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# tan help needed

1. theta is an acute angle and sin theta =1/4 find the exact value of tan

isnt tan=costheta/sintheta?

2. If sin = 1/4

then opp=1 and hyp=4

hence what is tan
3. --- mistake
4. (Original post by TenOfThem)
If sin = 1/4

then opp=1 and hyp=4

hence what is tan
tan is o/a

so i know what opposite is as its 1

but i dont have the adjacent
5. (Original post by TenOfThem)
If sin = 1/4

then opp=1 and hyp=4

hence what is tan
is the h square root of 5
6. (Original post by dongonaeatu)
tan is o/a

so i know what opposite is as its 1

but i dont have the adjacent
but you do have 2 sides so surely you know how to find adj
7. (Original post by dongonaeatu)
is the h square root of 5
no, h = 4
8. (Original post by TenOfThem)
but you do have 2 sides so surely you know how to find adj
is it square root of 5
9. (Original post by TenOfThem)
no, h = 4
sorry, is adjacent square root of 5
10. (Original post by dongonaeatu)
is it square root of 5
no

how do you get that
11. (Original post by dongonaeatu)
theta is an acute angle and sin theta =1/4 find the exact value of tan

isnt tan=costheta/sintheta?

sin(T) = opp/hyp

So draw out the triangle. Use Pythagoras (hyp^2 = adj^2 x opp^2) to calculate adj

12. (Original post by TenOfThem)
no

how do you get that
is it 5^2
13. (Original post by dongonaeatu)
is it 5^2
What are you doing to get these numbers

You have a right angled triangle

The Hyp = 4

The Opp = 1

You are looking for the third side

Remind me what exam level you are studying for
14. (Original post by lukas1051)
sin(T) = opp/hyp

So draw out the triangle. Use Pythagoras (hyp^2 = adj^2 x opp^2) to calculate adj

doesnt it equal 5 then?
15. (Original post by TenOfThem)
What are you doing to get these numbers

You have a right angled triangle

The Hyp = 4

The Opp = 1

You are looking for the third side

Remind me what exam level you are studying for
okay so pythagorus therom a^2+b^2=c^2
4^2+1^2=c^2
16+1=c^2
c= square root of 17
so the a = square root of 17?
AS level c2
16. I don't know how much trigonometry you know, but I'll try and be as simple as possible. sin is opposite over hypotenuse, tangent is opposite over adjacent. Since sinx=1/4, we assume the opposite is 1 and the hypotenuse is 4. Pythagoras' theorem tells us that the adjacent length is sqrt(15). Hence, tanx=1/sqrt(15).
17. so from sin(x) = 1/4 we know that:

opp= 1 hyp= 4

so by Pythagoras' Theorem: opp^2 + adj^2= hyp^2

rearranging to get adj = sqrt(hyp^2-opp^2)

so adj = sqrt(4^2 - 1^2) = sqrt(16 - 1) = sqrt(15)

so tan(x) = 1/sqrt(15)
18. (Original post by dongonaeatu)
okay so pythagorus therom a^2+b^2=c^2
4^2+1^2=c^2
16+1=c^2
c= square root of 17
so the a = square root of 17?
AS level c2
Remember c is the longest side! So you have 1+b^2=16.
b=sqrt(15)
19. (Original post by dongonaeatu)
okay so pythagorus therom a^2+b^2=c^2
4^2+1^2=c^2
16+1=c^2
c= square root of 17
so the a = square root of 17?
AS level c2
c = hypotenuse..

so a^2 + b^2 = c^2

a^2 + 1^2 = 4^2
20. (Original post by dongonaeatu)
okay so pythagorus therom a^2+b^2=c^2
4^2+1^2=c^2
16+1=c^2
c= square root of 17
so the a = square root of 17?
AS level c2
No. The hypotenuse, c = 4. You are looking for the adjacent side (a or b)

4^2 = 1^2 + b^2

4^2 - 1^2 = b^2

16 - 1 = b^2

15 = b^2

You now how the opposite and adjacent sides. What does tan(T) now equal?

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