1. \text{If} \; \sum_{x=\pi-^{10}C_7}^{x= \pi + ^{10}C_r} \; \sin(x^o) = 0 , then value of r is ?
2. Side lengths of the triangle are 3 consecutive integers, and one of the angles in twice that of other, then no. of such triangle is :
3. In a triangle ABC, if BC =1 , sin (A/2) = x1 , sin (B/2)=x2 , cos (A/2) = y1 , cos(B/2)=y2 and (x1/x2)2006 = (y1/y2)2007 , then the length of AC is :
1057for the 2nd question....
using sine rule....
ratio of 2 sides of the triangle shud be sinAsin2A=12cosA
now since the sides are integers....cosA can only be 1(it cant be 1/2...since then the ratio of the sides will be one....which is not possible)
so now...
12=aa+1 or a+1a+2(this one has no solution) or aa+2 ....
we get two triangles...
first one of sides 1,2 and 3... and second of 2,3 and 4....
1708\hspace{-16}$Here $\bf{\sum_{x=\pi-\binom{10}{7}}^{x=\pi+\binom{10}{r}}\sin (x)=0}$\\\\\\ Now $\bf{\binom{10}{7}=\binom{10}{3}\;\;,\binom{10}{8}=\binom{10}{2}\;\;,\binom{10}{9}=\binom{10}{1}\;\;,\binom{10}{10}=\binom{10}{0}}$\\\\\\ So $\bf{\sum_{x=\pi-\binom{10}{3}}^{x=\pi+\binom{10}{r}}\sin (x)=0}$\\\\\\ $\bf{\sin \left(\pi-\binom{10}{3}\right)+\sin \left(\pi-\binom{10}{2}\right)+\sin \left(\pi-\binom{10}{1}\right)+\sin \left(\pi-\binom{10}{0}\right)+\sin \left(\pi+\binom{10}{0}\right)+.....+\sin \left(\pi+\binom{10}{r}\right)=0}$\\\\\\\\ So $\bf{\sin \left(\binom{10}{3}\right)+\sin \left(\binom{10}{2}\right)+\sin \left(\binom{10}{1}\right)+\sin \left(\binom{10}{0}\right)-\sin \left(\binom{10}{0}\right)-\sin \left(\binom{10}{1}\right)-\sin \left(\binom{10}{2}\right)-\sin \left(\binom{10}{3}\right)=0}$\\\\\\ So $\bf{r=3}$.
1708\hspace{-16}$Here $\bf{A+B+C=\pi\Leftrightarrow \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}}$\\\\ Now \underline{\underline{\bf{Using Sin Rule}}}::\\\\\\ $\bf{\frac{BC}{\sin (A)}=\frac{AC}{\sin (B)}\Leftrightarrow AC=\frac{\sin (B)}{\sin (A)}=\frac{2.\sin \left(\frac{B}{2}\right).\cos \left(\frac{B}{2}\right)}{2.\sin \left(\frac{A}{2}\right).\cos \left(\frac{A}{2}\right)}}$\\\\\\ So $\bf{AC=\left(\frac{x_{2}}{x_{1}}\right).\left(\frac{y_{2}}{y_{1}}\right)=\left(\frac{y_{2}}{y_{1}}\right)^{\frac{2007}{2006}+1}=\left(\frac{y_{2}}{y_{1}}\right)^{2+\frac{1}{2006}}=\left(\frac{x_{2}}{x_{1}}\right)^{2-\frac{1}{2007}}}$\\\\\\ Now Here $\bf{\frac{A}{2}\;\;,\frac{B}{2}<\frac{\pi}{2}}$, Then $\bf{x_{1}\;\;,x_{2}>0\;\;,y_{1}\;\;,y_{2}>0}$\\\\\\ If $\bf{\frac{x_{1}}{x_{2}}>1}\;,$ Then $\bf{\frac{y_{1}}{y_{2}}<1}\;,$\\\\\\ Similarly If $\bf{\frac{x_{1}}{x_{2}}<1}\;,$ Then $\bf{\frac{y_{1}}{y_{2}}>1}\;,$\\\\\\ So We can say that $\bf{}$
