#4**+3 **

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

CPhill
Oct 24, 2017

#6**+2 **

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

hectictar
Oct 25, 2017

#7**+2 **

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

ISmellGood
Oct 26, 2017

#8**+2 **

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

#9**+2 **

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

helperid1839321
Oct 26, 2017