3070.11 – Algebra with Angles

If twice ∠A\angle A is subtracted from the supplement of ∠A\angle A, then the remaining angle exceeds the complement of ∠A\angle A by 4 degrees. Find the size of ∠A\angle A.

Solution

Remember that ∠A\angle A and its supplement ∠S\angle S always sum to 180 degrees, and ∠A\angle A plus its complement ∠C\angle C make 90 degrees. Thus,

∠S=180−∠A,∠C=90−∠A.\begin{aligned} \angle S &=& 180 - \angle A, \\ \angle C &=& 90 - \angle A.\\ \end{aligned}

Using the information given in the problem, we know that:

∠S−2∠A=∠C+4.\angle S - 2\angle A = \angle C + 4.

Substituting and solving algebraically, we get:

(180−∠A)−2∠A=(90−∠A)+4,180−3∠A=94−∠A,86=2∠A,∠A=42.\begin{aligned} (180 - \angle A) - 2\angle A &=& (90 - \angle A) + 4, \\ 180 - 3\angle A &=& 94 - \angle A, \\ 86 &=& 2\angle A, \\ \angle A &=& 42 . \end{aligned}

We have arrived at the solution.