# Given a right triangle, where one cathetus is barA=11 m, the hypothenuse i

Given a right triangle, where one cathetus is $$\displaystyle\overline{{A}}={11}$$ m, the hypothenuse is $$\displaystyle\overline{{C}}={15}$$ m, and $$\displaystyle\angle{c}={90}^{\circ}$$ . Find all missing sides $$\displaystyle{\left(\overline{{B}}\right)}{\quad\text{and}\quad}\angle{s}{\left(\angle{a}{\quad\text{and}\quad}\angle{b}\right)}$$.

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davonliefI
first, to find $$\displaystyle\overline{{B}}$$, use the Pythagorean Theorem:
$$\displaystyle\overline{{A}}^{{2}}+\overline{{B}}^{{2}}=\overline{{C}}^{{2}}$$
Isolate $$\displaystyle\overline{{B}}^{{2}}$$
$$\displaystyle\overline{{B}}^{{2}}=\overline{{C}}^{{2}}-\overline{{A}}^{{2}}$$
$$\displaystyle\overline{{B}}^{{2}}={15}^{{2}}-{11}^{{2}}$$
$$\displaystyle\overline{{B}}=\sqrt{{104}}$$
$$\displaystyle\overline{{B}}\approx{10.2}$$ m
To find $$\displaystyle\angle{a}$$, using the cosine ratio (assuming $$\displaystyle\overline{{A}}$$ is adj to $$\displaystyle{a}{g}{n}\le{a}$$):
$$\displaystyle{\cos{\angle}}{a}=\frac{\overline{{A}}}{\overline{{C}}}=\frac{{11}}{{15}}$$
$$\displaystyle\angle{a}={\cos{{\left(-{1}\right)}}}{\left(\frac{{11}}{{15}}\right)}$$
$$\displaystyle\approx{43}^{\circ}$$
To find $$\displaystyle\angle{b}$$, subtract both other angles from $$\displaystyle{180}^{\circ}$$ (since the sum of all angles in any triangle is $$\displaystyle{180}^{\circ}$$):
$$\displaystyle\angle{b}={180}^{\circ}-\angle{a}-\angle{c}$$
$$\displaystyle\angle{b}={180}^{\circ}-{43}^{\circ}-{90}^{\circ}$$
$$\displaystyle\angle{b}={47}^{\circ}$$
$$\displaystyle\overline{{B}}\approx{10.2}{m};\angle{a}\approx{43}^{\circ};\angle{b}\approx{47}^{\circ}$$