Law of Cosines Calculator

Solve the triangle, if $$$a = 7$$$, $$$b = 14$$$, $$$C = 60^0$$$.

According to the law of cosines: $$$c^{2} = a^{2} + b^{2} - 2 a b \cos{\left(C \right)}$$$.

In our case, $$$c^{2} = 7^{2} + 14^{2} - \left(2\right)\cdot \left(7\right)\cdot \left(14\right)\cdot \left(\cos{\left(60^0 \right)}\right) = 147$$$.

Thus, $$$c = 7 \sqrt{3}$$$.

According to the law of cosines: $$$a^{2} = b^{2} + c^{2} - 2 b c \cos{\left(A \right)}$$$.

In our case, $$$7^{2} = 14^{2} + \left(7 \sqrt{3}\right)^{2} - \left(2\right)\cdot \left(14\right)\cdot \left(7 \sqrt{3}\right)\cdot \left(\cos{\left(A \right)}\right)$$$.

Thus, $$$\cos{\left(A \right)} = \frac{\sqrt{3}}{2}$$$.

Hence, $$$A = 30^0$$$.

The third angle is $$$B = 180^0 - \left(A + C\right)$$$.

In our case, $$$B = 180^0 - \left(30^0 + 60^0\right) = 90^0$$$.

The area is $$$S = \frac{1}{2} a b \sin{\left(C \right)} = \left(\frac{1}{2}\right)\cdot \left(7\right)\cdot \left(14\right)\cdot \left(\sin{\left(60^0 \right)}\right) = \frac{49 \sqrt{3}}{2}.$$$

The perimeter is $$$P = a + b + c = 7 + 14 + 7 \sqrt{3} = 7 \left(\sqrt{3} + 3\right)$$$.

$$$a = 7$$$A

$$$b = 14$$$A

$$$c = 7 \sqrt{3}\approx 12.124355652982141$$$A

$$$A = 30^0$$$A

$$$B = 90^0$$$A

$$$C = 60^0$$$A

Area: $$$S = \frac{49 \sqrt{3}}{2}\approx 42.435244785437494$$$A.

Perimeter: $$$P = 7 \left(\sqrt{3} + 3\right)\approx 33.124355652982141$$$A.