SOLUTION: A triangle is right- angled at B. D is the point on AC such that BD is perpendicular to AC. Let angle BAC=theta b i. Given that 6AD+BC=5AC show that 6costheta+tantheta=5sectheta

Algebra ->  Trigonometry-basics -> SOLUTION: A triangle is right- angled at B. D is the point on AC such that BD is perpendicular to AC. Let angle BAC=theta b i. Given that 6AD+BC=5AC show that 6costheta+tantheta=5sectheta       Log On


   

Question 1027374: A triangle is right- angled at B. D is the point on AC such that BD is perpendicular to AC. Let angle BAC=theta
b
i. Given that 6AD+BC=5AC show that 6costheta+tantheta=5sectheta
ii. Deduce that 6sin^2theta-sintheta-1=0
I have already done part i, I just found what AD, BC AND AC are in terms of their trignometric rations. I then subbed them into the equation and received it.
I don't know how to do question ii
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
6 cos theta+tantheta = 5sectheta

6 cos theta+tantheta = 5/costheta

multiply by costheta
6cos^2theta +tantheta*costheta = 5
6cos^2theta+sintheta =5
6((1-sin^2theta) +sintheta=5

6- 6sin^2theta +sintheta=5
rearrange
6sin^2theta -sintheta-1=0