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

Given a right triangle, where one cathetus is $\stackrel{―}{A}=11$ m, the hypothenuse is $\stackrel{―}{C}=15$ m, and $\mathrm{\angle }c={90}^{\circ }$ . Find all missing sides $\left(\stackrel{―}{B}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\angle }s\left(\mathrm{\angle }a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\angle }b\right)$.
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davonliefI
first, to find $\stackrel{―}{B}$, use the Pythagorean Theorem:
${\stackrel{―}{A}}^{2}+{\stackrel{―}{B}}^{2}={\stackrel{―}{C}}^{2}$
Isolate ${\stackrel{―}{B}}^{2}$
${\stackrel{―}{B}}^{2}={\stackrel{―}{C}}^{2}-{\stackrel{―}{A}}^{2}$
${\stackrel{―}{B}}^{2}={15}^{2}-{11}^{2}$
$\stackrel{―}{B}=\sqrt{104}$
$\stackrel{―}{B}\approx 10.2$ m
To find $\mathrm{\angle }a$, using the cosine ratio (assuming $\stackrel{―}{A}$ is adj to $agn\le a$):
$\mathrm{cos}\mathrm{\angle }a=\frac{\stackrel{―}{A}}{\stackrel{―}{C}}=\frac{11}{15}$
$\mathrm{\angle }a=\mathrm{cos}\left(-1\right)\left(\frac{11}{15}\right)$
$\approx {43}^{\circ }$
To find $\mathrm{\angle }b$, subtract both other angles from ${180}^{\circ }$ (since the sum of all angles in any triangle is ${180}^{\circ }$):
$\mathrm{\angle }b={180}^{\circ }-\mathrm{\angle }a-\mathrm{\angle }c$
$\mathrm{\angle }b={180}^{\circ }-{43}^{\circ }-{90}^{\circ }$
$\mathrm{\angle }b={47}^{\circ }$
$\stackrel{―}{B}\approx 10.2m;\mathrm{\angle }a\approx {43}^{\circ };\mathrm{\angle }b\approx {47}^{\circ }$
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