Two angles are complementary if the sum of their measures is exactly $90^\circ$.
Let the measure of one angle be $x$. Since the angles are complementary, the measure of the other angle is $90^\circ - x$.
The problem states that the measure of an angle is three times the measure of its complement. We can write this relationship as an equation:
$x = 3 \times (90^\circ - x)$
We found one angle to be $x = 67.5^\circ$.
The measure of its complement is $90^\circ - x = 90^\circ - 67.5^\circ = 22.5^\circ$.
So, the two angles are $22.5^\circ$ and $67.5^\circ$.
We can check: Is $67.5^\circ$ three times $22.5^\circ$? Yes, $3 \times 22.5^\circ = 67.5^\circ$.
This matches option B.
What is the sum of the angle complementary to $15^\circ$ and the angle supplementary to $125^\circ$?
In the given figure, AB and CD are parallel lines. O is a point such that angle CDO = $70^\circ$ and angle DOB = $100^\circ$. Find angle ABO.
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In the triangle, if AB = AC and ∠ABC = 72°, then ∠BAC is:
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In ΔABC, D is a point on side BC such that ∠ADC = 2∠BAD. If ∠A = 80° and ∠C = 38°, then what is the measure of ∠ADB?