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# Idk how to do this

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Idk how to do this

Oct 24, 2017

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Never mind what I posted here is no longer relevant.

Oct 24, 2017
edited by helperid1839321  Oct 24, 2017
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There are no mods online

ISmellGood  Oct 24, 2017
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nm it didn't need it

helperid1839321  Oct 24, 2017
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Even though ISG didn't need any help on this problem, it might be an interesting one to look at.

Area of ABC  can be expressed as

(1/2)xy sin (70)  =  30

xy  =  60 / sin(70)

So....the area of PQR  =

(1/2)xy sin (110)        subbing for xy, we have

(1/2)[ 60 / sin (70)] * sin(110)  ⇒    note that sin(70)  =  sin(110)  .....so....

Area of  PQR  =  (1/2) 60  =  30

Oct 24, 2017
edited by CPhill  Oct 24, 2017
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I still don't really understand...

ISmellGood  Oct 25, 2017
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Let the base of triangle ABC  be  " x ".

Now, let's draw a height from side  x  to angle  C .

This makes a right triangle with the hypotenuse of  y .

sin 70°  =  opposite / hypotenuse

sin 70°  =  height / y

y sin 70°  =   height

area of triangle ABC  =  (1/2)(base)(height)

area of triangle ABC  =  (1/2)(x)(y sin 70°)       This equals 30 because the problem says so.

Now let's look at triangle  PQR. Let the base be  x  .

This makes its height  =  y sin 110°

And the area of PQR  =  (1/2)(x)(y sin 110°)

Let's say

30  =  (1/2)(x)(y sin 70°)

a  =  (1/2)(x)(y sin 110°)        We want to know what  a  equals.

30 / a  =  [ (1/2)(x)(y sin 70°) ] / [ (1/2)(x)(y sin 110°) ]

30 / a  =  sin 70° / sin 110°

30 / a  =  1

30  =  a

Oct 25, 2017
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Thanks so so sooo much! Just one question, how did you know what sin 70 divided by sin 110 is?

ISmellGood  Oct 26, 2017
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sin(110°) = sin(180° - 70°)  →  sin(180°)*cos(70°) - sin(70°)*cos(180°)  →  0*cos(70°) - sin(70°)*(-1)  → sin(70°)

So sin(70°)/sin(110°) = 1

.

Alan  Oct 26, 2017
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Now that I think about it, 70 and 110 are both 20 degrees from 90, so their sines are equal.

helperid1839321  Oct 26, 2017