Let ABC be a right triangle with its right angle at C and side AC=3. Let point D be a point on side BC such that BD=5 and angle BAD=45 degrees.
How long is CD?
My first solution yields a quartic equation:
AD^2 = x^2 + 3^2
AB^2 = (x+5)^2 + 3^2
Then use law of cosines on triangle BAD
simplify to
x^4+10x^3+43x^2+90x-144=0
It works but is messy and gives positive solution x=1
I want to find an even simpler one than this, by my second just gives a quadratic:
Extend AD so that BE is perpendicular.
Triangle AEB is isosceles and right.
Triangles ACD and BED are similar
AD=sqrt(x^2+9)
DE=5a/sqrt(x^2+9)
BE=15/sqrt(x^2+9)
Setting AD+DE=BE quickly gives the quadratic
x^2+5x-6=0
(x-1)(x+6)=0
x=1
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Posted by Jer
on 2024-11-25 16:36:31 |
2 Solutions
