Can you help me here please?
For (a), what do you think?
Ok
Nah?
nah? What do you mean?
Hmm... nothing, sorry, what do you think for (a)?
For this, Should I find the reference angle?
Hmm, for these kind of question, you need to consider angle in each quadrant. I mean which quadrant(s) give(s) you a negative value for sine ratio
Quadrant 3 and 4
So, how many possible value for theta?
2 right?
yes :)
and you've also got the answer for (b), do you know why?
because the solutions lie in q3 and q4
How do I explain answer a though? Like I get why it would be 2solutions but is there a better explanation?
@DhananjayMaurya, Thank you for the medal:)
because there are 2 quadrants give negative values for sine ratio, so there are 2 solutions :P Not sure if this is correct
Ok, that kind of makes sense:). For question 3,is the reference angle by any chance -0.4468?
if -1≦sin theta ≦1 , then it is possible
What formula would I normally use to find the reference angle? I forget
sin theta = sth theta = sin^-1 sth
yeah. Thats what i put. Sin-1 (-0.4321) So reference angle -0.4468 is correct right>?
nope.. sin^-1 is arc sin
Ok, so my reference angle is 25.60 degrees right?
25.60 is in quadrant I, you need the angle in quadrant III and IV~
I would make it -25 but isnt reference angle always the absolute value?
-25.60 = 360-25.60 = angle in quadrant IV :)
so reference angle is not +25 but -25?
One of the reference angle is 360-25.60
And the two possible solutions are theta1 = 180 + 25.6 = 205.6 degrees theta2 = 360-25.6 = 334.4 degrees
right?
Yup~~~
Thank you. Sorry to bother you, Im sure youre busy. Thank you:)
I'm sorry.... really sorry that I didn't respond right after you had!!!
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