The distance of the red supergiant Betelgeuse is approximately 643 light-years. If it were to explode as a supernova, it would be one of the brightest stars in the sky. Right now, the brightest star other than the Sun is Sirius, with a luminosity of 26LSun and a distance of 8.6 light-years. How much brighter than Sirius would the Betelgeuse supernova be in our sky if it reached a maximum luminosity of 1x 10^10 LSun?
Express your answer using three significant figures. Apparent brightness of Betelgeuse Apparent brightness of Sirius

Step-by-step explanation:

Hey, there!!

Let's Simply solve it, alright.

Let's look back in "CAST rule".

In cast rule, you should learn various things.It carries vast knowledge. The most important thing changing of ratios and sign like plus and minus.

It's very difficult to let you understand in language here.

Fistly let me introduce you about "CAST"

C = cos, sec

A= all

S = sin, cosec

T = Tan, cot

As i have shown in figure, the vertical and horizontal lines, vertical lines are always odd and horizontal lines are always even here. And even ratios donot change but if they are odd then the ratios changes. Likewise, if the ratios are not in their quadrant then, their sign changes.

They change their ratios in

So, at first before understanding let me solve your question, to make you clear so that you will not have problem on solving again,

your question is : sin(-1200)

now,

Divide 1200/90

you will get 13.3333

or, you can say it, (13 × 90 +30) { because 13×90=1170 plus 30= 1200}.

After this step, it will be like,

sin(13×90+30)

As 13 is odd it changes so, its ratio changes into cos and it lies in 4th quadrant, and it is the right quadrant of cos and sec, so, its sign donot change,

That means,

Therefore, sin(-1200)= root 3/ 2.

Hope it helps...