Answer
Ans. Sinθ=√5/3 tanθ=√5/2
Work Step by Step
Given:-cosθ=2/3
We know that
Cosθ=B/H
Applying Pythagoras theorem in ∆PQR
$H^2=P^2+B^2$
Putting H=3 B=2
we get P=√5
Sinθ=√5/3 tanθ=√5/2
Ans.
Calculus, 10th Edition (Anton)
by Anton, Howard
Published by
Wiley
ISBN 10:
0-47064-772-8
ISBN 13:
978-0-47064-772-1
Appendix B - Trigonometry Review - Exercise Set B - Page A24: 8
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