If \angle ACP=\gamma, then what is \tan \gamma equal to?
Let ABC be a triangle right-angled at B. Let P be the point on BC such that BP=PC. If AB = 10 cm, \angle BAP=45^\circ and \angle CAP=\theta<br>(use \tan(\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta})
- A. \frac{1}{2} ✓
- B. \frac{1}{3}
- C. \frac{2}{3}
- D. 1
Correct Answer: A. \frac{1}{2}
Explanation
Angle ACP is simply \angle C of \triangle ABC. In right-angled \triangle ABC, \tan \gamma = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BC} = \frac{10}{20} = \frac{1}{2}.
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