trig problem again sorry (1 Viewer)

[Aznmichael92]

sorry i have another trig problem. This time somewhat different

Triangle ABC and Triangle BCD are similar triangles and are both isosceles. Angle ABC = Angle ACB = 72 degrees. AB = AC = 1. BD bisects Angle ABC BC = x.

1)Find the exact value of x
2) Hence, find exact value of cos 36

 

[adnan91]

gimmie the answers to make sure mine r right

 

[Aznmichael92]

i dont have the answers thats why i am asking?

 

[adnan91]

u might be asking for the method of solving them not just the answer . Anway i found x= 0.618 and exact value of cos 36= 2-(.618)^2 divded by 2. Using the sin rule for x and cosine rule for cos36

 

[Aznmichael92]

u might be asking for the method of solving them not just the answer . Anway i found x= 0.618 and exact value of cos 36= 2-(.618)^2 divded by 2. Using the sin rule for x and cosine rule for cos36

isnt that in decemial places?

 

[Azreil]

1) A/sina = B/sinb
x/sin36 = 1/sin72
x = sin36/sin72
= 0.618
2) cos a = (B^2 + C^2 - A^2)/2BC
cos 36 = (1^2 + 1^2 - 0.48^2)/2*1*1
= 1.618/2
= 0.809

 

[conics2008]

hey just use cosine rule for both parts.

part i use triangle abc

x^2 = 1^2 + 1^2 - 2cos36

x= 0.618

hence cos36 is like this

(0.618)^2 = 2-2cos34

carry the 2cos36 to the other side and take the 0.618^2 to the other side

you get

2cos36 = 2-(0.618)^2
cos36= 2-(0.618)^2/2

 

[Aznmichael92]

but its not exact values

 

[bored of sc]

i) x/sin36 = 1/sin72

x= sin36/sin72 --- make x the subject
= sin144/sin72 --- change sin36 to supplemantary value i.e. second quadrant (180-36)
= sin(2*72)/sin72 --- recognise relationship between 72 and 144 i.e. 72*2 = 144 (denominator has 72)
= 2sin72cos72/sin72 --- recognise double angle formula i.e. sin2@
= 2cos72 --- cancel sin72 from numerator and denominator

x = 2cos72

ii) cos36 = sin54 --- sin is complemantary to cos
= sin(108/2) --- express as a half angle i.e. 't' results
= t/rt(1+t²) --- express in terms of t where tan54 = t

This is way off but it was fun mucking around with trig stuff.

 

[Aznmichael92]

i) x/sin36 = 1/sin72

x= sin36/sin72 --- make x the subject
= sin144/sin72 --- change sin36 to supplemantary value i.e. second quadrant (180-36)
= sin(2*72)/sin72 --- recognise relationship between 72 and 144 i.e. 72*2 = 144 (denominator has 72)
= 2sin72cos72/sin72 --- recognise double angle formula i.e. sin2@
= 2cos72 --- cancel sin72 from numerator and denominator

x = 2cos72

ii) cos36 = sin54 --- sin is complemantary to cos
= sin(108/2) --- express as a half angle i.e. 't' results
= t/rt(1+t²) --- express in terms of t where tan54 = t

This is way off but it was fun mucking around with trig stuff.

i get part i) but not ii). Is that the t method?

 

[kaz1]

i get part i) but not ii). Is that the t method?

Not quite. t=tan(theta/2)

 

[Aznmichael92]

Not quite. t=tan(theta/2)

i dont think i have learnt that rule though. I feel so stupid now