FP1 Help please

I don't understand how to do the final part, how I get the area in EXACT form.

Area=(1/2)(2.5)(7.5)sin[cos^-1(-1/3)]

I can't get this exact form from my calculator. Any ideas?

Thanks a lot :smile:

Reply 1

BabyMaths

Original post by krisshP

I don't understand how to do the final part, how I get the area in EXACT form.

Area=(1/2)(2.5)(7.5)sin[cos^-1(-1/3)]

I can't get this exact form from my calculator. Any ideas?

Thanks a lot :smile:

Probably the easiest way is to imagine a line perpendicular to the x axis through P.

The area you need is the difference between the areas of two other triangles. Sketch and I'm sure you'll see what I mean.

Reply 2

ztibor

10

Original post by krisshP

I don't understand how to do the final part, how I get the area in EXACT form.

Area=(1/2)(2.5)(7.5)sin[cos^-1(-1/3)]

I can't get this exact form from my calculator. Any ideas?

Thanks a lot :smile:

\displaystyle \sin (arc \cos (x))=\sqrt{1- \left (cos \left (arc \cos (x) \right )\right )^2}=\sqrt{1-x^2}

Reply 3

brianeverit

9

Original post by krisshP

I don't understand how to do the final part, how I get the area in EXACT form.

Area=(1/2)(2.5)(7.5)sin[cos^-1(-1/3)]

I can't get this exact form from my calculator. Any ideas?

Thanks a lot :smile:

Do you know OF?
Can you find

\sin\theta

?
Area of triangle

=\frac{1}{2} \times OF \times FP \times \sin\theta

Reply 4

krisshP

OP

14

Original post by ztibor

\displaystyle \sin (arc \cos (x))=\sqrt{1- \left (cos \left (arc \cos (x) \right )\right )^2}=\sqrt{1-x^2}

It works

as well, but how did you get/derive that?

Thanks for helping

(edited 10 years ago)

Reply 5

krisshP

OP

14

Original post by brianeverit

Do you know OF?
Can you find

\sin\theta

?
Area of triangle

=\frac{1}{2} \times OF \times FP \times \sin\theta

OF=2.5

That's what I did though. I can't get an exact answer from my calculator though if I put that in :frown:

Reply 6

krisshP

OP

14

Original post by BabyMaths

Probably the easiest way is to imagine a line perpendicular to the x axis through P.

The area you need is the difference between the areas of two other triangles. Sketch and I'm sure you'll see what I mean.