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Circle AA, with a radius of 9, has a horizontal chord BC with a length of 10. From point C, a vertical line extends to point D. From point D, a horizontal line extends to point E on the circle's circumference. Line segment \DE has a length of 3. From point E, another vertical line extends to point F on the circle's circumference.

Points B andDD connect to form a line segment, and so do points D and F. What is the angle (in degrees) of ∠BDF?

 

 Jul 1, 2020
 #1
avatar+25574 
+4

Circle A, with a radius of 9, has a horizontal chord BC with a length of 10.
From point C, a vertical line extends to point D.
From point D, a horizontal line extends to point E on the circle's circumference.
Line segment DE has a length of 3.
From point E, another vertical line extends to point F on the circle's circumference.
Points B and D connect to form a line segment, and so do points D and F.
What is the angle (in degrees) of \(\angle BDF\)?

 

\(\begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle AEN:} \\ \hline \dfrac{y^2}{4} + 8^2 &=& 9^2 \\\\ \dfrac{y^2}{4} &=& 9^2- + 8^2 \\\\ \dfrac{y^2}{4} &=& 17 \\\\ \mathbf{y^2} &=& \mathbf{68} \\ \mathbf{y} &=& \mathbf{2\sqrt{17}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle DEF:} \\ \hline v^2 &=& 3^2 + y^2 \\ v^2 &=& 9+68 \\ \mathbf{v^2} &=& \mathbf{77} \\ \mathbf{v} &=& \mathbf{\sqrt{77}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle BMA:} \\ \hline 5^2+z^2 &=& 9^2 \\ z^2 &=& 81-25 \\ z^2 &=& 56 \\ \mathbf{z} &=& \mathbf{2\sqrt{14}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline x + \dfrac{y}{2} &=& z \\\\ x &=& z- \dfrac{y}{2} \\\\ x &=& 2\sqrt{14}- \dfrac{2\sqrt{17}}{2} \\\\ \mathbf{x} &=& \mathbf{2\sqrt{14}- \sqrt{17}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle BCD:} \\ \hline u^2 &=& 10^2+x^2 \\ u^2 &=& 100+ \left( 2\sqrt{14}- \sqrt{17} \right)^2 \\ u^2 &=& 100+ 4*14-4\sqrt{14*17} + 17 \\ \mathbf{u^2} &=& \mathbf{173-4\sqrt{238}} \\ \mathbf{u} &=& \mathbf{\sqrt{173-4\sqrt{238}}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle BPF:} \\ \hline w^2 &=& 13^2 +(x+y)^2 \\ w^2 &=& 13^2 +\left( 2\sqrt{14}- \sqrt{17}+2\sqrt{17} \right)^2 \\ w^2 &=& 13^2 +\left( 2\sqrt{14}+\sqrt{17} \right)^2 \\ w^2 &=& 13^2 +4*14+4\sqrt{14*17} + 17 \\ \mathbf{w^2} &=& \mathbf{242+4\sqrt{238}} \\ \hline \end{array}\)

 

cos-rule:

\(\begin{array}{|rcll|} \hline \mathbf{\text{In }\triangle BDF:} \\ \hline \mathbf{w^2} &=& \mathbf{u^2+v^2-2uv\cos(\theta)} \\\\ \cos(\theta) &=& \dfrac{u^2+v^2-w^2}{2uv} \\\\ \cos(\theta) &=& \dfrac{173-4\sqrt{238}+77-(242+4\sqrt{238})}{2\sqrt{173-4\sqrt{238}}\sqrt{77}} \\\\ \cos(\theta) &=& \dfrac{8-8\sqrt{238}} {2\sqrt{13321-308\sqrt{238}}} \\\\ \cos(\theta) &=& \dfrac{4(1-\sqrt{238})} {\sqrt{13321-308\sqrt{238}}} \\\\ \cos(\theta) &=& \dfrac{-57.7089944822} {92.5710938948} \\\\ \cos(\theta) &=& -0.62340188556 \\ \theta &=& \arccos(-0.62340188556) \\ \mathbf{\theta} &=& \mathbf{128.564985997^\circ} \\ \hline \end{array}\)

 

The angle (in degrees) of \(\angle BDF\) is \(\mathbf{128.564985997^\circ}\)

 

laugh

 Jul 1, 2020
 #2
avatar+1387 
+2

r = 9       BC = 10      DE = 3

 

CI = sqrt( BI2 - BC2 ) = 14.96662955

 

AH = sqrt( AE2 - HE2 ) = 4.123105626

 

EF = AH * 2 = 8.246211251

 

CD = CI/2 - AH = 3.360209149

 

∠BDH = arctan( CD / BC ) = 18.57342528º

 

∠FDI = arctan( DE / EF ) = 19.99156073º

 

∠BDF = ∠BDH + 90º + ∠FDI  = 128.564986º   smiley

 

 Jul 1, 2020

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