Question

In the given fid ABC is right D at B such that AB = 3cm and AC = 6 cm. Determine $\angle ACB$.

Answer 1

From pythagoras theorem:

${AB}^2+{BC}^2={AC}^2$

$⟹3^2+{BC}^2=6^2$

$⟹{BC}^2=36-9$

$⟹{BC}^2=27$

$⟹BC=3\sqrt{3}cm$

Using cosine rule in $△ABC$

$\cos \theta =\frac{{BC}^2+{AC}^2-{AB}^2}{2.BC.AC}$

$=\frac{27+36-9}{2\times 3\sqrt{3}\times 6}$

$=\frac{54}{36\sqrt{3}}$

$=\frac{3}{2\sqrt{3}}$

$⟹\theta ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\therefore \angle ACB={30}^{\circ }$