WILL GIVE MEDAL!!!! If tan(theta) = 3/4 and theta is an acute angle, find cos(theta)
HInt;- \[\sec^2(\theta) - \tan^2(\theta) = 1\] from above determine sec(theta)
remember \[\cos(\theta) = \frac{ 1 }{ \sec(\theta) }\]
I'm just lost. I don't even know what answer I am supposed to have
in ques value of tan(theta) is given from this u can determine sec(theta)
hmmm how would I do that? find sin?
@niksva
no tan(theta) = 3/4 is given sec^2(theta)-tan^2(theta) = 1 now plug in the value of tan(theta) and see what will come?
x=sqrt19/4?
@niksva
sec^2(theta) - (3/4)^2 = 1 sec^2(theta) - 9/16 = 1 sec^2(theta) = 1 +(9/16) sec^2(theta) = (16+9)/16 sec^2(theta) = 25/16 sec(theta) = 5/4 now u can determine cos(theta) ??????????
cos(theta) = 4/5? So 4/5 is my answer
@niksva
yeah!!!
What about: If sin(theta)=12/13 and theta is a acute angle, find cot(theta)
i have shown all the steps of previous answer now u have to show me your work first
I know that cot=1/tan and that tan=sin/cos. so tan=(12/13)/cos
from sin(theta) , determine cosec(theta) first? then use cosec^2(theta) - cot^2(theta) = 1
this is all new to me. I don't know how to do that
@niksva why can't you just show me step by step?
i have shown u the steps of the previous ques u need to try it out first because i m here to help only and this time show your steps systematically
but I have no clue how to do it. you showed me steps to finding cos, not cot
@niksva
value of sin(theta) is given tell me what is cosec(theta) ?
csc=13/12
@pgpilot326 please help
\[\tan \theta=\frac{ 3 }{ 4 }=\frac{ perpendicular }{ base }\]
hypotenuse= root(perpendicular ^2 + base^2)=root(3^2+4^2)=root(9+16)=root25 =5 \[\cos \theta = \frac{ base }{ hypotenuse}=\frac{ 4 }{ 5}\]
@washcaps
hey @gorv I actually answered the first question already as 4/5 is my answer. I need help on the second
draw a pic, then it's easier to see|dw:1393613761753:dw|
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