If \(0^\circ < x^\circ < 90^\circ\) and \(\sin x = \frac{1}{2}\), then \(\cos x = ?\)

Sinx = opps/hyp = 1/2
2^2 = 1^2 + x^2 x = sqrt(3) = adj
Cosx = adj/hyp = sqrt(3)/2

Answered by wblanco4600 | 2024-06-10

Value of** cos x = ( √3 / 2)** for 0° < x° < 90°.
What are trigonometric function?
" **Trigonometric function **is defined as the relation between sides and angles of a right angled triangle ."
Formula used

(Hypotenuse)² = ( Opposite side)² + ( Adjacent side)²

sinθ = (Opposite side) / (Hypotenuse)
cosθ = (Adjacent side) / (Hypotenuse)

According to the question,
Given,
0° < x° < 90°
sin x = 1 / 2
Substitute the value in the formula we get,

Opposite side = 1
Hypotenuse = 2
(2)² = (1 )² + ( Adjacent side )²
⇒( Adjacent side )² = 4 - 1
⇒ Adjacent side = ±√3
0° < x° < 90°
** cos x** >0
Therefore,
**cos x = √3 / 2 **

Hence, value of** cos x = ( √3 / 2)** for 0° < x° < 90°.
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Answered by Yogeshkumari | 2024-06-16

If sin x = 2 1 ​ and 0 ∘ < x ∘ < 9 0 ∘ , then x = 3 0 ∘ . Therefore, cos x = 2 3 ​ ​ .
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Answered by Yogeshkumari | 2024-08-07