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J
4
NV
4
HP
27 tháng 3 2021
1.
\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{1}{2}\Rightarrow\widehat{A}=60^o\)
\(S=\dfrac{1}{2}bc.sinA=\dfrac{1}{2}.8.5.sin60^o=10\sqrt{3}\)
\(S=\dfrac{1}{2}a.h_a=\dfrac{1}{2}.7.h_a=10\sqrt{3}\Rightarrow h_a=\dfrac{20\sqrt{3}}{7}\)
\(2R=\dfrac{a}{sinA}=\dfrac{7}{\dfrac{\sqrt{3}}{2}}=\dfrac{14\sqrt{3}}{3}\Rightarrow R=\dfrac{7\sqrt{3}}{3}\)
\(S=pr=\dfrac{a+b+c}{2}.r=10r=10\sqrt{3}\Rightarrow r=\sqrt{3}\)
\(m_a^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{129}{4}\Rightarrow m_a=\dfrac{\sqrt{129}}{2}\)
HP
27 tháng 3 2021
6.
a, Công thức trung tuyến:
\(AM^2=c^2=\dfrac{b^2+c^2}{2}-\dfrac{a^2}{4}=\dfrac{2b^2+2c^2-a^2}{4}\Rightarrow a^2=2\left(b^2-c^2\right)\)
b, \(a^2=2\left(b^2-c^2\right)\Rightarrow\dfrac{2\left(b^2-c^2\right)}{a^2}=1\)
\(\Leftrightarrow2\left(\dfrac{b^2}{a^2}-\dfrac{c^2}{a^2}\right)=1\)
\(\Leftrightarrow2\left(\dfrac{b^2}{a^2}.sin^2A-\dfrac{c^2}{a^2}.sin^2A\right)=sin^2A\)
\(\Leftrightarrow2\left(sin^2B-sin^2C\right)=sin^2A\)
Hay \(sin^2A=2\left(sin^2B-sin^2C\right)\)
4 tháng 5 2022
did - go
went - didn't enjoy
take
takes
goes
playing
am wanting
used to play - was
did - ever heard
will - travel
will be taking
will spend
haven't seen
7 tháng 5 2022
she is interested in reading comic
Tomorrow in this time I wil be lieing on the beach
My car isn't as expensive as his
She has started live in Paris since 2005
A new dam will build near the rive next year
9 tháng 5 2022
1. used to
2. will be taking
3. am watching
4. will go
5. will be washing
6. will be replaced
7. do
8. has visited
9. haven't bought
10. swim
9 tháng 5 2022
1. used to
2. will be taking
3. am watching
4. will go
5. will be washing
6. will be replaced
7. do
8. has visited
9. haven't bought
10. swim
9 tháng 5 2022
1. used to
2. will be taking
3. am watching
4. will go
5. will be washing
6. will be replaced
7. do
8. has visited
9. haven't bought
10. swim
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