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Q.

ΔABCis a right triangle, right-angled atC. If $∠ A = 30^{\circ}$ and $AB = 40$ units, find the remaining two sides and $∠ B in △ ABC$

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a

AC = 1 O \sqrt{3}

units,

BC = 50

units and

∠ B = 90^{\circ}

b

AC = 4 O \sqrt{3}

units,

BC = 30

units and

∠ B = 30^{\circ}

c

AC = 3 O \sqrt{3}

units,

BC = 80

units and

∠ B = 10^{\circ}

d

AC = 2 O \sqrt{3}

units,

BC = 20

units and

∠ B = 60^{\circ}

answer is D.

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Detailed Solution

Given,

AB = 40

∠ A = 30^{\circ}

∠ C = 90^{\circ}

(right angle)
By angle sum property of triangle,

∠ A + ∠ B + ∠ C = 180^{\circ}

Put the values,

30^{\circ} + ∠ B + 90^{\circ} = 180^{\circ}

∠ B = 180^{\circ} - 120^{\circ}

∠ B = 60^{\circ}

Question Image

Now, in the figure using the trigonometry property,

\cos A = \frac{AC}{AB}

\cos 30^{\circ} = \frac{AC}{40}

\frac{\sqrt{3}}{2} = \frac{AC}{40}

AC = \frac{\sqrt{3}}{2} \times 40

AC = 20 \sqrt{3} units

Similarly we have,

\sin A = \frac{BC}{AB}

\sin 30^{\circ} = \frac{BC}{40}

\frac{1}{2} = \frac{BC}{40}

40 \times \frac{1}{2} = 20 units

Thus, the required sides are

AC = 2 O \sqrt{3}

units,

BC = 20

units and required angle

∠ B = 60^{\circ}

Option 4 is correct.

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