Question

set up a system of linear equations and solve for the indicated quantities. in a right triangle, one of the acute angles is 16° more than the other acute angle. find the measure of each acute angle. part: 0 / 2 part 1 of 2 the measure of the smaller acute angle is °.

Explanation:

Step1: Set up equations

Let the smaller acute - angle be $x$ and the larger acute - angle be $y$. We know two facts:

1. The sum of the two acute angles in a right - triangle is $90^{\circ}$, so $x + y=90$.
2. One of the acute angles is $16^{\circ}$ more than the other, so $y=x + 16$.

Step2: Substitute the second equation into the first

Substitute $y=x + 16$ into $x + y=90$. We get $x+(x + 16)=90$.
Expand the left - hand side: $x+x + 16=90$.
Combine like terms: $2x+16 = 90$.
Subtract 16 from both sides: $2x=90 - 16=74$.
Divide both sides by 2: $x=\frac{74}{2}=37$.

Step3: Find the value of $y$

Substitute $x = 37$ into $y=x + 16$. Then $y=37+16 = 53$.

Answer:

The measure of the smaller acute angle is $37^{\circ}$ and the measure of the larger acute angle is $53^{\circ}$.