Trigonometry

for part i) I have no idea on how to find angle BAC, I was thinking about trying to find the length BC or AC, but im not sure thats what im supposed to do.

(edited 6 years ago)

Reply 1

Radioactivedecay

16

Part i) It is a root therefore the y valur there is 0. Pluggin lambda in we get:
2 (2cos (lam)-sin (lam))sin (lam)=0
Cancelling the and sinlam we get:
2cos (lam )-sin (lam)=0
2cos (lam)=sin (lam)

Dividing by coslam to both sides:
Sinlam/coslam=2
Tan(lam)=2.

Reply 2

znx

OP

10

Original post by Radioactivedecay

Part i) It is a root therefore the y valur there is 0. Pluggin lambda in we get:
2 (2cos (lam)-sin (lam))sin (lam)=0
Cancelling the and sinlam we get:
2cos (lam )-sin (lam)=0
2cos (lam)=sin (lam)

Dividing by coslam to both sides:
Sinlam/coslam=2
Tan(lam)=2.

Haha thanks mate, Not sure whats happened but I cant seem to get rid of that image, it for a different question.

Reply 3

znx

OP

10

Fixed the OP, now the image corresponds to the actual question :smile:

Reply 4

Radioactivedecay

16

Original post by znx

Haha thanks mate, Not sure whats happened but I cant seem to get rid of that image, it for a different question.

Oh haha ok then. Thats fine.

for part i of the attachement, just look at the point A. you know that the sum of all angles must equal 360 as it goes around one entire circle, so to speak. So you get
BAC +120+90-theta+90=360

then you simply rearrange. :smile:

Reply 5

znx

OP

10

Original post by Radioactivedecay

Oh haha ok then. Thats fine.

for part i of the attachement, just look at the point A. you know that the sum of all angles must equal 360 as it goes around one entire circle, so to speak. So you get
BAC +120+90-theta+90=360

then you simply rearrange. :smile:

Thank you mate

Trigonometry

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