Let the circle inscribed in \(\Delta ABC\) touch the sides AB, BC, and AC at points P, Q, and R, respectively.
According to the properties of tangents drawn from a point to a circle, we have:
Let \(AR = AP = x'\), \(BP = BQ = y'\), and \(CQ = CR = z'\).
The side lengths of the triangle can be expressed in terms of these segments:
We are given the following differences:
From these, we can express \(x'\) and \(y'\) in terms of \(z'\):
The perimeter of \(\Delta ABC\) is given as 60 cm.
Perimeter \(= AB + BC + AC = (x' + y') + (y' + z') + (x' + z') = 2(x' + y' + z')\)
\(60 = 2(x' + y' + z')\)
The semi-perimeter, \(s = x' + y' + z' = \frac{60}{2} = 30\) cm.
Substitute the expressions for \(x'\) and \(y'\) into the semi-perimeter equation:
\((z' + 3) + (z' + 2) + z' = 30\)
\(3z' + 5 = 30\)
\(3z' = 30 - 5 = 25\)
\(z' = \frac{25}{3}\) cm
Now, find \(x'\) and \(y'\):
We need to find the value of \(PB + AR\).
\(PB = y'\) and \(AR = x'\).
\(PB + AR = y' + x'\)
\(PB + AR = \frac{31}{3} + \frac{34}{3} = \frac{31 + 34}{3} = \frac{65}{3}\) cm.
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In the following figure, RS ∥ TU. Find the value of ∠PQR − ∠PRQ, if ∠QTU + ∠PRQ = 130° and ∠RPQ = 80°.

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