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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+26398 
+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 BDF?

 

In AEN:y24+82=92y24=92+82y24=17y2=68y=217In DEF:v2=32+y2v2=9+68v2=77v=77In BMA:52+z2=92z2=8125z2=56z=214

 

x+y2=zx=zy2x=2142172x=21417In BCD:u2=102+x2u2=100+(21417)2u2=100+41441417+17u2=1734238u=1734238

 

In BPF:w2=132+(x+y)2w2=132+(21417+217)2w2=132+(214+17)2w2=132+414+41417+17w2=242+4238

 

cos-rule:

In BDF:w2=u2+v22uvcos(θ)cos(θ)=u2+v2w22uvcos(θ)=1734238+77(242+4238)2173423877cos(θ)=88238213321308238cos(θ)=4(1238)13321308238cos(θ)=57.708994482292.5710938948cos(θ)=0.62340188556θ=arccos(0.62340188556)θ=128.564985997

 

The angle (in degrees) of BDF is 128.564985997

 

laugh

 Jul 1, 2020
 #2
avatar+1490 
+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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