How do you use the law of cosines or law of sines if you are given a = 4, b = 3 and c = 6?

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How do you use the law of cosines or law of sines if you are given a = 4, b = 3 and c = 6?

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Answer 1 · 2015-06-20T21:25:44

You can use the Law of Cosines to calculate the angles of the triangle.

$A \approx 36.3$ $A~~36.3$ degrees
$B \approx 117.3$ $B~~117.3$ degrees
$C \approx 26.4$ $C~~26.4$ degrees

Explanation:

In this case the only thing you need to solve this triangle is to calculate the angles.

To calculate angle $A$ $A$ you use:

$a^{2} = b^{2} + c^{2} - 2 \cdot b \cdot c \cdot cos A$ $a^2=b^2+c^2-2*b*c*cosA$
so you calculate:

$4^{2} = 3^{2} + 6^{2} - 2 \cdot 3 \cdot 6 \cdot cos A$ $4^2=3^2+6^2-2*3*6*cosA$
$16 = 9 + 36 - 36 cos A$ $16=9+36-36cosA$
$36 cos A = 9 + 36 - 16$ $36cosA=9+36-16$
$36 cos A = 29$ $36cosA=29$
$cos A = \frac{29}{36}$ $cosA=29/36$
$A \approx 36.3$ $A~~36.3$ degrees

The same way you can calculate C

$c^{2} = a^{2} + b^{2} - 2 \cdot a \cdot b$ $c^2=a^2+b^2-2*a*b$
$6^{2} = 3^{2} + 4^{2} - 2 \cdot 3 \cdot 4 \cdot cos C$ $6^2=3^2+4^2-2*3*4*cosC$
$36 = 9 + 16 - 24 cos C$ $36=9+16-24cosC$
$36 - 9 - 16 = - 24 cos C$ $36-9-16=-24cosC$
$11 = - 24 cos C$ $11=-24cosC$
$cos C = - \frac{11}{24}$ $cosC=-11/24$
$C \approx 117.3$ $C~~117.3$ degrees.

Now you can calculate B either using the Law of Cosines again or using the fact, that the sum of angles in triangle equals to 180 degrees.

$A + B + C = 180$ $A+B+C=180$
$36.3 + B + 117.3 = 180$ $36.3+B+117.3=180$
$B = 180 - 117.3 - 36.3$ $B=180-117.3-36.3$
$B \approx 26.4$ $B~~26.4$

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How do you use the law of cosines or law of sines if you are given a = 4, b = 3 and c = 6?

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