1. m∠QPR = 2. m∠QTP = 3. RP = Show all the work

Mathematics

1 answer:

alex41 [277]3 years ago

7 0

Answer:

1. 30°

2.90°

3. 12 units

Step-by-step explanation:

I can't believe there's nothing confirming that this is a parallelogram/a rhombus?! Assuming is awful, and I wish you could say you can't know for sure lol but for the sake of this answer, let's just call it a rhombus. (There was probably some context elsewhere that you didn't put over here, hopefully.)

The reason I say this is: in a rhombus, the diagonals bisect the angles. This means that the diagonals split the angles they meet into two equal parts. That way, it would make sense. m∠QPR=m∠SPR=30°.

If it is a rhombus, the diagonals are perpendicular to each other, so m∠QTP should be 90°.

Diagnonals in a rhombus (and in any parallelogram) bisect each other, so PT=TR=6, and RP=PT+TR=12 units.

Sorry if this is all dreadfully wrong, and I hope I helped you!

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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,

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