In an acute triangle ΔABC, a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. If ∠B = 45°, c = 3 and b = \sqrt{6} . Then

Engineering

Mathematics

Basic Rules of Properties of Triangle

Question

In an acute triangle ΔABC, a, b, c are the lengths of its sides and A, B, C are the angles of triangle ABC. If ∠B = 45°, c = 3 and b = $\sqrt{6}$ . Then

∠C is equal to 75°

area of ΔABC is $\frac{3 (3 + \sqrt{3})}{4}$

circumradius R of ΔABC is $\sqrt{3}$

a is equal to $\frac{3 + \sqrt{3}}{\sqrt{2}}$

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$\frac{\sin (135^{\circ} - θ)}{3} = \frac{\sin 45^{\circ}}{\sqrt{6}}$

sin(135° – θ) = $\frac{\sqrt{3}}{2}$

135° – θ = 60° or 120°

θ = 75° or 15°

θ = 15° (not possible as the Δ is acute)

∴ θ = 75°.

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