Question

In triangle ABC . Angle of A =50⁰,Angle B =70⁰ AB =6 cm . a). Find the diameter of its circum circle . b) Find the lengths of other two sides AC and BC. [Sin 60⁰= 0.87,sin 50⁰ = 0.77,sin 70⁰=0.94]

Answer 1

Draw diameter AD passing through center O and connect DB

In ΔABD, ∠D = 60° (Angle subtended by the same chord AB)

∠DBA = 90° (Angle subtended by the diameter on the circumference)

In ΔABD

\(\sin 60^\circ = \frac{AB}{AD}\)

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

Now, in triangle ABC, using sine rule

\(\frac{BC}{\sin 50^\circ} =\frac{AB}{\sin60^\circ}\)

\(\Rightarrow \frac{BC}{\sin 50^\circ} = d = 6.9 \ cm \quad \text{from (1)}\)

\(BC = 6.9 \times \sin 50^\circ = 6.9 \times 0.77 = 5.313 \ cm\)

Similarly

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

\(\frac{AC}{\sin 70^\circ} = d = 6.9\ cm \quad \text{from (1)}\)

\(AC = d \sin 70^\circ\)

\(AC = 6.9 \times 0.94 = 6.486 \ cm\)

a) Diameter of its circum circle = 6.9 cm

b) Lengths of AC and BC = 5.313 cm and 6.486 cm