Point D is lies on side BC of triangle ABC such that CD=2BD. It turned out that angle ADC=60 and angle ABC=45. Find angle BAC.
wlog, set B at the origin, C at(3,0), D at (1,0).
A is along the line y = x, since BA is at a 45 degree angle from horizontal BC
The line y = √3(x - 1) passes through D and A
At point A: y = x = √3(x - 1);
x = √3 / (√3 - 1)
x = (3 + √3)/2
Point A is at ( (3 + √3)/2 , (3 + √3)/2 )
Recall Point C is at (3,0)
Line AC: (y-0)/(x-3) = ((3 + √3)/2) / ((3 + √3)/2 - 3)
y = (x-3) * ( (√3 + 3)/(√3 - 3) )
y = (x-3) * ((12+6√3)/(-6))
y = -(2+√3)x + (6+3√3)
The slope of BC is 0
The slope of AB is 1
The slope of AC is -(2+√3)
The angle at B is arctan(1) = 45 degrees
For angle C, the arctan(-(2+√3)) = -75 degrees. Add 180.
This is the same at 105 degrees, but this is the exterior angle.
For the interior angle subtract this from 180 and we're back at 75 degrees
Angle BAC is 180 - 45 - 75 = 60 degrees
Diagram in Desmos Geometry
https://www.desmos.com/geometry/lpne6xurci
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Posted by Larry
on 2023-12-09 16:50:55 |
Cartesian solution
