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

fortdefruitI 2021-10-23 Answered
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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Expert Answer

davonliefI
Answered 2021-10-24 Author has 14012 answers
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 \(\displaystyle{180}^{\circ}\)):
\(\displaystyle\angle{b}={180}^{\circ}-\angle{a}-\angle{c}\)
\(\displaystyle\angle{b}={180}^{\circ}-{43}^{\circ}-{90}^{\circ}\)
\(\displaystyle\angle{b}={47}^{\circ}\)
Answer:
\(\displaystyle\overline{{B}}\approx{10.2}{m};\angle{a}\approx{43}^{\circ};\angle{b}\approx{47}^{\circ}\)
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