Question

In $\Delta ABC,\angle A={50}^{\circ },\angle B={70}^{\circ },AB=6cm.$ The length of 2 sides BC and AC are $[sin{50}^{\circ }=0.77,sin{60}^{\circ }=0.87sin{70}^{\circ }=0.94]$

• A. 6.2 cm, 7.5 cm
• B. 5.9 cm, 7.2 cm
• C. 5.313 cm, 6.486 cm
• D. 5.1 cm, 4.9 cm

Answer 1

The correct option is C

5.313 cm, 6.486 cm

Draw diameter AD passing through center O and connect DB

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

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

In $\Delta ABD$

$sin{60}^{\circ }=\frac{AB}{AD}.$

$AD=d=\frac{AB}{sin{60}^{\circ },}=\frac{6}{0.87}=6.9cm...\left(1\right)$

Now, in triangle ABC, using sine rule

$\frac{BC}{sin50}=\frac{AB}{sin{60}^{\circ }}$

$\Rightarrow \frac{BC}{sin{50}^{\circ }}=d=6.9cmfrom...\left(i\right)$

$BC=6.9\times sin{50}^{\circ }=6.9\times 0.77=5.313cm$

Similarly

$\frac{AC}{sin{70}^{\circ }}=\frac{AB}{sin{60}^{\circ }}$

$\frac{AC}{sin{70}^{\circ }}=d=6.9cmfrom...\left(i\right)$

$AC=dsin{70}^{\circ }$

$AC=6.9\times 0.94=6.486cm$