If one of the complementary angles is twice the other, the two angles are
a) $60^o, 30^o$
b) $20^o, 60^o$
c) $40^o, 40^o$
d) $10^o, 70^o$
Given:
One of the complementary angles is twice the other.
To do:
We have to find the two angles.
Solution:
Let the angles be $x$ and $2x$.
We know that,
Sum of two complementary angles is $90^o$.
This implies,
$x + 2x = 90^o$
$3x = 90^o$
$x=30^o$
The two angles are $30^o$ and $2(30)^o=60^o$.
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