Question

If the outer angles nat three vertices of a triangle are (2x+30)+(3x-10)+100. what is the value of x​? find all the angles?

Answer 1

Step-by-step explanation:

The sum of all interior angles of a triangle is always 180 degrees

So the above angles will sum up to 180 i.e

(2x+30) + (3x-110) + (5/2x+20) = 180

->By taking LCM of above equation

(2x+30) (2x+20)+ (3x-110) (2x+20)+ 5 = 180(2x+20)

4x.x + 40x + 60x + 600 + 6x.x + 60x - 220x - 2200 + 5 = 360x + 3600

10x.x - 60x - 2195 = 360x + 3600

Now, reducing above equation

10x.x - 420x - 5795 = 0

Which is a linear equation. Solving using

x = (−b ± √(b2 − 4ac))/2a

Where a= 10, b= -420 and c= -5795

x= [-(-420) - √((-420).(-420)-4(10)(-5795))]/2(10)

x=[420 - √(176400)-(231800))]/20

x=[420 - √(-55400)]/20

Since √(-1) = i (iota) gives √(-55400)=235i (approx)

x=[420 - 235i]/20

Hence x = 21 - 11.35i

Answer 2

Answer:

x=12

Step-by-step explanation:

. 2x+30+3x-10+100=180

:. 2x+3x+30-10+100=180

:. 5x+20+100=180

:. 5x+120=180

:. 5x=60

:. x=12