# Trigonometry Perpendicular line help!?

• Oct 21st 2012, 12:50 PM
Benja303
Trigonometry Perpendicular line help!?
8. In a triange ABC, AB = 4.8cm, BC = 3.8 cm and A ^B C =42 degrees

P is the foot of the perpendicular from A to BC.
i. Calculate the area of triangle ABC

ii. Deduce the length of AP.

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Okay, I don't understand what they mean by "P is the foot of the perpendicular from A to BC"
• Oct 21st 2012, 01:03 PM
skeeter
Re: Trigonometry Perpendicular line help!?
did you make a sketch ... ?

from vertex A to side BC , AP is the altitude from A to BC.

also ...

Quote:

... and A ^B C =42 degrees
what angle is this?
• Oct 22nd 2012, 11:20 AM
Benja303
Re: Trigonometry Perpendicular line help!?
Okay thanks.

soo I worked out the area to be 61.02 usuing 1/2*a*b*sin C

now for part II i'm not sure what to do.. I know the angle is 90. can u give me a hint or guide me a bit? :S
• Oct 22nd 2012, 11:55 AM
MarkFL
Re: Trigonometry Perpendicular line help!?
Your area is 10 times greater than it should be. To find the altitude, simply use:

$\displaystyle \sin(42^{\circ})=\frac{\bar{AP}}{\bar{AB}}$
• Oct 22nd 2012, 04:58 PM
Benja303
Re: Trigonometry Perpendicular line help!?
But i don't know what AP is!
• Oct 22nd 2012, 05:14 PM
Benja303
Re: Trigonometry Perpendicular line help!?
oKAY I used Area= 1/2*base*height

and then i rearranged theformula to get the height as 3.7

correct?
• Oct 22nd 2012, 06:20 PM
MarkFL
Re: Trigonometry Perpendicular line help!?
You have:

$\displaystyle A=\frac{1}{2}\bar{BC}\cdot\bar{AB}\sin(42^{\circ})$

and:

$\displaystyle A=\frac{1}{2}\bar{BC}\cdot\bar{AP}$

Hence:

$\displaystyle \bar{AP}=\bar{AB}\sin(42^{\circ})$ (which you would get from my previous post)

$\displaystyle \bar{AP}\approx3.2\text{ cm}$