# Need this for a presentation please help

ABCDE is inscribed in a circle with AB=BC=CD=DE=4 and AE=1. Compute

(1-cosB)(1-cosACE)

## 1 Answer

Using Cosine law on the trianles ABC we have

\[AC^2=AB^2+BC^2-2AB \cdot BC \cos B=4^2+4^2-2\cdot 4\cdot 4 \cos B.\]

Hence

\[AC^2=32(1-\cos B). (1)\]

Using Cosine law on the trianles ACE we have

\[AE^2=AC^2+CE^2-2AC \cdot CE\cos (ACE),\]

since $AC=CE$ we get

\[1^2=AC^2+AC^2-2AC^2 \cos (ACE)=2AC^2(1-\cos (ACE)),\]

or

\[\frac{1}{2AC^2}=(1-\cos (ACE)). (2)\]

Multiplying (1) and (2) we get

\[\frac{1}{2}=32(1-\cos B)(1-\cos (ACE)).\]

Hence

\[(1-\cos B)(1-\cos (ACE))=\frac{1}{64}\]

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Use the Cosine Theorem for triangles ABC and ACE. Note that AC=CE. If you want full solution, it deserves a bounty.