In the diagram O and O' are the centres of the circles radii 15cm and 8cm respectively. If PQ = 12cm, find |OO'|.
A.
8.46cm
B.
19.04cm
C.
20.81cm
D.
26.16cm
Correct answer is B
In \(\Delta\) POQ,
\(12^2 = 15^2 + 15^2 - 2(15)(15) \cos < POQ\)
\(144 = 450 - 450\cos < POQ\)
\(450 \cos < POQ = 450 - 144 = 306\)
\(\cos <POQ = \frac{306}{450} = 0.68\)
\(< POQ = 47.2°\)
In \(\Delta\) PO'Q,
\(12^2 = 8^2 + 8^2 - 2(8)(8) \cos <PO'Q\)
\(144 - 128 = -128 \cos < PO'Q\)
\(\cos < PO'Q = - 0.125\)
\(< PO'Q = 97.2°\)
In \(\Delta\) POQ,
\(\cos 23.6 = \frac{x}{15}\)
\(x = 15 \times \cos 23.6\)
= 13.75 cm
In \(\Delta\) PO'Q,
\(\cos 48.6 = \frac{y}{8}\)
\(y = 8 \times \cos 48.6\)
= 5.29 cm
\(\therefore\) OO' = x + y = 13.75 + 5.29
= 19.04 cm