How do you use the law of cosines or law of sines if you are given C = 105 degrees, a = 10, b = 4.5?

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How do you use the law of cosines or law of sines if you are given C = 105 degrees, a = 10, b = 4.5?

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Answer 1 · 2018-03-28T09:44:23

$c = 11.98, ˆ A = {53.73}^{\circ}, ˆ B = {21.27}^{\circ}, A_{t} = 21.73 sq units$ $c = 11.98, hat A = 53.73^@, hat B = 21.27^@, A_t = 21.73 " sq units"$

Explanation:

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Given : $a = 10, b = 4.5, ˆ C = 105$ $a = 10, b = 4.5, hat C = 105$

First let's apply cosine law to find side c.

$c = \sqrt{a^{2} + b^{\circ} - 2 a b cos C}$ $c = sqrt(a^2 + b^@ -2 a b cos C)$

$c = \sqrt{10^{2} + {4.5}^{2} - 2 \cdot 10 \cdot 4.5 \cdot cos (105) \approx 11.98}$ $c = sqrt(10^2 + 4.5^2 - 2 * 10 * 4.5 * cos (105) ~~ 11.98$

Now let's apply sine law to find the remaining two angles.

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

$ˆ A = {sin}^{- 1} (\frac{10 \cdot sin 105}{11.98}) = {53.73}^{\circ}$ $hat A = sin ^-1( (10 * sin 105) / 11.98 )= 53.73^@$

$ˆ B = {sin}^{_} 1 (\frac{4.5 \cdot sin 105}{1} `, 98) = {21.27}^{\circ}$ $hat B = sin ^_1 ((4.5 * sin 105)/1`,98) = 21.27^@$

$Area of triangle A_{t} = (\frac{1}{2}) a b sin C = (\frac{1}{2}) \cdot 10 \cdot 4.5 \cdot sin 105 = 21.73 sq units$ $"Area of triangle " A_t = (1/2) a b sin C = (1/2) * 10 * 4.5 * sin 105 = 21.73 " sq units"$

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How do you use the law of cosines or law of sines if you are given C = 105 degrees, a = 10, b = 4.5?

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