One of the 2 complementary angles is 6 less than twice the other.

Find the measure of the larger angle.

$58^{\circ}$

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$38^{\circ}$

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$32^{\circ}$

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$52^{\circ}$

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Solution

The correct option is A
$58^{\circ}$

Let the complementary angles be $x^{\circ} a n d (90 - x)^{\circ}$

$\Rightarrow x = 2 (90 - x) - 6$

$\Rightarrow x = 180 - 2 x - 6$

$\Rightarrow x + 2 x = 180 - 6$

$\Rightarrow 3 x = 174$

$\Rightarrow x = 58$

$\Rightarrow (90 - x) = 90 - 58 = 32 °$

$∴ L a r g e r a n g l e = 58^{\circ}$

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