a Use the knowledge you have of the relationship of the angles.
B
b Do not forget what the domain of the functions is.
A
a See solution.
B
b See solution.
Practice makes perfect
a The exercise states that the measures of the three angles are m∠ 1=x^(∘), m∠ 2=y_1^(∘), and m∠ 3=y_2^(∘). Let's look at the complementary angles and the supplementary angles separately.
Complementary Angles
If ∠ 1 and ∠ 2 are complementary, then the following is true.
m∠ 1 + m∠ 2 = 90^(∘)
With the substitutions m∠ 1 = x^(∘) and m∠ 2=y_1^(∘), we can write another equation.
x + y_1 = 90^(∘)
To get an equation for y_1 as a function of x, we can rewrite the equation.
y_1 = 90^(∘) - x
Both angles must have a measure greater than 0^(∘).
m∠ 1&: &&0^(∘) < x
m∠ 2&: 0^(∘) < 90^(∘) - x ⇒ &&x < 90^(∘)
Domain&: 0^(∘) < x < 90^(∘)
Supplementary Angles
The exercise states that ∠ 1 and ∠ 3 are supplementary.
m∠ 1 + m∠ 3 = 180^(∘).
With the substitutions m∠ 1 = x^(∘) and m∠ 3=y_2^(∘) we can rewrite this equation.
x + y_2 = 180^(∘)
To get an equation for y_2, we solve the equation for y_2.
y_2 = 180^(∘) - x
Both angles must have a measure greater than 0^(∘).
m∠ 1&: &&0^(∘) < x
m∠ 3&: 0^(∘) < 180^(∘) - x ⇒ &&x < 180^(∘)
Domain&: 0^(∘) < x < 180^(∘)
But we have already know that x<90 and that hasn't changed. Therefore we constrict the domain.
Domain&: 0^(∘) < x < 90^(∘)
b We have two functions y_1 = 90^(∘) - x and y_2 = 180^(∘) - x, both with the domain 0^(∘)
