How do you solve the triangle given α = 15.6 degrees, b = 10.25, and c = 5.5?

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How do you solve the triangle given α = 15.6 degrees, b = 10.25, and c = 5.5?

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Answer 1 · 2015-11-08T17:18:48

$a = 5.17$ $a=5.17$
$B = {147.7}^{o}$ $B=147.7^o$
$C = {16.67}^{o}$ $C=16.67^o$

Explanation:

This is a side-angle-side triangle so we must use the laws of cosine.
First we'll solve for side a $a^{2} = b^{2} + c^{2} - 2 a c cos A$ $a^2=b^2+c^2-2ac cosA$

$a^{2} = {10.25}^{2} + {5.5}^{2} - (2 \cdot 10.25 \cdot 5.5 \cdot cos 15.6)$ $a^2=10.25^2+5.5^2-(2*10.25*5.5*cos15.6)$ which gives 26.716. Take the square root of that and we have $a = 5.17$ $a=5.17$

From there we can use the laws of sine $\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $sinA/a=sinB/b=sinC/c$ or we can stick with the laws of cosine $cos B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}$ $cosB=(a^2+c^2-b^2)/(2ac)$

I stuck with cosine $cos B = \frac{{5.17}^{2} + {5.5}^{2} - {10.25}^{2}}{2 \cdot 5.17 \cdot 5.5}$ $cosB=(5.17^2+5.5^2-10.25^2)/(2*5.17*5.5)$ which gives us -0.846. Then we take ${cos}^{- 1} (- 0.846) = {147.73}^{o}$ $cos^-1(-0.846)=147.73^o$

We know that triangles have $180^{o}$ $180^o$ so $180 - 15.6 - 147.73 = {12.27}^{o}$ $180-15.6-147.73=12.27^o$

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How do you solve the triangle given α = 15.6 degrees, b = 10.25, and c = 5.5?

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