Question

In $\Delta ABC,AB=10cm,AC=6cm,\angle A={70}^{\circ }$. The length BC is equal to. $[cos{70}^{\circ }=0.34,sin{70}^{\circ }=0.94]$

• A. 8.25 cm
• B. 9.75 cm
• C. 7.5 cm
• D. 10.5 cm

Answer 1

The correct option is B

9.75 cm

Draw CD perpendicular to AB.

In right angle triangle ACD

$sin{70}^{\circ }=\frac{CD}{AC}$

$CD=AC.sin{70}^{\circ }$

$CD=6\times 0.94=5.64cm$

$cos{70}^{\circ }=\frac{AD}{AC}$

$AD=ACcos{70}^{\circ }$

$AD=6\times 0.34=2.04cm$

$BD=AB-AD=(10-2.04)=7.96cm$

In right angled triangle BCD

$B{C}^2=B{D}^2+C{D}^2$

$BC=\sqrt{(7.96{)}^2+(5.64{)}^2}$

$BC=\sqrt{31.81+63.36}$

$BC=\sqrt{95.17}=9.75cm.$