D be a point inside triangle ABC.
The following are given:
Angle DAC = 30° and Angle DCA = 40° , AB = BD + DC, BD = AC
Find angle DBC.
Let DC=1
Then by law of sines on triangle ADC
AC=BD=2sin110=2sin70
AD=2sin40
AB=1+2sin70
We can now use law of cosines on triangle ADB
cos(ADB)=(4sin^2(40)-4sin(70)-1)/(8sin(40)sin(70))
[I didn't simplify this but calculator says]
angleADB=130
so angleBDC=120
by law of cosines, then law of sines on triangle BDC
BC^2=4sin(70)^2+2sin(70)+1
sin(DBC)=sqrt(3)/(2sqrt(4sin(70^2+2sin(70)+1)))
[I didn't simplify this but calculator says]
angleDBC=20
Remark: In drawing this on Desmos Geometry
https://www.desmos.com/geometry/knhawfwkeu
I noticed D is the incenter of triangle ABC.
If I had noticed this first, I may have tried proving this fact instead. It may have been simpler.
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Posted by Jer
on 2024-05-29 08:41:55 |
Solution
