Question

(b) In the diagram below, ABC is an isos-celes triangle. /AB/=/AC/= 5 cm and/BC/= 8 cm. Calculate correct to thenearest degree, ZBAC.AB5 cm8 cmX 5 cmC

Answer 1

Step 1

Given;

Step 2

From the triangle;

Using cosine rule;

$c^2=a^2+b^2-2(a)(b)cos(C)$

$\begin{gathered} 8^2=5^2+5^2-2(5)(5)cos(y) \\ m\angle BAC=y \\ 64-50=-50(cos(y)) \\ \\ \end{gathered}$$\begin{gathered} \frac{14}{-50}=cosy \\ y\approx106^o \end{gathered}$

Answer;

$m\angle BAC\approx106^o\text{ to the nearest degree}$