What is $sin(90-θ)$ and $cos(90-θ)$?

Solution :

In the given triangle ABC,

$\angle B=90°$, $C = \theta$.

We know that the sum of all the angles in a triangle is 180°.

$\angle A+ \angle B + \angle C=180°$

$\angle A+ 90°+ \theta = 180°$

$\angle A = 180° - (90°+ \theta)$

$\angle A = 180° - 90°- \theta$

$\angle A = 90°- \theta$

$\angle C = \theta$

$\angle A + \angle C = 90°$

Therefore, $\angle A$ and $\angle C$ are complementary.

$sin C = sin \theta = \frac{AB}{AC} $

$cos C = cos \theta = \frac{BC}{AC} $

$sin A = sin (90-\theta) = \frac{BC}{AC}= cos C = cos \theta $

$cos C = cos (90-\theta) = \frac{AB}{AC}=sin C = sin \theta  $

Therefore, $sin (90-\theta) $ is $cos \theta$ and $cos (90-\theta)$ is $sin \theta$.

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Updated on: 10-Oct-2022

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