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Solution to puzzle 132: Triangular angle

Skip restatement of puzzle.In triangleABC, draw AD, where D is the midpoint of BC.  If angleACB = 30° and angleADB = 45°, find angleABC.


Extend CB to O, so that AOperpendicular toCO.  (See remark 2, below.)
Without loss of generality, let BD = DC = 1.
Let AO = h.
Then, since 1 = cot 45° = OD/h, we obtain OB = h − 1.

Triangle ABC, with median AD, and angles and lengths as described above.

root 3 = cot 30° = (h + 1)/h = 1 + (1/h).
Hence 1/h = root 3 − 1.
From here, we could note that cot OBA = (h − 1)/h = 1 − (1/h) = 2 − root 3.

This essentially solves the problem, as angleOBA = cot−1(2 − root 3), and then angleABC = 180° − angleOBA, and we have expressed angleABC in terms of known quantities.  However, if we hope to find an exact value in degrees for angleABC, it is not obvious that cot−1(2 − root 3) = 75°.  Rather than work backwards, by calculating cot 75° or tan 75°, we pursue an alternative approach below.

Since 1/h = root 3 − 1, rationalizing the denominator, we obtain h = ½(1 + root 3).

Then AB2 = h2 + (h − 1)2, by Pythagoras' Theorem
 = 2h(h − 1) + 1
 = ½(1 + root 3)(−1 + root 3) + 1
 = 2

Now note that AB:DB = BC:BA = root 2:1.
Since angleABC is contained within triangles ABC and DBA, it follows that these two triangles are similar.
Hence angleBCA = angleDAB = 30°.
Hence angleABD = 180° − (45° + 30°) = 105°.

Therefore angleABC = 105°.


Remarks

  1. A generalization of the above approach yields: cot OBA = 2 cot ODA − cot OCA.
  2. In the above diagram, we have assumed that angleABC is an obtuse angle, so that O lies on an extension of DB.  If instead we assume that angleABC is acute, so that O lies within line segment DB, we obtain the same value for angleABC as above.  (We would find that OB is negative, indicating that O lies on the other side of B.)  Finally, assuming that angleABC is a right angle, so that O coincides with B, gives rise to an inconsistency, indicating that this configuration is impossible for the given angles.

Source: Angle ABC, on flooble :: perplexus

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