Question

In triangle ABC, $\angle CAB={40}^{\circ }$, AC = 6 cm, AB = 7 cm. Find the length BC. $[sin{40}^{\circ }=0.67,cos{40}^{\circ }=0.76]$

Answer 1

Draw CD$\perp AB$

In $△ADC\sin {40}^{\circ }=\frac{CD}{AC}0.64=\frac{CD}{6}CD=6\times 0.64=3.85m\cos {40}^{\circ }=\frac{AD}{AC}0.76=\frac{AD}{6}AD=6\times 0.76AD=4.56m\text{From figure, we can see that}DB=AB-ADDB=7-4.56DB=2.44m\text{In right-angled triangle CDB}$

$B{C}^2=B{D}^2+D{C}^{2}B{C}^2={\left(2.44\right)}^2+{\left(3.85\right)}^{2}B{C}^2=5.95++17.82B{C}^2=20.77BC=4.55m$