Cho ∆ABC biết A=90° và tính a)a=15cm,b=10cm b)b=12cm,c=7cm c),a=18cm,góc B=60°

Đáp án:

$\begin{array}{l}
a)Theo\,Pytago:\\
{a^2} = {b^2} + {c^2}\\
 \Rightarrow {c^2} = {15^2} – {10^2} = 125\\
 \Rightarrow c = 5\sqrt 5 \left( {cm} \right)\\
\sin B = \dfrac{b}{a} = \dfrac{{10}}{{15}} = \dfrac{2}{3}\\
 \Rightarrow \widehat B = {41^0} \Rightarrow \widehat C = {49^0}\\
b)b = 12;c = 7\\
 \Rightarrow {a^2} = {12^2} + {7^2} = 144 + 49 = 193\\
 \Rightarrow a = \sqrt {193} \left( {cm} \right)\\
 \Rightarrow \sin \widehat B = \dfrac{b}{a} = \dfrac{{12}}{{\sqrt {193} }}\\
 \Rightarrow \widehat B = {30^0} \Rightarrow \widehat C = {60^0}\\
c)a = 18;\widehat B = {60^0}\\
 \Rightarrow \widehat C = {30^0}\\
 \Rightarrow c = \dfrac{1}{2}a = 9\left( {cm} \right)\\
 \Rightarrow b = 9\sqrt 3 \left( {cm} \right)
\end{array}$