Hello everybody!I am the poster of the following quetions.
http://web2.0calc.com/questions/i-get-no-diea
THe anwsers is very helpful for me.I learn more skills from that.
And I also find out the angle C and angle B.
in (1) we got angle A=120 degrees=2$${\mathtt{\pi}}$$/3
B+C=$${\mathtt{\pi}}$$-A=$${\mathtt{\pi}}$$/3$$\Rightarrow$$ B=$${\mathtt{\pi}}$$/3-C
iN question （2） we have sin B+ sin C =1 $$\Rightarrow$$sin ($${\mathtt{\pi}}$$/3-C)+sinC=1
the addition formula for the sine function
change it to $${\mathtt{sinC}} = {\mathtt{2}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{cosC}}$$,and then square both sides of the equation
sin^2C=4-4*cosC+3cos^2C , substitute it into $${{\mathtt{sinC}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{cosC}}}^{{\mathtt{2}}} = {\mathtt{1}}$$
then we have $${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{cosC}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{cosC}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}} = {\mathtt{0}}$$,
0<C<$${\frac{{\mathtt{\pi}}}{{\mathtt{3}}}}$$, C=30 dgrees=$${\frac{{\mathtt{\pi}}}{{\mathtt{6}}}}$$, $${\mathtt{B}} = {\frac{{\mathtt{\pi}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\mathtt{C}} = {\frac{{\mathtt{\pi}}}{{\mathtt{3}}}}{\mathtt{\,-\,}}{\frac{{\mathtt{\pi}}}{{\mathtt{6}}}} = {\frac{{\mathtt{\pi}}}{{\mathtt{6}}}}$$, so B=C. it is a isosceles triangle.
I was showing how I solve the question.I really learned a lot skills from you all.
Like Melody did
$${\mathtt{0}} = {\frac{{\mathtt{SinA}}}{{\mathtt{a}}}}{\mathtt{\,\times\,}}\left[\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{c}}\right){\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{c}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right){\mathtt{\,\times\,}}{\mathtt{c}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}\right]$$,and sinA and a can be 0.
then you get 0=（2b+c)*b+(2c+b)*c-2a^2
i realized that 0 divede any number except 0 equal to 0.
This skills really helpful for me to solve the similar questions in my practices.
i post my question on the forum because i dont really know how to solve the questions in sometimes.
sometimes,i just believe that the question is impossible to find out the soloution,but you guys can.That is amazing.But is sometimes, i get the anwsers ,but I am not sure.So i post it in this forum to check my anwser.
Anyway,this website is very helpful for me!Thanks all!
http://web2.0calc.com/questions/same-image-but-different-question
same poster
in question (1) we got B=$${\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{3}}}}$$, so cos B= -1/2
cosB =(a^2+c^2-b^2)/2ac $$\Rightarrow$$$${\mathtt{cosB}} = {\frac{\left[{\left({\mathtt{a}}{\mathtt{\,\small\textbf+\,}}{\mathtt{c}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{ac}}\right]}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{ac}}\right)}} = {\frac{\left[{{\mathtt{4}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\sqrt{{\mathtt{13}}}}}^{\,{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{ac}}\right]}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{ac}}\right)}} = {\frac{\left({\mathtt{3}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{ac}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{ac}}\right)}} = {\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$
3-2ac=-ac$$\Rightarrow$$ac=3
the area of any triangles (S) ,
so the area of the triangle ABC=1/2*ac*sinB=3/2*sin(2$${\mathtt{\pi}}$$/3)=$${\frac{{\mathtt{3}}{\mathtt{\,\times\,}}{\sqrt{{\mathtt{3}}}}}{{\mathtt{4}}}}$$ units square.
+118687 Are you showing us what you have learned. If so that is really good.
OR
Is there are question here?
I was showing how I solve the question.I really learned a lot skills from you all.
Like Melody did
$${\mathtt{0}} = {\frac{{\mathtt{SinA}}}{{\mathtt{a}}}}{\mathtt{\,\times\,}}\left[\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\small\textbf+\,}}{\mathtt{c}}\right){\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{c}}{\mathtt{\,\small\textbf+\,}}{\mathtt{b}}\right){\mathtt{\,\times\,}}{\mathtt{c}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{a}}}^{{\mathtt{2}}}\right]$$,and sinA and a can be 0.
then you get 0=（2b+c)*b+(2c+b)*c-2a^2
i realized that 0 divede any number except 0 equal to 0.
This skills really helpful for me to solve the similar questions in my practices.
i post my question on the forum because i dont really know how to solve the questions in sometimes.
sometimes,i just believe that the question is impossible to find out the soloution,but you guys can.That is amazing.But is sometimes, i get the anwsers ,but I am not sure.So i post it in this forum to check my anwser.
Anyway,this website is very helpful for me!Thanks all!
+118687 Thank you anon,
That is really nice of you, I am really glad that we can help you.
Why don't you join up properly. It is really easy and you do not have to give any personal information.
You just have to choose a username and a password.
You get your own watchlist which is really helpful and you can access the message centre and best of all we will get to know you. :))
.....I have a username.I used to post question as a menber.But i didnt post question in this forum.so, do you welcome me back to this forum as a menber?
