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Q.

In a right $△ ABC$ right angled at C, if $∠ B = 60^{\circ}$ and AB = 15 units. Find the remaining angles and sides.

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a

{Angles - 90}^{o}, 30^{o}

and

Sides - 7.5 units \frac{15 \sqrt{3}}{2}

Units

b

{Angles - 100}^{o}, 20^{o}

and

Sides - 3.5 units \frac{7 \sqrt{3}}{5}

Units

c

{Angles - 45}^{o}, 85^{o}

and

Sides - 2.5 units \frac{15 \sqrt{2}}{4}

Units

d

{Angles - 60}^{o}, 60^{o}

and

Sides - 7 units \frac{5 \sqrt{3}}{2}

Units

answer is A.

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Detailed Solution

Given,
Right angle at C in the triangle

ABC

. In this triangle,

∠ B = 60^{\circ}

∠ C = 90^{\circ}

AB = 15 Now, in triangle ABC,

∠ A + ∠ B + ∠ C = 180^{\circ}

[Sum of the all angles in a triangle = 180°]

60^{\circ} + ∠ A + 90^{\circ} = 180^{\circ}

∠ A = 30^{\circ}

C:UserssoftberryDownloadsP0041 JPEGQ.24.jpg

Now from the trigonometry property,

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

\sin 30^{\circ} = \frac{CB}{15}

\frac{1}{2} \times 15 = CB

[Since,

si n 30^{\circ} = 1 / 2]

CB = 7.5 units

Now,

A B^{2} = {AC}^{2} + {BC}^{2}

[By using of Pythagoras theorem]

15^{2} = {AC}^{2} + {(\frac{15}{2})}^{2}

AC = 225 - (\frac{225}{4})

{AC}^{2} = 225 (\frac{3}{4})

AC = \frac{15 \sqrt{3}}{2}

Units
Correct option is 1.

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