The circle inscribed in the triangle ABC touches its sides at points M, K, and P. Find the angles of the triangle ABC if the angles of the triangle MKP are 38 °, 78 °, and 64 °
∠KMP = 38 °
∠MKP = 78 °
∠KPM = 64 °
Consider the triangle AMK.
AM = AK (by the second tangent property)
Therefore, AMK is an isosceles triangle, then, by the property of an isosceles triangle:
∠AMK = ∠AKM
Note that both of these angles span the arc MK, and are therefore equal to half its degree measure (by the property of the angles on the circle).
∠MPK is an angle inscribed in a circle and rests on the same arc, therefore it is equal to half the degree measure of this arc.
It turns out that:
∠AMK = ∠AKM = ∠MPK = 64 °
Applying the theorem on the sum of the angles of a triangle:
180 ° = ∠AMK + ∠AKM + ∠MAK
180 ° = 64 ° + 64 ° + ∠MAK
∠MAK = 52 °
Similarly, for the other two triangles we get:
∠BKP = ∠BPK = ∠PMK = 38 °
∠KBP = 180 ° -38 ° -38 ° = 104 °
∠CPM = ∠CMP = ∠MKP = 78 °
∠PCM = 180 ° -78 ° -78 ° = 24 °
Answer: 52 °, 104 ° and 24 °
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