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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.

Feb 23, 2015
 #2
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Feb 23, 2015

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