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.
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?
2 answers
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.
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.
1 answer