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a)
A B C 100*
=> Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}\) = 180o
100o + \(\widehat{B}+\widehat{C}\) = 180o
\(\widehat{B}+\widehat{C}\) = 180o - 100o
\(\widehat{B}+\widehat{C}\) = 80o
Góc B = (80o+50o):2 = 65o
=> \(\widehat{C}\) = 65o - 50o = 15o
Vậy \(\widehat{B}\) = 65o ; \(\widehat{C}\) = 15o
b)
80* A B C
Ta có : \(\widehat{3A}+\widehat{B}+\widehat{2C}\) = 180o
\(\widehat{3A}+\widehat{2C}\) = 180o - 80o
\(\widehat{3A}+\widehat{2C}\) = 100o
=> \(\widehat{A}\) = 100o:(3+2).3 = 60o
\(\widehat{C}\) = 100o - 60o = 40o
Vậy \(\widehat{A}\) = 60o ; \(\widehat{C}\) = 40o
a) góc A = 70o, => B + C = 110o
=> B =(110 + 10) : 2 = 60
C = 60 - 10 = 50
b) góc A = 100 , => B + C = 80
=> B = (80 + 50) : 2 = 65
C = 65 - 50 = 15
c) B = 2C => 180 - 60 = 3C = 120
=> C = 40
=> B = 40 . 2 = 80
a, \(3\widehat{A}=4\widehat{B}\Leftrightarrow\dfrac{3\widehat{A}}{12}=\dfrac{4\widehat{B}}{12}\Rightarrow\dfrac{\widehat{A}}{4}=\dfrac{\widehat{B}}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{\widehat{A}}{4}=\dfrac{\widehat{B}}{3}=\dfrac{\widehat{A}-\widehat{B}}{4-3}=\dfrac{20^0}{1}=20^0\)
+)\(\dfrac{\widehat{A}}{4}=20^0\Rightarrow\widehat{A}=20^0.4=80^0\)
+)\(\dfrac{\widehat{B}}{3}=20^0\Rightarrow\widehat{B}=20^0.3=60^0\)
Xét △ABC có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\\ 80^0+60^0+\widehat{C}=180^0\\ \widehat{C}=180^0-80^0-60^0=40^0\)
Vậy \(\Delta ABC\) có \(\widehat{A}=80^0;\widehat{B}=60^0;\widehat{C}=40^0\)
a) Gọi số đo các góc lần lượt là x,y ( x,y > 0 )
Theo bài ra ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}\) và \(x-y=20^0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{x-y}{4-3}=\dfrac{20^0}{1}=20^0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=20^0\Rightarrow x=80^0\\\dfrac{y}{3}=20^0\Rightarrow x=60^0\end{matrix}\right.\)
Xét \(\Delta ABC\) có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
mà \(\widehat{A}=80^0;\widehat{B}=60^0\)
\(\Rightarrow80^0+60^0+\widehat{C}=180^0\)
\(\Rightarrow140^0+\widehat{C}=180^0\)
\(\Rightarrow\widehat{C}=180^0-140^0\)
\(\Rightarrow\widehat{C}=40^0\)
Vậy ........................
a) ΔABC có:
\(\widehat{A}\) + \(\widehat{B}\) + \(\widehat{C}\) = 180o hay 100o + \(\widehat{B}\) + \(\widehat{C}\) = 180o
\(\Rightarrow\) \(\widehat{B}\) + \(\widehat{C}\) = 180o - 100o = 80o
Ta có: \(\widehat{B}\) + \(\widehat{C}\) = 80o(cm trên) ; \(\widehat{B}\) - \(\widehat{C}\) = 50o (gt)
\(\Rightarrow\) \(\widehat{B}\) = (80o + 50o ) : 2 = 65o
\(\widehat{C}\) = (80o - 50o) : 2 = 15o
b) ΔABC có:
\(\widehat{B}\) + \(\widehat{A}\) + \(\widehat{C}\) = 180o hay 80o + \(\widehat{A}\) + \(\widehat{C}\) = 180o
\(\Rightarrow\) \(\widehat{A}\) + \(\widehat{C}\) = 180o - 80o = 100o
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
\(\Rightarrow\) \(\begin{cases}\widehat{A}=40^o\\\widehat{C}=60^o\end{cases}\)
a) Ta có: \(\)\(\widehat{A}+\widehat{B}+\widehat{C}=180^{\circ}\) (Tổng ba góc trong tam giác)
<=> \(\left.\begin{matrix} \widehat{B}+\widehat{C}=180-\widehat{A}=180^{\circ}-100^{\circ}=80^{\circ} & & \\ \widehat{B}-\widehat{C}=30^{\circ} & & \end{matrix}\right\}\)
=> \(2\widehat{B}=110^{\circ}\)
=> \(\widehat{B}=55^{\circ}\)
=> \(\widehat{C}=25^{\circ}\)
P/s: câu b tương tự
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 \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{B}+\widehat{C}=110^0\Leftrightarrow4\widehat{C}=110^0\Rightarrow\widehat{C}=27,5^0\)
\(\widehat{B}=3\widehat{C}=27,5^0.3=82,5^0\)
b: Vì góc ABC là góc ngoài cua ΔAHB
nên góc ABC=góc AHB+góc HAB=90 độ+góc HAB
Xét ΔHAC vuông tại H có góc HAC+góc ACB=90 độ
=>góc ACB=90 độ-góc HAC
c: 1/2(góc ABC-góc ACB)
=1/2(180 độ-góc ABH-90 độ+góc HAC)
=1/2(90 độ-góc ABH+góc HAC)
=góc DAH
a: \(\widehat{BAC}=180^0-70^0-30^0=80^0\)
=>\(\widehat{CAD}=40^0\)
\(\widehat{ADC}=180^0-40^0-30^0=110^0\)
b: \(\widehat{B}-\widehat{C}=40^0\)
nên \(\widehat{B}=\widehat{C}+40^0\)
Ta có: \(\widehat{ABD}+\widehat{ADB}+\widehat{BAD}=\widehat{ACD}+\widehat{ADC}+\widehat{CAD}\)
\(\Leftrightarrow\widehat{C}+40^0+\widehat{ADB}=\widehat{C}+\widehat{ADC}\)
\(\Leftrightarrow\widehat{ADB}-\widehat{ADC}=-40^0\)
mà \(\widehat{ADB}+\widehat{ADC}=180^0\)
nên \(-2\cdot\widehat{ADC}=\dfrac{-40^0-180^0}{2}=-110^0\)
hay \(\widehat{ADC}=55^0\)
Bài làm
a) Tổng số đo của góc B và C là:
180o - 80o = 100o
Số đo góc B là:
( 100o + 30o ) : 2 = 65o
Số đo góc C là:
100o - 65o = 35o
Vậy góc B = 65o
góc C = 35o
# Học tốt #
Bài làm
Ta có: \(5.\widehat{A}=3.\widehat{B}\)
=> \(\frac{\widehat{A}}{3}=\widehat{\frac{B}{5}}\Rightarrow\frac{7.\widehat{A}}{21}=\frac{4.\widehat{B}}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{7.\widehat{A}}{21}=\frac{4.\widehat{B}}{20}=\frac{7\widehat{A}-4\widehat{B}}{21-20}=\frac{15^0}{1}=15^0\)
Do đó: \(\hept{\begin{cases}\widehat{\frac{A}{3}}=15^0\\\widehat{\frac{B}{5}}=15^0\end{cases}\Rightarrow\hept{\begin{cases}A=45^0\\B=75^0\end{cases}}}\)
Vậy A = 45o , B = 75o
# Học tốt #