Toán Lớp 9: Tam giác ABC vuông tại A, biết:
a, b=10cm, góc C =30°
b, b=28cm,C=21cm
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Toán Lớp 9: Tam giác ABC vuông tại A, biết:
a, b=10cm, góc C =30°
b, b=28cm,C=21cm
Comments ( 1 )
\Delta ABC:\widehat A = {90^0}\\
a)\widehat B + \widehat C = {90^0}\\
\Leftrightarrow \widehat B = {60^0}\\
+ \sin \widehat B = \dfrac{b}{a}\\
\Leftrightarrow \sin {60^0} = \dfrac{{10}}{a} = \dfrac{{\sqrt 3 }}{2}\\
\Leftrightarrow a = \dfrac{{20\sqrt 3 }}{3}\left( {cm} \right)\\
Theo\,Pytago:\\
{b^2} + {c^2} = {a^2}\\
\Leftrightarrow c = \sqrt {{a^2} – {b^2}} = \dfrac{{10\sqrt 3 }}{3}\left( {cm} \right)\\
Vậy\,\widehat B = {60^0};a = \dfrac{{20\sqrt 3 }}{3}cm;c = \dfrac{{10\sqrt 3 }}{3}cm\\
b)\\
{a^2} = {b^2} + {c^2} = {28^2} + {21^2}\\
\Leftrightarrow a = 35\left( {cm} \right)\\
\sin \widehat B = \dfrac{b}{a} = \dfrac{{28}}{{35}} = \dfrac{4}{5}\\
\Leftrightarrow \widehat B = {53^0}\\
\Leftrightarrow \widehat C = {37^0}\\
Vậy\,a = 35cm;\widehat B = {53^0};\widehat C = {37^0}
\end{array}$