One thing I don't understand, if the original question already stated that cos A + cos B = 1/root 2 AND A + B = 120, by commutative law doesn't it already imply that if (A,B) = (15,105) then (B,A) can also be (15,105)? i.e. (A,B) can also be (105,15)?

That saves the trouble of going to the negative region of the C quadrant.

]]>Thanks for pointing this out!

Many of us are so used to the anti-clockwise circuit through the quadrants as the majority of such trigo questions deal with the range of 0° ≤ x ≤ 360°.

But in this case, a clockwise route needs to be considered as well since the range has a negative region. Miss Loi has updated the explanation in the original comment.

It's amazing how everyone (including yours truly) missed this out in 2008! So it's not so straightforward to get into EJC afterall eh?

]]>Your answer is only partially correct. (A-B)/2 = 45 OR -45.

[*not* 45 OR 315, since (A-B)/2 ranges between -60 and 60]

Hence, the complete solution set is [A=15, B=105] OR [A=105,B=15].

Plug both solutions into the given info and see for youself that they both work!

]]>Do continue to work hard in your quest towards your mastery of the *Four As Guaranteed Explosive Mugging Technique* one day.

Since = cos (45°),

Thus

= 45°, 315°

Yes, whenever you see phrases like "*find the values of x ... given that A° < x < B°*" blah blah blah ... you'll know you're dealing with the

You've already proved in Part 1 that and so the **basic angle** of , `α` = cos^{-1} = 45°

And since 1/(√2) is *positive*, it follows from the SATC diagram (**A** and **C** quadrants) above that = 45° or (360-45)° = 315°

**IMPORTANT:** We only did *one* 'round' of the 360° circuit since the 'coefficient' of (A-B) is ≤ 1. More 'rounds' are usually required when the 'coefficient' > 1 - refer to your textbook if you're lost or see this example for a rough idea.

**2012 Update: NOT SO FAST!**

Thanks to Ryan for pointing out, there's a tricky part to this question which apparent everyone has missed in 2008!!!

It's vital to consider the *range* of when deciding its possible values.

In this case the max value of = 60° (`A`_{max} − `B`_{min}/2) and its min value is −60° (`A`_{min} − `B`_{max}/2), so

−60° ≤ ≤ 60°

Many of us are so used to the anti-clockwise circuit as the majority of such trigo questions has 0° ≤ `x` ≤ 360°, but in this case the range goes into a *negative* region. So while 315° lies outside the range of , its 'corresponding' angle of −45° in the same quadrant does lie inside its valid range, since cos(−`θ`) = cos `θ`. Thus it's imperative you consider going around the clockwise (negative direction) when you see a range in the negative region!

So ...

~~(A-B) = 90°, 630°~~

~~A = 90°+B, 630°+B --- (ii)~~

= 45°, −45°

(A-B) = 90°, −90°

A = 90°+B, −90°+B --- (ii)

Since A+B = 120° --- (i),

Sub (i) into (ii)

~~(90+B) + B = 120°, (630+B) + B = 120°~~

~~90 + 2B = 120°, 630 + 2B = 120°~~

~~B = 15, -255 --- (iii)~~

(90°+B) + B = 120°, (−90°+B) + B = 120°

90° + 2B = 120°, −90° + 2B = 120°

B = 15, 105 --- (iii)

Sub (iii) into (ii)

~~A = 105°, 375°~~

Since A & B lies between 0° - 120°,

Thus the values are:

~~A = 105° & B = 15°.~~

`A` = 105° & `B` = 15° or `A` = 15° & `B` = 105°.

OMG **Pai Mei**?! Thought you're already dead?! *scampers away to save her eyeball*