Question 3 If tan(A+B) = 3^0.5 and tan(A-B) = 1/3^0.5; 0° < A + B ≤ 90°, A > B find A and B.
Class 10 - Math - Introduction to Trigonometry Page 187
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tan(A+B)=√3. A+B=60°
tan(A-B)=1/√3. A-B=30°
A=45°. B=15°
tan(A-B)=1/√3. A-B=30°
A=45°. B=15°
Answered by
1
tan ( A + B ) =
tan ( 45° + 15°) =tan60° = √3
tan ( A - B )
tan ( 45° - 15°) =tan30° = 1/√3
so A = 45° and B = 15°
tan ( 45° + 15°) =tan60° = √3
tan ( A - B )
tan ( 45° - 15°) =tan30° = 1/√3
so A = 45° and B = 15°
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