Solution to puzzle 132: Triangular angle

In ABC, draw AD, where D is the midpoint of BC. If ACB = 30° and ADB = 45°, find ABC.

Extend CB to O, so that AOCO. (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.

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

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

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

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

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

Therefore ABC = 105°.

Remarks

A generalization of the above approach yields: cot OBA = 2 cot ODA − cot OCA.
In the above diagram, we have assumed that ABC is an obtuse angle, so that O lies on an extension of DB. If instead we assume that ABC is acute, so that O lies within line segment DB, we obtain the same value for ABC as above. (We would find that OB is negative, indicating that O lies on the other side of B.) Finally, assuming that ABC 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