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MatroZZZ [7]
3 years ago
11

Given: y ll z

Mathematics
2 answers:
emmainna [20.7K]3 years ago
7 0

Answer:

As per the properties of parallel lines and interior alternate angles postulate, we can prove that:

m\angle 5+m\angle 2+m\angle 6=180^\circ

Step-by-step explanation:

<u>Given:</u>

Line y || z

i.e. y is parallel to z.

<u>To Prove:</u>

m\angle 5+m\angle 2+m\angle 6=180^\circ

<u>Solution:</u>

It is given that the lines y and z are parallel to each other.

m\angle 5, m\angle 1 are <em>interior alternate angles </em>because lines y and z are parallel and one line AC cuts them.

So, m\angle 5= m\angle 1 ..... (1)

Similarly,

m\angle 6, m\angle 3 are <em>interior alternate angles </em>because lines y and z are parallel and one line AB cuts them.

So, m\angle 6= m\angle 3 ...... (2)

Now, we know that the line y is a straight line and A is one point on it.

Sum of all the angles on one side of a line on a point is always equal to 180^\circ.

i.e.

m\angle 1+m\angle 2+m\angle 3=180^\circ

Using equations (1) and (2):

We can see that:

m\angle 5+m\angle 2+m\angle 6=180^\circ

<em>Hence proved.</em>

Finger [1]3 years ago
3 0

Answer:

Step-by-step explanation:

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General solutions of sin(x-90)+cos(x+270)=-1<br> {both 90 and 270 are in degrees}
mixer [17]

Answer:

\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

Step-by-step explanation:

Given:

\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1

First, note that

\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x

So, the equation is

-\cos x+\sin x= -1

Multiply this equation by \frac{\sqrt{2}}{2}:

-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}

The general solution is

x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.

4 0
3 years ago
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