Question

∠a and ∠b are complementary. find m∠a and m∠b

1. ( mangle a = x^circ ) ( mangle b = (x - 30)^circ )
2. ( mangle a = (5x + 4)^circ ) ( mangle b = (7x - 10)^circ )
3. ( mangle a = (4x - 2)^circ ) ( mangle b = (11x + 17)^circ )
4. ( mangle a = (6x - 9)^circ ) ( mangle b = (8x + 1)^circ )

Explanation:

Let's solve problem 28 first (we can solve others similarly as they follow the same logic of complementary angles, i.e., sum to \(90^\circ\)).

Problem 28:

Step 1: Recall complementary angles property

Complementary angles sum to \(90^\circ\), so \(m\angle A + m\angle B = 90^\circ\). Substitute \(m\angle A = x^\circ\) and \(m\angle B=(x - 30)^\circ\):
\(x+(x - 30)=90\)

Step 2: Solve for \(x\)

Simplify the left - hand side: \(2x-30 = 90\).
Add 30 to both sides: \(2x=90 + 30=120\).
Divide both sides by 2: \(x=\frac{120}{2}=60\)

Step 3: Find \(m\angle A\) and \(m\angle B\)

\(m\angle A=x^\circ = 60^\circ\)
\(m\angle B=(x - 30)^\circ=(60 - 30)^\circ = 30^\circ\)

Problem 29:

Step 1: Use complementary angles property

\(m\angle A+m\angle B = 90^\circ\), substitute \(m\angle A=(5x + 4)^\circ\) and \(m\angle B=(7x-10)^\circ\):
\((5x + 4)+(7x-10)=90\)

Step 2: Solve for \(x\)

Simplify the left - hand side: \(12x-6 = 90\).
Add 6 to both sides: \(12x=90 + 6 = 96\).
Divide both sides by 12: \(x=\frac{96}{12}=8\)

Step 3: Find \(m\angle A\) and \(m\angle B\)

\(m\angle A=(5x + 4)^\circ=(5\times8 + 4)^\circ=(40 + 4)^\circ = 44^\circ\)
\(m\angle B=(7x-10)^\circ=(7\times8-10)^\circ=(56 - 10)^\circ = 46^\circ\)

Problem 30:

Step 1: Apply complementary angles rule

\(m\angle A+m\angle B = 90^\circ\), substitute \(m\angle A=(4x - 2)^\circ\) and \(m\angle B=(11x + 17)^\circ\):
\((4x-2)+(11x + 17)=90\)

Step 2: Solve for \(x\)

Simplify the left - hand side: \(15x + 15 = 90\).
Subtract 15 from both sides: \(15x=90 - 15 = 75\).
Divide both sides by 15: \(x=\frac{75}{15}=5\)

Step 3: Find \(m\angle A\) and \(m\angle B\)

\(m\angle A=(4x - 2)^\circ=(4\times5-2)^\circ=(20 - 2)^\circ = 18^\circ\)
\(m\angle B=(11x + 17)^\circ=(11\times5 + 17)^\circ=(55 + 17)^\circ = 72^\circ\)

Problem 31:

Answer:

s:

• Problem 28: \(m\angle A = 60^\circ\), \(m\angle B = 30^\circ\)
• Problem 29: \(m\angle A = 44^\circ\), \(m\angle B = 46^\circ\)
• Problem 30: \(m\angle A = 18^\circ\), \(m\angle B = 72^\circ\)
• Problem 31: \(m\angle A = 33^\circ\), \(m\angle B = 57^\circ\)