.
1. tan(A) = 5 implies = 25 implies = = = implies sin(A) = = .
The sign is "-" (minus) at the square root, since A is in QIII.
Then cos(A) = = = .
The sign is "-" (minus) at the square root, since A is in QIII.
2. sin(B) = implies cos(B) = = = = = .
The sign is "-" (minus) at the square root, since B is in QII.
3. Now
sin(A-B) = sin(A)*cos(B) - cos(A)*sin(B) = - = + = = . = =
= = 0.861626 (approximately).
cos(A-B) = cos(A)*cos(B) + sin(A)*sin(B) = + = - = = . = =
= = -0.50754 (approximately).
Since sin(A-B) is positive, while cos(A-B) is negative, the angle A-B lies in QII.
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To see more solved similar problems on calculating trig functions, look into the lessons
- Calculating trigonometric functions of angles
- Advanced problems on calculating trigonometric functions of angles
- Evaluating trigonometric expressions
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Trigonometry: Solved problems".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
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