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if sin 50 = p

then sin 140 = ?

Guest Oct 5, 2014

Best Answer 

 #2
avatar+91436 
+10

I did this a little differently (the answer is the same although Chris's is not quite finished.)

The co in cosine stands for complement of sine.  Complementary angles add to 90 degrees.

So 

$$\boxed{sin\theta=cos(90-\theta)}\\\\$$

 

also, 140 degrees is in the 2nd quadrant.  Sine is positive in the 2nd quad.  so

$$\\sin140^0=sin(180-140)=sin40\\\\
sin40=cos50\\
so \\
sin140=cos50$$

 

This is the same as CPhill's answer.

Now we know that sin50=p=p/1

Draw a right angled triangle and label on of the acute angles as 50 degrees.

The opposite side is p

The hypotenuse is 1

So using pythagoras' Theorum the adjacent side must be     $$\sqrt{1-p^2}$$

 

$$\\cos50=\frac{adj}{hyp}=\frac{\sqrt{1-p^2}}{1}\\\\
cos50^0=\sqrt{1-p^2}$$

Melody  Oct 5, 2014
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4+0 Answers

 #1
avatar+80933 
+5

Using an additive identity (and assuming degrees), we have

sin 140 =

sin(90 + 50) =

sin90cos50 + sin50cos90 =

1*cos50 + p*0 =

cos50

 

CPhill  Oct 5, 2014
 #2
avatar+91436 
+10
Best Answer

I did this a little differently (the answer is the same although Chris's is not quite finished.)

The co in cosine stands for complement of sine.  Complementary angles add to 90 degrees.

So 

$$\boxed{sin\theta=cos(90-\theta)}\\\\$$

 

also, 140 degrees is in the 2nd quadrant.  Sine is positive in the 2nd quad.  so

$$\\sin140^0=sin(180-140)=sin40\\\\
sin40=cos50\\
so \\
sin140=cos50$$

 

This is the same as CPhill's answer.

Now we know that sin50=p=p/1

Draw a right angled triangle and label on of the acute angles as 50 degrees.

The opposite side is p

The hypotenuse is 1

So using pythagoras' Theorum the adjacent side must be     $$\sqrt{1-p^2}$$

 

$$\\cos50=\frac{adj}{hyp}=\frac{\sqrt{1-p^2}}{1}\\\\
cos50^0=\sqrt{1-p^2}$$

Melody  Oct 5, 2014
 #3
avatar+80933 
0

Thanks, Melody...I like your answer better, too.....

 

CPhill  Oct 5, 2014
 #4
avatar+91436 
0

Melody  Oct 6, 2014

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