A geometry problem by A Former Brilliant Member

Geometry Level 3

If sin ⁡ A + sin ⁡ B = 1 2 \sin A+\sin B=\dfrac{1}{2} and cos ⁡ A + cos ⁡ B = 5 4 \cos A+\cos B=\dfrac{5}{4} , then tan ⁡ A + tan ⁡ B = p q \tan A+\tan B = \dfrac{p}{q} , where p p and q q are relatively prime positive integers. Find p + q p+q .


The answer is 373.

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

a a and b b , in either order are tan ⁡ − 1 ( 58 − 5 1015 2 1015 + 145 ) and tan ⁡ − 1 ( 5 1015 + 58 145 − 2 1015 ) \tan ^{-1}\left(\frac{58-5 \sqrt{1015}}{2 \sqrt{1015}+145}\right)\text{ and }\tan ^{-1}\left(\frac{5 \sqrt{1015}+58}{145-2 \sqrt{1015}}\right) . tan ⁡ ( a ) + tan ⁡ ( b ) \tan (a)+\tan (b) simplifies to 256 117 \frac{256}{117} .

0 Helpful 1 Interesting 0 Brilliant 1 Confused

Got a proof?

Pi Han Goh - 2 years ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...