How do you solve the right triangle given b = 4, B = 10 degrees?

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How do you solve the right triangle given b = 4, B = 10 degrees?

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Answer 1 · 2018-03-26T10:15:12

$a = 22.69, c = 23.04, ˆ A = 80^{\circ}, A_{t} = 45.38 sq units$ $color(brown)(a = 22.69, c = 23.04, hat A = 80^@, A_t = 45.38 " sq units"$

Explanation:

$it's a right triangle with b = 4, ˆ B = 10^{\circ}$ $"it's a right triangle with " b = 4, hat B = 10^@$

![http://blog.cardinalec.com/act-math-finding-an-angle-measure-in-a-right-triangle/]( useruploads.socratic.org )

$ˆ A = 180 - 90 - 10 = 80^{\circ}$ $hat A = 180 - 90 - 10 = 80^@$

Applying Law of Sines,

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

$a = \frac{b sin A}{sin B} = \frac{4 \cdot sin 80}{sin 10} = 22.69$ $a = (b sin A) / sin B = (4 * sin 80) / sin 10 = 22.69$

$c = \frac{b sin C}{sin B} = \frac{4 \cdot sin 90}{sin 10} = 23.04$ $c = (b sin C) / sin B = (4 * sin 90) / sin 10 = 23.04$

Verification by applying Pythagoras theorem,

$c = \sqrt{a^{2} + b^{2}} = \sqrt{{22.69}^{2} + 4^{2}} = 23.04$ $c = sqrt(a^2 + b^2) = sqrt(22.69^2 + 4^2) = 23.04$

$Area of the rt triangle A_{t} = (\frac{1}{2}) \cdot a \cdot b = (\frac{1}{2}) \cdot 22.69 \cdot 4 = 45.38$ $"Area of the rt triangle " A_t = (1/2) * a * b = (1/2) * 22.69 * 4 = 45.38$

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How do you solve the right triangle given b = 4, B = 10 degrees?

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