If $0 < x < 45 °$ and $45 ° < y < 90 °$ , then which one of the following is correct?

$\sin x = \sin y$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\sin x < \sin y$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\sin x > \sin y$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\sin x \leq \sin y$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$\sin x < \sin y$

Explanation for the correct option:
Comparing $s i n x$ and $s i n y$
As it is given that $0 < x < 45 °$ , and we know that
$s i n 0 ° = 0 s i n 30 ° = \frac{1}{2}$
Similarly, it is given that $45 ° < y < 90 °$
$s i n 60 ° = \frac{\sqrt{3}}{2}$
So, we can observe that
$\sin 0 ° < \sin 30 ° < \sin 60 °$
$\Rightarrow \sin x < \sin y$
Hence, the correct answer is option B.

Suggest Corrections

Similar questions

tan 30^{0} = x, tan 45^{0} = y, tan 60^{0} = z,

s i n (x - y) = \frac{1}{2}

s i n (x + y) = \frac{\sqrt 3}{2}

0 < y < x < 90^{\circ}

sin θ = cos (2 θ - 45^{\circ}), 0 < (2 θ - 45^{\circ}) < 90^{\circ},

tan θ

2 {sin}^{2} θ + {cos}^{2} 45^{0} = tan 45^{0} a n d 0 \leq θ \leq 90^{0},

θ

If $c o t 3 x = s i n 45^{\circ} c o s 45^{\circ} + c o s 60^{\circ}$ and $0 < x \leq 90^{\circ}$ , then $x$ = __________________

Join BYJU'S Learning Program