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Let πœƒ be an angle in quadrant II such that tan πœƒ = βˆ’47. Find sin πœƒ and sec

 Dec 16, 2020
 #1
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In Q2, y  will be positive and x will be negative

 

tan (theta)  = y  / x  =   47 /  -1

 

r  =sqrt  ( x^2 + y^2)  =  sqrt   [  (-1)^2  +  47^2  ] =  sqrt [ 2210 ] 

 

sin (theta)    = y/r  =  47  /sqrt (2210)  = 47 sqrt (2210)  /2210

 

sec (theta)  =  r/x  =  sqrt (2210)  / -1  =    -sqrt (2210)

 

cool cool cool

 Dec 16, 2020
 #2
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bless ur heart

Guest Dec 16, 2020

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