Question

If 3y –15, 4y –20 and 5y –25 are three angles of a triangle, find each of them.

Answer 1

Step-by-step explanation:

let <A = 3y-15° ,<B = 4y-20° ,<C = 5y-25°

By angle some property.

sum of all angles óf triangle is 180°.

<A+<B+<C = 180°

3y-15°+4y-20°+5y-25°= 180°

12y-60°= 180°

12y = 180°+60°

12y = 240°

y = 20°

So,

<A = 3y-15° = 3×20°-15° = 60°-15° = 45°

<B = 4y-20° = 4×20°-20° = 80°-20° = 60°

<C = 5y-25° = 5×20°-25° = 100°-25° = 75°

Answer 2

Answer:

Here is u r answer

Step-by-step explanation:

By angle sum of a triangle

(3y –15)+(4y –20) + (5y –25)=180 degree

3y-15+4y-20+5y-25=180

12y-60=180

12y=180+60

12y=240

y=240/12

=20

Therefore angles are

3y-15=3×20-15=60-15=45 degree

4y-20=4×20-20=80-20=60 degree

5y-25=5×20-25=100-25=75 degree

I hope it will help u

And dont forget to mark me as brainlist❤ ꈍᴗꈍ