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oh sorry triangle abc bc is 12 and cos aBc is 2/3 ac is 2cm bigger then ab
find the length od ab and ac
find the length od ab and ac
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#5
(Original post by louise18152)
oh sorry triangle abc bc is 12 and cos aBc is 2/3 ac is 2cm bigger then ab
find the length od ab and ac
oh sorry triangle abc bc is 12 and cos aBc is 2/3 ac is 2cm bigger then ab
find the length od ab and ac
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#8
(Original post by louise18152)
i mean i don't know what you do with the cos
i mean i don't know what you do with the cos
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#9
(Original post by louise18152)
i mean i don't know what you do with the cos
i mean i don't know what you do with the cos
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(Original post by louise18152)
yeah not right angle
yeah not right angle
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#12
(Original post by louise18152)
i still dont understand
i still dont understand
https://www.mathsisfun.com/algebra/trig-cosine-law.html
Last edited by mqb2766; 1 month ago
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#14
let AB = x and AC = x + 2
Use the cosine rule:
cos(A) = (b^2 + c^2 - a^2)/2bc
let A = 2/3 (make sure its in radians)
a = x+2
b = x
c = 12
So:
cos(2/3)= (x^2 + 12^2 - (x+2)^2)/(2 x x^2 x 12^2)
x = (70)/ (12 cos(2/3) + 1)
is what i got aka 5.345
Use the cosine rule:
cos(A) = (b^2 + c^2 - a^2)/2bc
let A = 2/3 (make sure its in radians)
a = x+2
b = x
c = 12
So:
cos(2/3)= (x^2 + 12^2 - (x+2)^2)/(2 x x^2 x 12^2)
x = (70)/ (12 cos(2/3) + 1)
is what i got aka 5.345
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#15
(Original post by louise18152)
no
no
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(Original post by potatohouse)
let AB = x and AC = x + 2
Use the cosine rule:
cos(A) = (b^2 + c^2 - a^2)/2bc
let A = 2/3 (make sure its in radians)
a = x+2
b = x
c = 12
So:
cos(2/3)= (x^2 + 12^2 - (x+2)^2)/(2 x x^2 x 12^2)
x = (70)/ (12 cos(2/3) + 1)
is what i got aka 5.345
let AB = x and AC = x + 2
Use the cosine rule:
cos(A) = (b^2 + c^2 - a^2)/2bc
let A = 2/3 (make sure its in radians)
a = x+2
b = x
c = 12
So:
cos(2/3)= (x^2 + 12^2 - (x+2)^2)/(2 x x^2 x 12^2)
x = (70)/ (12 cos(2/3) + 1)
is what i got aka 5.345
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reply
(Original post by potatohouse)
let AB = x and AC = x + 2
Use the cosine rule:
cos(A) = (b^2 + c^2 - a^2)/2bc
let A = 2/3 (make sure its in radians)
a = x+2
b = x
c = 12
So:
cos(2/3)= (x^2 + 12^2 - (x+2)^2)/(2 x x^2 x 12^2)
x = (70)/ (12 cos(2/3) + 1)
is what i got aka 5.345
let AB = x and AC = x + 2
Use the cosine rule:
cos(A) = (b^2 + c^2 - a^2)/2bc
let A = 2/3 (make sure its in radians)
a = x+2
b = x
c = 12
So:
cos(2/3)= (x^2 + 12^2 - (x+2)^2)/(2 x x^2 x 12^2)
x = (70)/ (12 cos(2/3) + 1)
is what i got aka 5.345
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#19
(Original post by louise18152)
why did you square the second part?
why did you square the second part?
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