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# trig question

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In right triangle ABC, with B = 90 degrees, we have sin A = 2 cos A + tan A. What is tan A?

Apr 16, 2022

#1
+23250
+1

Since no one else has posted an answer, I'll post what I have ...

Using 'a' as the side opposite angle(A), 'b' as the side opposite angle(B), and 'c' as the side opposite angle(C):

Since angle(B) is a right angle:  sin(A) = a/b     cos(A) = c/b     tan(A) = a/c     b2 = a2 + c2

sin(A)  =  2·cos(A) + tan(A)     --->     a/b  =  2(c/b) + a/c

multiplying by  bc          --->      ac  =  2c2 + ab

Solving for  c  by using the quadratic formula:  ac  =  2c2 + ab

2c2 - ac + ab  =  0

c  =  [ a +/- sqrt( a2 - 8ab ) ] /4

Since the answer is a real number:  a2 - 8ab  >= 0

a2  >=  8ab

Since  a  must be positive:                         a  >=  8b

which means that  a  must be larger than  b  but  b  is the hypotenuse!

So, either the problem, as stated, is impossible or I messed up!  Help!

Apr 16, 2022

#1
+23250
+1

Since no one else has posted an answer, I'll post what I have ...

Using 'a' as the side opposite angle(A), 'b' as the side opposite angle(B), and 'c' as the side opposite angle(C):

Since angle(B) is a right angle:  sin(A) = a/b     cos(A) = c/b     tan(A) = a/c     b2 = a2 + c2

sin(A)  =  2·cos(A) + tan(A)     --->     a/b  =  2(c/b) + a/c

multiplying by  bc          --->      ac  =  2c2 + ab

Solving for  c  by using the quadratic formula:  ac  =  2c2 + ab

2c2 - ac + ab  =  0

c  =  [ a +/- sqrt( a2 - 8ab ) ] /4

Since the answer is a real number:  a2 - 8ab  >= 0

a2  >=  8ab

Since  a  must be positive:                         a  >=  8b

which means that  a  must be larger than  b  but  b  is the hypotenuse!

So, either the problem, as stated, is impossible or I messed up!  Help!

geno3141 Apr 16, 2022