# Trigonometry Problem: BP1 = BP2

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• Nov 12th 2011, 12:50 PM
JohnTerry
Trigonometry Problem: BP1 = BP2
http://www.mathhelpforum.com/math-he...1&d=1321130450

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

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

-----------------------

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)?

Thanks in advance!
• Nov 12th 2011, 02:32 PM
scounged
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?
• Nov 13th 2011, 03:48 AM
sbhatnagar
Re: Trigonometry Problem: BP1 = BP2
If $\angle B = 90 ^{\circ}$ then $\tan{x}=\frac{BP_2}{BP_1}=1$ because it is given that $BP_1=BP_2$.

Quote:

Find $\angle x$.
http://www.mathhelpforum.com/math-he...untitled-2.jpg

$\tan{x}=1$

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

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

Quote:

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

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

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

Quote:

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

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

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

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

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

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

Quote:

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

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