What is \tan \theta 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{1}{4}
- D. \frac{1}{5}
Correct Answer: B. \frac{1}{3}
Explanation
In \triangle ABP, \tan 45^\circ = \frac{BP}{10} \implies BP = 10. Since BP=PC, BC = 20. In \triangle ABC, \tan(45^\circ + \theta) = \frac{20}{10} = 2. Using the expansion formula: \frac{1 + \tan \theta}{1 - \tan \theta} = 2 \implies 1 + \tan \theta = 2 - 2\tan \theta \implies 3\tan \theta = 1 \implies \tan \theta = \frac{1}{3}.
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