OCR C2 Trigonometry

Since theta is obtuse, the triangle is an ambiguous case and therefore, I can't use trig because when I rearrange for sin C I get 1.12 and not between 0 and 1.

So how would I solve this... Somehow I imagine multiplying something by 2.

Reply 1

boromir9111

17

I would imagine you use

A =\displaystyle\frac{1}{2}r^{2} \theta

Unless I am wrong, feel free to correct me anyone!

(edited 13 years ago)

Reply 2

I'm clever

OP

6

Original post by boromir9111

I would imagine you use

A =\displaystyle\frac{1}{2}r^{2} \theta

Unless I am wrong, feel free to correct me anyone!

Still I get theta = 0.64

???

Reply 3

boromir9111

17

Original post by I'm clever

Still I get theta = 0.64

???

What's the answer supposed to be?

Reply 4

I'm clever

OP

6

Original post by boromir9111

What's the answer supposed to be?

If I knew that why would I be asking the question... -.-

Reply 5

boromir9111

17

Original post by I'm clever

If I knew that why would I be asking the question... -.-

Then my method is correct!

Reply 6

I'm clever

OP

6

Original post by boromir9111

Then my method is correct!

Ok I think I have it:

Since theta is obtuse I CAN use the Sin rule. I'll have to do 180 - theta.

A= \frac{1}{2}r^2 \theta => 8 = \frac{1}{2} \times 5^2 \theta

\theta = 0.64

\theta = sin^{-1} (0.64) =39.79

180-39.79 = 140

Original post by I'm clever

Ok I think I have it:

Since theta is obtuse I CAN use the Sin rule. I'll have to do 180 - theta.

A= \frac{1}{2}r^2 \theta => 8 = \frac{1}{2} \times 5^2 \theta

\theta = 0.64

\theta = sin^{-1} (0.64) =39.79

180-39.79 = 140

I didn't even read the question but you should be working in radians.

Reply 8

Farhan.Hanif93

18

Original post by I'm clever

A= \frac{1}{2}r^2 \theta => 8 = \frac{1}{2} \times 5^2 \theta

That's the formula for the area of sector OAB rather than the triangle OAB. You don't know what the area of the sector is so you can't deduce that step above.

What I would do is use formula for the area of the triangle

(A=\frac{1}{2}ab\sin C)

to find that

\frac{1}{2}5^2 \sin \theta = 8

and just solve for

\sin \theta

. Then recall that

\sin \theta = \sin (\pi - \theta)

.

Original post by boromir9111

I would imagine you use

A =\displaystyle\frac{1}{2}r^{2} \theta

Unless I am wrong, feel free to correct me anyone!

And the area of the triangle is

\displaystyle\frac{1}{2}r^{2} \sin \theta

Reply 10

I'm clever

OP

6

Original post by Mr M

And the area of the triangle is

\displaystyle\frac{1}{2}r^{2} \sin \theta

A=\frac{1}{2}absinC

8=\frac{1}{2}(5^2)sinC

sinC=39.79

Since theta is obtuse 180-39.79=140=2.44 rad.

Is that correct?

(edited 13 years ago)

Reply 11

Core

6

Original post by I'm clever

A=\frac{1}{2}absinC

8=\frac{1}{2}(5^2)sinC

sinC=39.79

Since theta is obtuse 180-39.79=140

Is that correct?

Wait never mind thats good.

(edited 13 years ago)

Original post by I'm clever

A=\frac{1}{2}absinC

8=\frac{1}{2}(5^2)sinC

sinC=39.79

Since theta is obtuse 180-39.79=140

Is that correct?

I hope you don't mind if I shout:

RADIANS

Reply 13

I'm clever

OP

6

Original post by Mr M

I hope you don't mind if I shout:

RADIANS

EDIT: I updated my post before you posted this.

Reply 14

Farhan.Hanif93

18

Original post by I'm clever

A=\frac{1}{2}absinC

8=\frac{1}{2}(5^2)sinC

sinC=39.79

Since theta is obtuse 180-39.79=140

Is that correct?

No, that's the value of

\sin \theta

for this angle theta and think about it; if your angle is 140 radians (roughly 44.5 pi, which equates to roughly 8010 degrees!), then that's a little too big isn't it? So that's a tell-tale sign that you've made a mistake somewhere. :p:

You need to inverse sine both sides (using you calculator) and then, if you get an acute angle for theta, consider

\pi - \theta

so that it will be the acute version. I've explained why this works in my first post.

Original post by I'm clever

EDIT: I updated my post before you posted this.

Well it is still wrong to 3 sig figs which is the required accuracy.

Your calculator has a radians mode. Use it.

Reply 16

I'm clever

OP

6

Original post by Mr M

Well it is still wrong to 3 sig figs which is the required accuracy.

Your calculator has a radians mode. Use it.

sin^{-1}(0.64)=0.694 rad => \theta = \pi-0.694=2.45rad

Correct now?

Original post by I'm clever

sin^{-1}(0.64)=0.694 rad => \theta = \pi-0.694=2.45rad

Correct now?

Yes. Try to get used to working in radians as you will only occasionally use degrees from now on. Don't round answers prematurely - write down plenty of figures for intermediate steps and only round the final answer.

Reply 18

I'm clever

OP

6

Original post by Mr M

Yes. Try to get used to working in radians as you will only occasionally use degrees from now on. Don't round answers prematurely - write down plenty of figures for intermediate steps and only round the final answer.

Well I learnt how to store my answer in the calculator in Statistics so I can apply this to C2, or any question that requires multiple stages of working in maths.

Reply 19

Josh_Dey

16

But aren't you clever 'I'm clever'?

OCR C2 Trigonometry

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