Let AB be a diameter of a circle, and let C be a point on the circle such that AC=8 and BC=4 The angle bisector of ACB intersects the circle at point M Find CM.
+14903 Find CM.
Hello Guest!
CM is x.
\(MA=\sqrt{x^2+8^2-2\cdot 8x\cdot cos\ 45^{\circ}}\\ MB=\sqrt{x^2+4^2-2\cdot 4x\cdot cos\ 45^{\circ}}\\ MA+MB=\sqrt{4^2+8^2}=\sqrt{80}\\ \sqrt{x^2+8^2-2\cdot 8x\cdot cos\ 45^{\circ}}+\sqrt{x^2+4^2-2\cdot 4x\cdot cos\ 45^{\circ}}=\sqrt{80}\\ {\color{blue}CM=x=\dfrac{8\sqrt{2}}{3}}=3.771\ (WolframAlpha)\)
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