https://web2.0calc.com/questions/help-1_39

https://web2.0calc.com/questions/help-2_42

https://web2.0calc.com/questions/help-3_37

https://web2.0calc.com/questions/help-5_28

qwertyzz Jul 15, 2020

#1**+2 **

**https://web2.0calc.com/questions/help-1_39**

\(\begin{array}{|rcll|} \hline \cos(60^\circ) &=& \dfrac{CB}{12} \\ CB &=& 12*\cos(60^\circ) \quad | \quad \cos(60^\circ) = \dfrac12 \\ CB &=& 12*\dfrac12 \\ \mathbf{CB} &=& \mathbf{6} \\ \hline \tan{30^\circ} &=& \dfrac{XC}{CB} \\\\ \tan{30^\circ} &=& \dfrac{XC}{6} \\ XC &=& 6*\tan{30^\circ} \quad | \quad \tan{30^\circ} = \dfrac{\sin(30^\circ)} {\cos(30^\circ)}= \dfrac{\dfrac{1}{2}} {\dfrac{\sqrt{3}} {2} } = \dfrac{1}{\sqrt{3}} \\ XC &=& 6*\dfrac{1}{\sqrt{3}} \\ XC &=& 6*\dfrac{1}{\sqrt{3}}*\dfrac{\sqrt{3}}{\sqrt{3}} \\ \mathbf{XC} &=& \mathbf{2* \sqrt{3}} \\ \hline 2*\text{area}_{BXA} &=& 12*h \quad | \quad h = XC \\ 2*\text{area}_{BXA} &=& 12*XC \quad | \quad \mathbf{XC=2* \sqrt{3}} \\ 2*\text{area}_{BXA} &=& 12*2* \sqrt{3} \\ \mathbf{\text{area}_{BXA}} &=& \mathbf{12*\sqrt{3}} \\ \hline \end{array}\)

heureka Jul 15, 2020

#2**+3 **

**https://web2.0calc.com/questions/help-2_42**

\(\begin{array}{|rcll|} \hline 3CD &=& 12 \\ CD &=& \dfrac{12}{3} \\ \mathbf{CD} &=& \mathbf{4} \\ \hline AD &=& 3CD \\ AD &=& 3*4 \\ \mathbf{AD} &=& \mathbf{12} \\ \hline AC &=& AD+CD \\ AC &=& 12 + 4 \\ \mathbf{AC} &=& \mathbf{16} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline AB^2 + BC^2 &=& AC^2 \quad | \quad \mathbf{AC=16} \\ AB^2 + BC^2 &=& 16^2 \quad | \quad AB^2 = 12^2+BD^2 \\ 12^2+BD^2 + BC^2 &=& 16^2 \quad | \quad BC^2 = 4^2+BD^2 \\ 12^2+BD^2 + 4^2+BD^2 &=& 16^2 \\ 2BD^2 &=& 16^2 -12^2-4^2 \\ 2BD^2 &=& 256-144-16 \\ 2BD^2 &=& 96 \\ BD^2 &=& 48 \\ BD^2 &=& 16*3 \\ \mathbf{BD} &=& \mathbf{4\sqrt{3}} \\ \hline \end{array}\)

heureka Jul 15, 2020

#3**+2 **

**https://web2.0calc.com/questions/help-3_37**

\(\begin{array}{|rcll|} \hline 2*\text{area}_{PQR} &=& 12*12*\sin(120^\circ) \quad | \quad \sin(120^\circ)= \sin(60^\circ)=\dfrac{\sqrt{3}} {2} \\ 2*\text{area}_{PQR} &=& 12*12*\dfrac{\sqrt{3}} {2} \\ 2*\text{area}_{PQR} &=& 12*6* \sqrt{3} \\ \text{area}_{PQR} &=& 6*6* \sqrt{3} \\ \mathbf{\text{area}_{PQR}} &=& 36* \mathbf{\sqrt{3}} \\ \hline \end{array}\)

heureka Jul 15, 2020

#4**+3 **

**https://web2.0calc.com/questions/help-5_28**

\(\begin{array}{|rcll|} \hline \tan(A) &=& \dfrac{3}{2} \\\\ \tan(A) &=& \dfrac{x}{p} \\\\ \hline \dfrac{3}{2} &=& \dfrac{x}{p} \\ \dfrac{2}{3} &=& \dfrac{p}{x} \\ p &=& \dfrac{2}{3}*x \\ \hline \end{array} \begin{array}{|rcll|} \hline \tan(C) &=& \dfrac{2}{3} \\\\ \tan(C) &=& \dfrac{x}{q} \\\\ \hline \dfrac{2}{3} &=& \dfrac{x}{q} \\ \dfrac{3}{2} &=& \dfrac{q}{x} \\ q &=& \dfrac{3}{2}*x \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline p + x + q &=& AC \\\\ \dfrac{2}{3}*x + x+ \dfrac{3}{2}*x &=& AC \quad | \quad AC = \sqrt{13} \\\\ \dfrac{2}{3}*x + x+ \dfrac{3}{2}*x &=& \sqrt{13} \\\\ x \left( \dfrac{2}{3} +1+ \dfrac{3}{2} \right) &=& \sqrt{13} \\\\ x \left( \dfrac{4}{6} +\dfrac{6}{6}+ \dfrac{9}{6} \right) &=& \sqrt{13} \\\\ x * \dfrac{19}{6} &=& \sqrt{13} \\ \mathbf{x} &=& \mathbf{\dfrac{6}{19}* \sqrt{13}} \\ \hline \end{array}\)

The side length of the square is \(\mathbf{\dfrac{6}{19}* \sqrt{13}}\)

heureka Jul 15, 2020

#5**+2 **

qwertyzz, it's okay if you need help but dumping 4 questions for someone to do in a post is a bit too much. 1-2 is fine. From here it's clear that these are pretty much homework problems you need to submit before a due date.

gwenspooner85 Jul 15, 2020

#7**+1 **

I did! I have a lot of homework and over the past week these are the ones I needed help with. I posted them all in the same post because I couldn't figure them out and the only responses I got was the guest giving me rong answers.

qwertyzz
Jul 15, 2020