# In triangle MPK: PK = 2. Point A is marked on the side MK so that MA = AP = √3, AK = 1. Find the MPK angle.

By the cosine theorem:
PA2 = PK2 + AK2 – 2 * PK * AK * cos PKA;
cos PKA = (PK2 + AK2 – PA2) / 2 * PK * AK = (4 + 1 – 3) / (2 * 2 * 1) = 2/4 = 1/2;
RKA angle = arccos (1/2) = 60 °;
PK2 = PA2 + AK2 – 2 * RA * AK * cos PAK;
cos PAK = (PA2 + AK2 – PK2) / 2 * PA * AK = (3 + 1 – 4) / (2 * √3 * 1) = 0;
angle PAK = arccos 0 = 90 °.
The angles PAK and PAM are adjacent, their sum is 180 °, so the angle PAM = 180 – angle PAK = 180 ° – 90 ° = 90 °.
In a triangle MPA, the sides AM and RA are equal, therefore it is isosceles with the base MP, the angles at the base are equal, hence: angle AMP = angle MPA = (180 ° – angle PAM) / 2 = 45 °.
Knowing the magnitude of the RCA and AMR angles, we can find the MRC angle:
angle MRK = 180 ° – 60 ° – 45 ° = 75 °.

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