# Thread: Trigonometry Problem: BP1 = BP2

1. ## Trigonometry Problem: BP1 = BP2

How do you resolve this one?
What i got till now:

BP1= 16+12/tanx
BP2= 12+16tanx

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BP1=BP2 = 16+12/tanx = 12+16tanx = (...) = 4tanx - 16tanx2 +12=0

Is it correct?
All you have to do is resolve the equation, find tanx, then sinx and cosx and subsitute them in C(x)?

2. ## Re: Trigonometry Problem: BP1 = BP2

I think that you're making this more complicated than it has to be. BP1=BP2. Can't you just use this to show that tanx=1?

3. ## Re: Trigonometry Problem: BP1 = BP2

If $\displaystyle \angle B = 90 ^{\circ}$ then $\displaystyle \tan{x}=\frac{BP_2}{BP_1}=1$ because it is given that $\displaystyle BP_1=BP_2$.

Find $\displaystyle \angle x$.

$\displaystyle \tan{x}=1$

$\displaystyle \Rightarrow \sin{x}=\cos{x}$

$\displaystyle \Rightarrow x=45 ^{\circ}$

Find $\displaystyle P_1 A$
$\displaystyle \sin{x}=\frac{12}{P_1 A}$ ...................($\displaystyle \sin{x}=\frac{\text{Perpendicular}}{\text{ Hypotenuse }}$)

$\displaystyle \Rightarrow \sin{45^{\circ}}=\frac{12}{P_1 A}$

$\displaystyle \Rightarrow P_1 A=12\sqrt{2}$

Find $\displaystyle P_2 A$
$\displaystyle \angle B+x+\angle P_2 =180 ^{\circ}$ .......................(Angle Sum Property)

$\displaystyle \Rightarrow \90^{\circ}+45^{\circ}+\angle P_2 =180 ^{\circ}$

$\displaystyle \Rightarrow \angle P_2 =45^{\circ}$

$\displaystyle \sin{\angle P_2 }=\frac{16}{P_2 A}$ ...................($\displaystyle \sin{x}=\frac{\text{Perpendicular}}{\text{ Hypotenuse }}$)

$\displaystyle \Rightarrow \sin{45^{\circ}}=\frac{16}{P_2 A}$

$\displaystyle \Rightarrow P_2 A=16\sqrt{2}$

Find $\displaystyle P_1 P_2$
$\displaystyle P_1 P_2 = P_1 A + P_2 A =12\sqrt{2}+16\sqrt{2}=28\sqrt{2}$

Did you understand? In fact you could also solve this question without the use of trigonometry.