How do you find the missing sides of a triangle given sides are x and 6 root 3, angles are 60 and 90?

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Answer 1 · 2018-02-25T15:44:32

Missing sides and angles are

$x = a = 9, c = 5.1962, ˆ C = 30^{\circ}$ $color(indigo )(x = a = 9, c = 5.1962, hat C = 30^@$

Given : $a = x, b = 6 \sqrt{3} = 10.3923, ˆ A = 60^{\circ}, ˆ B = 90^{\circ}$ $a = x, b = 6sqrt3 = 10.3923, hatA = 60^@, hat B = 90^@$

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Applying law of sines,

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a / sin A = b / sin B = c / sin C$

$a = x = \frac{b sin A}{sin B} = \frac{6 \sqrt{3} \cdot sin (60)}{sin (90)}$ $a = x = (b sin A) / sin B = (6sqrt3 * sin (60) ) / sin (90)$

$a = \frac{6 \sqrt{3} \cdot \sqrt{3}}{2} = 9$ $a =( 6 sqrt3 * sqrt3) / 2 = 9$

Third angle $ˆ C = 180 - 90 - 60 = 30^{\circ}$ $hat C = 180 - 90 - 60 = 30^@$

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As the angles of the right triangle are $60 : 90 : 30$ $60 : 90 : 30$ , sides will be in the ratio $\sqrt{3} : 2 : 1$ $sqrt3 : 2 : 1$

Third side $c = (\frac{1}{2}) b = (\frac{1}{2}) \cdot 6 \sqrt{3} = 5.1962$ $c = (1/2) b = (1/2) * 6sqrt3 = 5.1962$