Question 14

$$AB$$ is a diameter of a semicircle of centre $$O$$. $$C$$ is the midpoint of the arc $$AB$$. $$AC$$ and the tangent at $$B$$ to the semicircle meet at $$P$$. $$D$$ is the midpoint of $$BP$$. If $$ACDO$$ is a parallelogram and $$\angle PAD=\theta$$, then $$\sin\theta$$ is

Solution

Take $$O=(0,0)$$, $$A=(-r,0)$$ and $$B=(r,0)$$, so the midpoint of the arc gives $$C=(0,r)$$. The line $$AC$$ meets the tangent $$x=r$$ at $$P=(r,2r)$$, so $$D=(r,r)$$. The slopes of $$AP$$ and $$AD$$ are $$1$$ and $$1/2$$ respectively, and the sine of the angle between these two lines is $$1/\sqrt{10}$$.

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$$AB$$ is a diameter of a semicircle of centre $$O$$. $$C$$ is the midpoint of the arc $$AB$$. $$AC$$ and the tangent at $$B$$ to the semicircle meet at $$P$$. $$D$$ is the midpoint of $$BP$$. If $$ACDO$$ is a parallelogram and $$\angle PAD=\theta$$, then $$\sin\theta$$ is

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