C-10-Q043mediumsingle_mcq

$\sin 60^\circ = ?$

a $\dfrac{1}{2}$
b $\dfrac{\sqrt{3}}{2}$
c $\dfrac{1}{\sqrt{2}}$
d $1$

ব্যাখ্যা

$\sin 60^\circ = \dfrac{\sqrt{3}}{2}$ is a standard angle value, the ratio of the opposite side to the hypotenuse in a $30^\circ$ – $60^\circ$ – $90^\circ$ triangle.

এই অধ্যায়ের আরও প্রশ্ন

C-10-Q044
C-10-Q045
C-10-Q046
C-10-Q047With the height of a tower, m the horizontal distance, and , the equation equals—
C-10-Q048Therefore the height satisfies—
C-10-Q049Numerically equals—
C-10-Q050After rationalizing, becomes—
C-10-Q051The approximate height of the tower in Example 1 is—
C-10-Q052To rationalize , the numerator and denominator are multiplied by—
C-10-Q053In Example 2, with tree m, foot distance , and , equals—

Chapter 10: Distance and Elevation — সব প্রশ্ন দেখুন (160টি) →

$\sin 60^\circ = ?$

\dfrac{1}{2}

\dfrac{\sqrt{3}}{2}

\dfrac{1}{\sqrt{2}}

1

ব্যাখ্যা

$\sin 60^\circ = \dfrac{\sqrt{3}}{2}$ is a standard angle value, the ratio of the opposite side to the hypotenuse in a $30^\circ$ – $60^\circ$ – $90^\circ$ triangle.

sin⁡60∘=?\sin 60^\circ = ?sin60∘=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

sin⁡60∘=?\sin 60^\circ = ?sin60∘=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

$\sin 60^\circ = ?$

$\sin 60^\circ = ?$