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2 years ago

Help me pls i’m gonna fail the class>3

Mathematics

1 answer:

Fed [463]2 years ago

3 0

What’s the question ?

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3 years ago

The volume of a cube is 27 cubic inches. What is the length of one edge of the cube?

hjlf

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2 years ago

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hram777 [196]

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2 years ago

Read 2 more answers

Someone please help me with this lol… have no idea what I’m doing

Sholpan [36]

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

$\sin^2\theta=1-\cos^2\theta$

$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

$\sin \theta=-\sqrt{1-\dfrac{9}{25}}$

$\sin \theta=-\sqrt{\dfrac{25-9}{25}}$

$\sin \theta=-\sqrt{\dfrac{16}{25}}$

$\sin \theta=-\dfrac{4}{5}$

Therefore, the correct option is B.

6 0

2 years ago