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Question 3, Exercise 10.2
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drawing the reference triangle as shown: {{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\cos \theta =-\dfrac{... drawing the reference triangle as shown: {{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\cos \theta =-\dfrac{
Question 1, Exercise 10.2
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drawing the reference triangle as shown: {{ :fsc-part1-kpk:sol:unit10:fsc-part1-kpk-ex10-2-q1.png?nolink |reference triangle}} we find $\sin \theta =\dfrac{1