 84,{3^0} \approx 19, 9$$