Question 1113953
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Angles A and B are acute angles such that tanA+tanB+(tanA)(tanB)=1. If A-B=41°, find the measure of angle A.
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It is nice non-standard Trigonometry problem.
I will show you how to solve it.


<pre>
You are given

tan(A) + tan(B) + (tan(A))*(tan(B)) = 1.


It implies

tan(A) + tan(B) = 1 - (tan(A))*(tan(B)),   and, after dividing both sides by the right side


{{{(tan(A) + tan(B))/(1-tan(A)*tan(B))}}} = 1.


The left side is nothing else as  tan(A+B),  according to the adding formula for tangent function.  So your equation takes the form

tan(A+B) = 1,

and, since the angles A and B are acute, it implies that

A + B = 45  degrees.


So, now you have two equations:

A + B = 45      (1)
A - B = 41      (2)
------------------------- Add the equations to eliminate B. You will get


2A = 45 + 41 = 86  ====>  A = {{{86/2}}} = 43.


Thus the angle A = 43°.

Then from equation (1),  B = 45° - 43° = 2°.


<U>Answer</U>.  Angle A = 43°.   Angle B = 2°.
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Solved.