Let \(\triangle MBT\) be a triangle with MB = 4 and MT = 7. Furthermore, let circle ω be a circle with center O which is tangent to MB at B and MT at some point on segment MT. Given OM = 6 and ω intersect BT at I \(\neq\) B, find the length of TI.
+26367 Let \(\triangle MBT\) be a triangle with MB = 4 and MT = 7.
Furthermore, let circle \(\omega\) be a circle with center O which is tangent to MB at B and MT at some point on segment MT.
Given OM = 6 and \(\omega\) intersect BT at \(I \ne B\), find the length of \(TI\).
\(\text{Let $TI = \color{red}x$}\)
\(\begin{array}{|rcll|} \hline \mathbf{r=\ ?} \\ \hline 4^2+r^2 &=& 6^2 \\ r^2 &=& 6^2 -4^2 \\ r^2 &=& 36-16 \\ r^2 &=& 20 \\ r^2 &=& 4*5 \\ \mathbf{r} &=& \mathbf{2\sqrt{5}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \cos(A) &=& \dfrac{4}{6} \\\\ \cos(A) &=& \dfrac{2}{3} \\\\ \mathbf{\cos^2(A)} &=& \mathbf{\dfrac{4}{9}} \\\\ \cos(2A) &=& \cos^2(A)-\sin^2(a) \\ \cos(2A) &=& \cos^2(A)-\left(1- cos^2(A) \right) \\ \mathbf{\cos(2A)} &=& \mathbf{2\cos^2(A)-1} \\\\ \cos(2A) &=& 2*\dfrac{4}{9}-1 \\\\ \cos(2A) &=& \dfrac{8}{9}-1 \\\\ \cos(2A) &=& \dfrac{8}{9}-\dfrac{9}{9} \\\\ \mathbf{\cos(2A)} &=& \mathbf{-\dfrac{1}{9}} \\ \hline \end{array}\)
cos-rule:
\(\begin{array}{|rcll|} \hline BT^2 &=& 4^2+7^2-2*4*7*\cos(2A) \\ BT^2 &=& 16+49-56\cos(2A) \quad | \quad \mathbf{\cos(2A)=-\dfrac{1}{9}} \\ BT^2 &=& 16+49+56*\dfrac{1}{9} \\ BT^2 &=& 65+\dfrac{56}{9} \\ BT^2 &=& \dfrac{9*65+56}{9} \\ BT^2 &=& \dfrac{641}{9} \\ \mathbf{BT} &=& \mathbf{\dfrac{\sqrt{641}}{3}} \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \mathbf{x=\ ?} \\ \hline 3^2 &=& x(x+BI) \quad | \quad BI = BT -x \\ 9 &=& x(x+BT -x) \\ 9 &=& x*BT \\ x &=& \dfrac{9}{BT} \quad | \quad \mathbf{BT=\dfrac{\sqrt{641}}{3}} \\\\ x &=& \dfrac{9}{\dfrac{\sqrt{641}}{3}} \\\\ x &=& \dfrac{3*9}{ \sqrt{641} } \\\\ x &=& \dfrac{27}{ \sqrt{641} } \\\\ x &=& \dfrac{27}{ 25.3179778023 } \\\\ \mathbf{x} &=& \mathbf{1.06643588247} \\ \hline \end{array}\)
The length of TI is \(\approx \mathbf{1}\)
