If \angle BAC=\theta then what is \sin \theta+\cos \theta equal to?
ABC is a triangle right-angled at B. Given that AC-AB=2 cm and BC=16 cm
- A. 1
- B. \frac{71}{65}
- C. \frac{73}{65}
- D. \frac{79}{65} ✓
Correct Answer: D. \frac{79}{65}
Explanation
By Pythagoras theorem, AC^2 - AB^2 = BC^2 \implies (AB+2)^2 - AB^2 = 16^2 \implies 4AB + 4 = 256 \implies AB = 63. Thus, AC = 65. Then, \sin \theta + \cos \theta = \frac{BC}{AC} + \frac{AB}{AC} = \frac{16}{65} + \frac{63}{65} = \frac{79}{65}.
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