if theta is an acute angle between the lines x²–7xy+12y²=0 then find 2cos theta+3sin theta÷4sin theta+5cos theta?
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Answer:
if theta is an acute angle between the lines x²–7xy+12y²=0 then find 2cos theta+3sin theta÷4sin theta+5cos theta?
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Given : x²–7xy+12y²=0
theta is an acute angle between the lines x²–7xy+12y²=0
To Find : (2cosθ + 3sinθ)/(4sinθ + 5cosθ)
Solution:
x²–7xy+12y²=0
=> x² - 3xy - 4xy + 12y² = 0
=> x(x - 3y) - 4y(x - 3y) = 0
=> (x - 4y)(x - 3y) = 0
x - 4y = 0
=> y = x/4 slope = 1/4
x - 3y = 0
=> y = x/3 slope = 1/3
Angle between two line = θ
tanθ = | (1/4 - 1/3) | / (1 + (1/4)(1/3)|
=> tanθ = | -1/13|
=> tanθ = 1/13
(2cosθ + 3sinθ)/(4sinθ + 5cosθ)
= (2 + 3tanθ)/(4 + 5tanθ)
= (2 + 3/13)( 4 + 5/13)
= 29/57
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