Idk how to do this

Never mind what I posted here is no longer relevant.

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

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