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# trig problem

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If A is an acute angle such that tan A + sec A = 3, then find cos A.

Dec 1, 2020

#1
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tan A  +  sec A  = 3

sin A            1

_____ +   _____   = 3

cos A        cos A

sin A  +  1

_________ =    3

cos  A

sin A + 1  = 3 cos A          square both sides

sin^2 A + 2sin A  +  1  =  9cos^2A

sin^2 A  + 2sinA  + 1  = 9 ( 1 -sin^2A)

sin^2 A  + 2sin A +  1  =  9 - 9sin^2 A      simplify as

10sin^2A  + 2sin A  - 8    = 0    divide through by 2

5sin^2 A  + sina  -  4 =    0      factor as

(5sinA  - 4) ( sin A + 1)   = 0

Setting each factor to 0 and solving for A  gives us that

sin A = 4/5       or   sin A   =  - 1

If sinA  = -1....A =  270°...but the tangent  and sec are undefine at the angle

So

sin A =4/5

And cos A   =  3/5

.

Dec 2, 2020
edited by CPhill  Dec 2, 2020
#2
+1

If A is an acute angle such that tan A + sec A = 3, then find cos A.

tanA ==> x

x + √(x2 + r2) = 3     ==>   x = 1.333333333     or     4/3

angle A = tan-1(4/3) = 53.13010235º

cos A = 3/5

Dec 2, 2020