Projects And Homeworks of a school Student

In a ∆ABC, CA=CB. D and E are points on AB such that AD=DE=EB. If C=120° find `\angle CED` First the figure is CA=CB and AD=DE=EB: Now, let's construct a line segment CP perpendicular to AB to C: Now, by the property of isoscleles triangle, its altitude is its median. So CP divided angle C in two halfs and also bisects AB. So AP=PB Now, in `\triangle CDP` and `\triangle CEP`: CP=CP (given) `\angle CPD `=`\angle CPE`=90° DP=PE (as AP=PB and AD=EB. So AP-AD=AD-EB hence DP=PE) `\therefore  \triangle CDP \cong \triangle CEP`...........(i) But DP+PE=DE Now let AD=DE=EB=2x Then DP=PE=x Also, `\angle PCA =\angle PCB =60`° `also ~ \angle PCD=\angle PCE=\alpha`......from(i) Now, let's focus on `\triangle PCB` as follows: Here, `tanC=\frac{PB}{PC}` `\rightarrow tan60`°=`\frac{3x}{PC}` `\rightarrow \frac{3x}{\sqrt{3}}=PC` Now in `\triangle PCE=` `\rightarrow tan \alpha =\frac{PE}{PC} ` `\rightarrow  tan \alpha =\frac{x}{\frac{3x}{\sqrt{3}}}` `\rightarrow tan \alpha =\frac{\sqrt{3} x}{3x}...

Euler's continued Fraction Formula: If `r_i` are complex numbers and `x` is defined by: ` x=1+\sum_{i=1}^{\infty}r_{1}r_{2}\cdot r_(i)=1+\sum_{i=1}^\infty  (\prod_{j=1}^{i}r_{j})` Then this equality can be proved by induction:  `x=\frac{1}{1-\frac{r_1}{1+r_1-\frac{r_2}{1+r_2-\frac{r_3}{1+r_3-\ddots}}}}` Its prove will be posted later...

  Find   x , if \displaystyle \begin{array}{|l}4^{\frac{x}{y}+\frac{y}{x}}=32\\ \log_3(x-y)+\log_3(x+y)=1\end{array} 4 y x ​ + x y ​ = 32 lo g 3 ​ ( x − y ) + lo g 3 ​ ( x + y ) = 1 ​ Solution: Checking if the system is defined for two variables is a hard task, so we shall find the eventual solutions to the system and check directly if the system is defined for them. We shall only write  \displaystyle \begin{array}{|l}x+y>0\\x-y>0\end{array} x + y > 0 x − y > 0 ​  for now. \displaystyle \begin{array}{|l}\frac{x}{y}+\frac{y}{x}=log_432\\log_3(x^2-y^2)=1\end{array} y x ​ + x y ​ = l o g 4 ​ 32 l o g 3 ​ ( x 2 − y 2 ) = 1 ​ \displaystyle \begin{array}{|l}\frac{x}{y}+\frac{y}{x}=\frac{1}{2}log_232\\x^2-y^2=3\end{array} y x ​ + x y ​ = 2 1 ​ l o g 2 ​ 32 x 2 − y 2 = 3 ​ \displaystyle \begin{array}{|l}\frac{x^2+y^2}{xy}=\frac{5}{2}\\x^2-y^2=3\end{array} x y x 2 + y 2 ​ = 2 5 ​ x 2 − y 2 = 3 ​ \displaystyle \begin{array}{|l}x^2+y^2=\frac{5}{2}xy\\x^2-y^2=3\end{a...