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Solve right triangle if $$$a = 7$$$, $$$b = 7$$$, $$$C = 90^{\circ}$$$
Your Input
Solve the triangle, if $$$a = 7$$$, $$$b = 7$$$, $$$C = 90^{\circ}$$$.
Solution
According to the Pythagorean theorem: $$$c^{2} = a^{2} + b^{2}$$$.
In our case, $$$c^{2} = 7^{2} + 7^{2} = 98$$$.
Thus, $$$c = 7 \sqrt{2}$$$.
According to the definition of the sine: $$$\sin{\left(A \right)} = \frac{a}{c}$$$.
Thus, $$$\sin{\left(A \right)} = \frac{\sqrt{2}}{2}$$$.
There are two possible cases:
$$$A = 45^{\circ}$$$
The third angle is $$$B = 180^{\circ} - \left(A + C\right)$$$.
In our case, $$$B = 180^{\circ} - \left(45^{\circ} + 90^{\circ}\right) = 45^{\circ}$$$.
The area is $$$S = \frac{1}{2} a b = \left(\frac{1}{2}\right)\cdot \left(7\right)\cdot \left(7\right) = \frac{49}{2}$$$.
The perimeter is $$$P = a + b + c = 7 + 7 + 7 \sqrt{2} = 7 \left(\sqrt{2} + 2\right)$$$.
$$$A = 135^{\circ}$$$
The third angle is $$$B = 180^{\circ} - \left(A + C\right)$$$.
In our case, $$$B = 180^{\circ} - \left(135^{\circ} + 90^{\circ}\right) = -45^{\circ}$$$.
This case is impossible, since the angle is nonpositive.
Answer
$$$a = 7$$$A
$$$b = 7$$$A
$$$c = 7 \sqrt{2}\approx 9.899494936611665$$$A
$$$A = 45^{\circ}$$$A
$$$B = 45^{\circ}$$$A
$$$C = 90^{\circ}$$$A
Area: $$$S = \frac{49}{2} = 24.5$$$A.
Perimeter: $$$P = 7 \left(\sqrt{2} + 2\right)\approx 23.899494936611665$$$A.