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High School Math Grades 9-12, Ages 13-18

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#1 h Apr 20, 2025, 5:48 PM

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OABC is a trapezium with OC // AB and ∠AOB = 37°. Furthermore, A, B, C all lie on the circumference of a circle centred at O. The perpendicular bisector of OC meets AC at D. If ∠ABD = x°, find last 2 digit of 100x.

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#2 h Apr 21, 2025, 10:52 AM

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Given the circle, midpoint $O(0,0)$ and radius $r$ , then $C(r,0),B(r\cos \alpha,r\sin \alpha),A(r\cos(\alpha+37^{\circ}),r\sin(\alpha+37^{\circ})\ )$ .
$AB // OC \Rightarrow r\sin \alpha=r\sin(\alpha+37^{\circ}) \Rightarrow 2\alpha=143^{\circ}$ .

The line $AC\ :\ y=-\cot \frac{\alpha+37^{\circ}}{2} \cdot (x-r)$ intersects the line $x=\frac{r}{2}$ in the point $D(\frac{r}{2},\frac{r\cot \frac{\alpha+37^{\circ}}{2}}{2})$ .

The slope of the line $BD\ :\ \tan \angle ABD=m_{BD}=\frac{\sin \alpha-\frac{\cot \frac{\alpha+37^{\circ}}{2}}{2}}{\cos \alpha-\frac{1}{2}}$ ,
$\tan \angle ABD=\frac{2\sin \alpha\sin \frac{\alpha+37^{\circ}}{2}-\cos \frac{\alpha+37^{\circ}}{2}}{2\sin \frac{\alpha+37^{\circ}}{2}\cos \alpha-\sin \frac{\alpha+37^{\circ}}{2}}$ ,
$\tan \angle ABD=\frac{\cos \frac{\alpha-37^{\circ}}{2}-\cos \frac{3\alpha+37^{\circ}}{2}-\cos \frac{\alpha+37^{\circ}}{2}}{\sin \frac{3\alpha+37^{\circ}}{2}+\sin \frac{-\alpha+37^{\circ}}{2}-\sin \frac{\alpha+37^{\circ}}{2}}$ ,

$\tan \angle ABD=\frac{\cos \frac{\alpha-37^{\circ}}{2}}{\sin \frac{-\alpha+37^{\circ}}{2}}$ ,
$\tan \angle ABD=\cot \frac{37^{\circ}-\alpha}{2}=\tan(90^{\circ}-\frac{37^{\circ}-\alpha}{2})=\tan \frac{180^{\circ}-37^{\circ}+\alpha}{2}=\tan 107.25^{\circ}$ .

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