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If cos Θ = negative two over five and tan Θ > 0, what is the value of sin Θ? negative square root / 21 negative square root / 21 over 5 square root / 21 over 5 square root of 21 over 2

2 Answers

4 votes

Answer:


sin\theta=-(√(21))/(5) Option B is the correct option.

Explanation:

It is given in the question
cos\theta =-(2)/(5)
and
tan\theta >0

We have to calculate the value of
sin\theta

We can see in the figure attached since cos∅ = -2/5

Here negative sign is showing that angle theta is lying either in quadrant 2nd or 3rd.

And
tan\theta >0
is showing as angle is either in quadrant 1 or 3rd.

Therefore it is confirm that angle is lying in quadrant 3 where sine is always negative.

Therefore
sin\theta=-(√(21))/(5)
will be the answer.

If cos Θ = negative two over five and tan Θ > 0, what is the value of sin Θ? negative-example-1
User Dylan El Bar
by
8.5k points
5 votes

Answer: The correct option is (B) negative square root 21 over 5.

Step-by-step explanation: Given that for an angle
\theta,


\cos \theta=-(2)/(5),~~~\tan \theta>0.

We are to find the value of
\sin \theta.

We know that

Cosine is negative in Quadrant II and Quadrant III.

Tangent is positive in Quadrant 1 and Quadrant III.

So, the given angle
\theta lies in the Quadrant III.

We will be using the following trigonometric identity:


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

We have


\cos \theta=-(2)/(5)\\\\\\\Rightarrow √(1-\sin^2\theta)=-(2)/(5)\\\\\\\Rightarrow 1-\sin^2\theta=(4)/(25)\\\\\\\Rightarrow \sin^2\theta=1-(4)/(25)\\\\\\\Rightarrow \sin^2\theta=(21)/(25)\\\\\\\Rightarrow \sin \theta=\pm(√(21))/(5).

Since the angle
\theta lies in Quadrant III and sine is negative in that quadrant, so


\sin\theta=-(√(21))/(5).

Thus, option (B) is correct.

User Masterfu
by
8.5k points