Question

Bài 1:

a) \(\Delta ABC\) có \(\widehat{A}\)= 100^o ; \(\widehat{B}\) - \(\widehat{C}\) = 50^o. Tính \(\widehat{B}\), \(\widehat{C}\)

b) \(\Delta ABC\) có \(\widehat{B}\) = 80^o và \(3.\widehat{A}\) = \(2.\widehat{C}\) . Tính \(\widehat{A}\), \(\widehat{C}\)

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Answer 1

a) ΔABC có:

\(\widehat{A}\) + \(\widehat{B}\) + \(\widehat{C}\) = 180^o hay 100^o + \(\widehat{B}\) + \(\widehat{C}\) = 180^o

\(\Rightarrow\) \(\widehat{B}\) + \(\widehat{C}\) = 180^o - 100^o = 80^o

Ta có: \(\widehat{B}\) + \(\widehat{C}\) = 80^o(cm trên) ; \(\widehat{B}\) - \(\widehat{C}\) = 50^o (gt)

\(\Rightarrow\) \(\widehat{B}\) = (80^o + 50^o ) : 2 = 65^o

\(\widehat{C}\) = (80^o - 50^o) : 2 = 15^o

b) ΔABC có:

\(\widehat{B}\) + \(\widehat{A}\) + \(\widehat{C}\) = 180^o hay 80^o + \(\widehat{A}\) + \(\widehat{C}\) = 180^o

\(\Rightarrow\) \(\widehat{A}\) + \(\widehat{C}\) = 180^o - 80^o = 100^o

Ta có: 3 . \(\widehat{A}\) = 2 . \(\widehat{C}\) => \(\frac{\widehat{A}}{2}\) = \(\frac{\widehat{C}}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{\widehat{A}}{2}\) = \(\frac{\widehat{C}}{3}\) = \(\frac{\widehat{A}+\widehat{C}}{2+3}\) = \(\frac{100}{5}\) = 20

\(\Rightarrow\) \(\begin{cases}\widehat{A}=40^o\\\widehat{C}=60^o\end{cases}\)