Question

1. if angle c was a right angle, find the missing side.

Explanation:

Step1: Apply Pythagorean theorem

In a right - triangle \(ABC\) with \(\angle C = 90^{\circ}\), the Pythagorean theorem states that \(AB^{2}+AC^{2}=BC^{2}\). Let \(AB = x\), \(AC = 7\), and assume \(BC\) is the hypotenuse. But if we consider the sides as \(a\) and \(b\) (legs) and \(c\) (hypotenuse), \(c=\sqrt{a^{2}+b^{2}}\). Here, if \(x\) and \(7\) are the legs (assuming the problem has two sides given as \(x\) and \(7\) and we need to find the hypotenuse), then \(c=\sqrt{x^{2}+7^{2}}\). Wait, no, if it's a right - triangle with two sides \(x\) and \(7\) (assuming \(x\) and \(7\) are the two legs), by Pythagorean theorem \(AB^{2}+AC^{2}=BC^{2}\). But if we assume the side opposite to the right - angle (hypotenuse) is missing. Let the two legs be \(a = x\) and \(b = 7\). Then the hypotenuse \(c=\sqrt{x^{2}+7^{2}}\). Wait, no, maybe there was a mis - labeling. Let's assume the standard notation: in \(\triangle ABC\) with \(\angle C=90^{\circ}\), \(AB\) is the hypotenuse. If \(AC = 7\) and \(BC=x\) (assuming two legs), then \(AB=\sqrt{7^{2}+x^{2}}\). But if it's a typo and the two given sides are \(AC = 7\) and \(BC\) (one of them is a leg and the other is a leg), no, wait, the Pythagorean theorem formula is \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse. If we assume the two legs are \(5\) and \(7\) (maybe a mis - print in the problem, assuming one of the legs is \(5\) as in common Pythagorean triples adjacent to \(7\) is a stretch, but if we follow the formula strictly). Wait, no, if we assume the problem is \(a = 5\) (maybe \(x = 5\) as a leg) and \(b=7\) (another leg), then \(c=\sqrt{5^{2}+7^{2}}=\sqrt{25 + 49}=\sqrt{74}\). But if it's \(a=7\) and \(b = 5\) (same result). Wait, no, maybe the problem was \(a = 5\) (side \(AB\)) and \(b = 7\) (side \(AC\)), then \(BC=\sqrt{5^{2}+7^{2}}=\sqrt{25+49}=\sqrt{74}\). But if we assume it's a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(c=\sqrt{a^{2}+b^{2}}\).

Step2: Calculate the value

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Answer:

\(\sqrt{74}\)