Posted by **HAYDEN4747** on Wednesday, August 31, 2011 at 8:48pm.

FIND THE SOLUTION FOR THE EQUATION. ASSUME THAT ALL ANGLES IN WHICH AN UNKNOWN APPEARS ARE ACUTE ANGLES.

COT ALPHA=TAN( ALPHA +40DEGREES)

HOW DO I SOLVE THIS?

- MATH -
**Reiny**, Wednesday, August 31, 2011 at 9:19pm
I will use A for alpha

cot A = tan(A + 40)

1/tanA = (tanA + tan40)/(1 - tan^2A)

let tanA = x

1/x = (x + .8391)/(1-x(.8391))

x^2 + .8391x = 1 - .8391x

x^2 + 1.6782x - 1 = 0

by the formula

x = (-1.6782 ± √6.816355)/2 = .4663 or -2.1445

so tanA = .4663, then A = 25° or 205°

if tanA = -2.1445 , A = 115° or 295°

I will check the 205°

LS = cot 205° = 2.1445

RS = tan(205+40) = tan 245 = 2.1445

YEAHHH!

I will let you check the others.

- MATH -
**Reiny**, Wednesday, August 31, 2011 at 11:06pm
Just came back from a walk, and I kept thinking about this question and why the answer came out to rather exact values of the angle.

Then I realized that I could have used the complementary angle property, that is,

cot A = tan(90°-A)

so in

cotA = tan(A+40)

tan(90-A) = tan(A+40)

so obviously

90-A = A + 40

50 = 2A

**A = 25 !!!!!!!!!!!!**

and of course since tanA = tan(180-A)

A is also equal to 180-25 = 115°

and since the period of tanA is 90° , adding or subtracting multiples of 90 would yield more answers

so 25+90 = 115

115+90 = 205

205+90 = 295

which is the same answer as the complicated solution I had before.

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