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# Geometry/Right Triangle

Question
in a figure triangle PQR is right angled at R and S is the mid point of hypotenuse PQ. If RS = 25 cm and PR = 48 cm find QR.

When a right triangle is put in a circle, the hypotenuse is the diameter of the circle.
This means that the triangle PSR is symmetric and the triangle RSQ is symmetric since
PS, SQ, and SR are all radii of the circle.

Using the law of cosines, the measure of the angle opposite the side of length 48 can be found.
Using the law of cosines again and the fact that the angle on the other triangle is the supplement of the angle on this triangle, the far side of that triangle can be found.

Law of Cosines: a^2 + b^2 - 2ab*cosC = c^2.

Using this, cosC = (a^2 + b^2 - c^2)/(2ab).

From there, angel C can be found.

Once that is known, the angle in question is the supplement of that angle, so it is 180-A.

Once that is found, we have two sides that are both 25 and the angle between them,
and those can be used as a, b, and C in the Law of Cosine equation to find c^2.
Once this has been done, find the square root to get the answer.  Note that since
it is a distance, we don't need to worry about the fact that square roots are both
positive and negative, for in this case, a negative value is not a distance.

Geometry

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