Which of the following measurements could be the side lengths of a right triangle? A. 66 in, 88 in, 110 in

1 answer:

5 0

So the way to figure out the sides of a right triangle, is basically

A^2 + B^2 = C^2

Letter A is the only one that would work, because the first number squared (times itself), added to the second number squared, would equal the third number squared. This is a basic principle of the pythagorean theorem, which I can explain in more detail if you'd like. :)

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If cosine theta equals negative square root of three over two and pi over two less than theta less than pi, what are the values

Natalka [10]

<h2>Answer:</h2>

The values of sinθ and tanθ are:

$\sin \theta=\dfrac{1}{2}$

and

$\tan \theta=-\dfrac{1}{\sqrt{3}}=-\dfrac{\sqrt{3}}{3}$

<h2>Step-by-step explanation:</h2>

It is given that:

$\cos \theta=-\dfrac{\sqrt{3}}{2}$

where

$\dfrac{\pi}{2}$

i.e. the angle θ lie in the second quadrant.

Also, we know that in the second quadrant sine and cosecant function is positive.

whereas cosine,secant, tangent and cotangent functions are negative.

Now, we know that:

$\cos (\dfrac{5\pi}{6})=-\dfrac{\sqrt{3}}{2}$

i.e. here

$\theta=\dfrac{5\pi}{6}$

Hence, we have:

$\sin (\theta)=\sin (\dfrac{5\pi}{6})\\\\i.e.\\\\\sin \theta=\dfrac{1}{2}$

Also,

$\tan \theta=\dfrac{\sin \theta}{\cos \theta}\\\\i.e.\\\\\tan \theta=\dfrac{\dfrac{1}{2}}{\dfrac{-\sqrt{3}}{2}}\\\\i.e.\\\\\tan \theta=\dfrac{1}{-\sqrt{3}}\\\\i.e.\\\\\tan \theta=-\dfrac{1}{\sqrt{3}}$