The solution escaped me for a bit, then I saw it staring me in the face. My own procedure is not as simple as what the teacher does in the video, so follow the teacher's solution. I'll leave my more cumbersome solution in the comments.
Saturday, August 19, 2023
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We have two triangles that overlap: ABF and ACE. ∠FAC is 2x degrees; ∠ACE is y degrees; ∠FDC is 3y degrees; and ∠BFA is x degrees. ∠BDC is 63º. Solve for x + y.
ReplyDeleteRight away, we know that
3y + 63 = 180, so
3y = 180 - 63 = 117, so
y = 39. So now, we know y.
Inside triangle BCD, we now know the three angles are 63, 39, and 3x. How do we know the third angle is 3x? Because we know ∠BCF equals the sum of ∠CAF and ∠AFB. That's (x + 2x).
63 + 39 + 3x = 180, and
102 + 3x = 180, and
3x = 180 - 102 = 78, so
x = 78/3 = 26
So if y is 39 and x is 26,
39 + 26 = 65º.
QED.
To further explain that 3x thing: there's a rule saying that, for triangles, any exterior angle is equal to the sum of the two opposite interior angles. So if, in this problem, we have a triangle with two interior angles of x and 2x, then the exterior angle opposite those two interior angles must be 3x.
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