Question 147592: Suppose that triangle ABC has a right angle at B, that BF is the altitude drawn from B to AC, and that BN is the median drawn from B to AC. Find angles ANB and NBF, given that angle C is 42 degrees.
Answer by orca(409)   (Show Source):
You can put this solution on YOUR website! SOLUTION:
Note that when < C is less than 45 degrees, N must lie between F and C.
In right triangle ABC, < A = 90 - 42 = 48.
In right triangle ABF, < ABF = 90 - 48 = 42.
AS BN is a median, triangle BCN is an isosceles triangle with NB = NC, < CBN = < C = 42
AS BN is a median, triangle ABN is also an isosceles triangle with NA = NB, So < ABN = < A = 48.
Thus in triangle ABN, < ANB = 180 - 48 - 48 = 84
< NBF = < ABN - < ABF = 48 - 42 = 6.
Note:
If in your drawing, F lies between N and C, you will obtain < NBF = -6. The fact that < NBF is negative suggests that the N and F are in wrong order.
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