Trig question (1 Viewer)

[BenBrownTown]

Hey can someone help with a question form a past paper 1997 Question 3 part (b) (i)
I dont get why it get the right solution I need to make angle POC as theta
thankyou
http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97MAT3U.PDF

 

[Carrotsticks]

Angle POC = 180 - theta.

We are considering the sin (Angle POC) = sin (180 - theta).

But recall that sin (180 - x) = sin(x)

Hence, sin (Angle POC) = sin (180 - theta) = sin (theta).

 

[Zenox]

You dont make angle POC as theta. Angle POA is theta, given from the diagram.

 

[BenBrownTown]

Angle POC = 180 - theta.

We are considering the sin (Angle POC) = sin (180 - theta).

But recall that sin (180 - x) = sin(x)

Hence, sin (Angle POC) = sin (180 - theta) = sin (theta).

Is sin (180 -x) = sin(x) because they are both positive in the 1st and 2nd quadrants hence why they are equal?
thankyou for helping

 

[karnbmx]

there is a VERY similar question to this in 2008 HSC.

 

[Carrotsticks]

Is sin (180 -x) = sin(x) because they are both positive in the 1st and 2nd quadrants hence why they are equal?
thankyou for helping

Exactly correct =)

 

[BenBrownTown]

Exactly correct =)

Cheers man appreciate it

 

[shongaponga]

Is sin (180 -x) = sin(x) because they are both positive in the 1st and 2nd quadrants hence why they are equal?
thankyou for helping

Exactly correct =)

You can verify it by using expansion of sin(180-x): = Sin(180)Cos(-x) + Cos(180)Sin(-x) = 0+(-1)[-Sin(x)] = Sin(x)