Trigonometry
Hi, I know when doing trigonometry that if the if a side is at the bottom of ten equation so AB\8 =sin 60 that you would times the 8 by sin 60 but what would you do if the unknown is at the bottom? Sorry if it's too vague
Do you have a picture of the question?
Original post by Foreverton
Do you have a picture of the question?
No sorry😁
Original post by Holidaylover647
No sorry😁
I don't really understand the question
Do you mean when you're looking for an unknown side and you have one side and one angle?
[QUOTE=Foreverton;60403481]Do you mean when you're looking for an unknown side and you have one side and one angle?
It's fine, I've finally found something on it! Thanks anyway☺️
It's fine, I've finally found something on it! Thanks anyway☺️
Original post by Holidaylover647
It's fine, I've finally found something on it! Thanks anyway☺️
Alright then
Feel free to ask about any other maths questions though!
Original post by Foreverton
Alright then
Feel free to ask about any other maths questions though!
Feel free to ask about any other maths questions though!
Hey is it possible for you to help me with this question?
(b) By writing the following equation as quadratics in tan θ/2, solve, in the
interval 0 ≤ θ ≤ 360°:
(i) sinθ + 2cosθ = 1
Original post by Lilly1234567890
Hey is it possible for you to help me with this question?
(b) By writing the following equation as quadratics in tan θ/2, solve, in the
interval 0 ≤ θ ≤ 360°:
(i) sinθ + 2cosθ = 1
(b) By writing the following equation as quadratics in tan θ/2, solve, in the
interval 0 ≤ θ ≤ 360°:
(i) sinθ + 2cosθ = 1
I have no idea sorry 😁
I'm only year 10
Posted from TSR Mobile
(edited 8 years ago)
Original post by Foreverton
OH sozzzz
Original post by Lilly1234567890
OH sozzzz
What year are you?
Posted from TSR Mobile
Original post by Lilly1234567890
Hey is it possible for you to help me with this question?
(b) By writing the following equation as quadratics in tan θ/2, solve, in the
interval 0 ≤ θ ≤ 360°:
(i) sinθ + 2cosθ = 1
(b) By writing the following equation as quadratics in tan θ/2, solve, in the
interval 0 ≤ θ ≤ 360°:
(i) sinθ + 2cosθ = 1
Well you would first work out what θ equals by using the trigonometric equations, then you would substitute θ into tan θ/2 to get the angle. I assume that is what you would do anyways, been a while since I've done trig.
Well when i checked on solutionbank,
the first step was to:
Let tan θ/2=t
Then it changed to:
2t/1+t^2 + 2(1-t^2)/1+t^2 = 1
I just dont get that
the first step was to:
Let tan θ/2=t
Then it changed to:
2t/1+t^2 + 2(1-t^2)/1+t^2 = 1
I just dont get that
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