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Ta có \(\hept{\begin{cases}\widehat{A}-\widehat{B}=22^0\\\widehat{B}-\widehat{C=22^0}\end{cases}}\) (*)
\(\Rightarrow\widehat{A}-\widehat{B}=\widehat{B}-\widehat{C}\)
\(\Leftrightarrow\widehat{A}+\widehat{C}=2\widehat{B}\) (1)
Và \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) (Vì 3 góc của tam giác)
\(\Rightarrow\widehat{A}+\widehat{C}=180^0-\widehat{B}\)(2)
Từ (1) và (2)
\(\Rightarrow2\widehat{B}=180^0-\widehat{B}\)
\(\Leftrightarrow3\widehat{B}=180^0\)
\(\Rightarrow\widehat{B}=\frac{180^0}{3}=60^0\)
Từ (*)
\(\Rightarrow\widehat{A}-\widehat{B}+\widehat{B}-\widehat{C}=22^0-22^0=0^0\)(3)
Từ (1) ;(3) và góc B = 60 độ
\(\hept{\begin{cases}\widehat{A}+\widehat{C}=2\cdot60^0=120^0\\\widehat{A}-\widehat{C}=0^0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\widehat{A}=60^0\\\widehat{C}=60^0\end{cases}}\)
Vậy, \(\widehat{A}=\widehat{B}=\widehat{C}=60^0\)
Vì \(\widehat{A}-\widehat{B}=\widehat{B}-\widehat{C}\) nên \(\widehat{A}-2\widehat{B}+\widehat{C}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{A}-2\widehat{B}+\widehat{C}=0^0\left(1\right)\\\widehat{A}+\widehat{B}+\widehat{C}=180^0\left(2\right)\end{matrix}\right.\)
Trừ \(\left(2\right)\) cho \(\left(1\right)\), ta được \(3\widehat{B}=180^0\Rightarrow\widehat{B}=60^0\)
\(\Rightarrow\widehat{A}+\widehat{C}=120^0\)
Vậy GTLN của \(\widehat{A}\) là \(119^0\) vì \(\widehat{C}>0\)
Bài 1:
\(\widehat{A}\div\widehat{B}\div\widehat{C}=1\div2\div3=\frac{\widehat{A}}{1}=\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}\)
Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\) (Tổng ba góc của một tam giác)
Áp dụng t/d dãy tỉ số bằng nhau, ta có: \(\frac{\widehat{A}}{1}=\frac{\widehat{B}}{2}=\frac{\widehat{C}}{3}=\frac{\widehat{A}+\widehat{B}+\widehat{C}}{1+2+3}=\frac{180^0}{6}=30\)
\(\Rightarrow\widehat{A}=30.1=30^0\)
\(\widehat{B}=30.2=60^0\)
\(\widehat{C}=30.3=90^0\)
Vậy .....
Bài 2:
Gọi số đo các góc của tam giác ABC lần lượt là: a;b;c (\(a;b;c\inℕ^∗\) )
Ta có: \(a-b=18^0\Rightarrow a=18+b\)
\(b-c=18^0\Rightarrow c=b-18\)
Trong tam giác ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
\(\Leftrightarrow a+b+c=180^0\)
\(\Leftrightarrow18+b+b+b-18=180^0\)
\(\Leftrightarrow3b=180^0\Rightarrow b=60\Rightarrow\widehat{B}=60^0\)
\(\Rightarrow\widehat{A}=18^0+\widehat{B}=18^0+60^0=78^0\)
\(\Rightarrow\widehat{C}=180^0-60^0-78^0=42^0\)
Vậy .....
a) ta có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{B}+\widehat{C}=100^0\Leftrightarrow\widehat{B}=100^0-\widehat{C}\)
mà \(\widehat{B}-\widehat{C}=20^0\Leftrightarrow100^0-\widehat{C}-\widehat{C}=20^0\Leftrightarrow\widehat{C}=40^0\)
vậy \(\widehat{B}=100^0-\widehat{C}=60^0\)
b) ta có \(\widehat{B}=3\widehat{C}\)
mà \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{B}+\widehat{C}=110^0\Leftrightarrow4\widehat{C}=110^0\Rightarrow\widehat{C}=27,5^0\)
\(\widehat{B}=3\widehat{C}=27,5^0.3=82,5^0\)
Do ΔABC cân tại B => A = C = \(\dfrac{180^o-80^o}{2}=50^o\)
=> góc BAI = 50o - 10o = 40o
góc BCI = 50o - 30o = 20o
=> \(IBC=\dfrac{1}{3}ABI\Rightarrow IBC=\dfrac{80^o}{3+1}=20^o;ABI=80^o-20^o=60^o\)
\(\Leftrightarrow AIB=180^o-40^o-60^o=80^o\)
Ta có : \(\Delta ABC=\Delta ACB=\Delta BCA\)
\(\Rightarrow AB=AC=BC;BC=CB=CA;AC=AB=AB\)
\(\Rightarrow\Delta ABC\)đều \(\Rightarrow\widehat{A}=\widehat{B}=\widehat{C}=60^o\)
\(\widehat{B}+\widehat{C}=140^0\)
\(\Leftrightarrow4\cdot\widehat{C}=140^0\)
\(\Leftrightarrow\widehat{C}=35^0\)
hay \(\widehat{B}=105^0\)
Vậy: ΔABC tù
\(1,\widehat{A}+\widehat{B}+\widehat{C}=180^0\\ \text{Mà }\widehat{A}=\widehat{B}=\widehat{C}\\ \Rightarrow\widehat{A}=\widehat{B}=\widehat{C}=\dfrac{180^0}{3}=60^0\\ 2,\widehat{A}+\widehat{B}+\widehat{C}=180^0\\ \Rightarrow\widehat{B}+\widehat{C}=180^0-\widehat{A}=110^0\\ \text{Mà }\widehat{B}-\widehat{C}=10^0\\ \Rightarrow\left\{{}\begin{matrix}\widehat{B}=\left(110^0+10^0\right):2=60^0\\\widehat{C}=60^0-10^0=50^0\end{matrix}\right.\)