Question

In a triangle ABC, $\angle A={40}^{\circ },\angle B={80}^{\circ },AB=8cm.$ as shown in figure. The length of the two sides AC $\&$ BC is equal to $\left[\begin{array}{c}sin{40}^{\circ }=0.64\\ sin{60}^{\circ }=0.86\\ sin{80}^{\circ }=0.98\end{array}\right]$

• A. 6.45 cm, 8.5 cm
• B. 5.95 cm, 9.11 cm
• C. 5.5 cm, 8.2 cm
• D. 6.5 cm, 9.5 cm

Answer 1

The correct option is B

5.95 cm, 9.11 cm

Draw diameter AD passing through centre O and connect DB

$\angle D={60}^{\circ }$ (Angle subtended by the same chord AB)

$\angle ABD={90}^{\circ }$ (Angle subtended by the diamter on the circumference)

In right - angled triangle ABD

$sin{60}^{\circ }=\frac{AB}{AD}AD=\frac{AB}{sin{60}^{\circ }}=\frac{8}{0.86}=9.3cm...\left(1\right)$

In $\Delta ABC,$ using sine rule,

$\frac{BC}{sin{40}^{\circ }}=\frac{AB}{sin{60}^{\circ }}$

$\frac{BC}{sin{40}^{\circ }}=d=9.3cmfrom...\left(1\right)$

$BC=9.3\times sin{40}^{\circ }$

$BC=9.3\times 0.64=5.952cm$

Similarly

$\frac{AC}{sin{80}^{\circ }}=\frac{AB}{sin{60}^{\circ }}$

$\frac{AC}{sin{80}^{\circ }}=d=9.3cmfrom...\left(1\right)$

$AC=9.3sin{80}^{\circ }$

$AC=9.3\times 0.98=9.114cm$