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3 years ago

Would this angle be name angle CUE or CUT ?

Mathematics

1 answer:

Goshia [24]3 years ago

6 0

Answer:

$\huge\boxed{\angle\text{CUE}}$

Step-by-step explanation:

We can model what both readings of this angle would be. Angle names are usually formed by taking the two lines that form to intersect this angle at a vertice, and these 3 make up the name.

Angle ∠CUT would be formed by the lines CU and UT. We can see that this is not the case, as the angle is formed by lines CU and CE.

Since they intersect at point U, that means the angle name would be CUE.

Hope this helped!

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Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

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Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

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

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

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

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

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3 years ago

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