Question

In a right angled triangle ABC, right angled at B, AB=x/2 , BC=x+2 and AC =x+3 . The value of x is​

Answer 1

Step-by-step explanation:

AB² = AC² + BC² - 2(AC)(BC)cos c

(x + 2)² = (x - 1)² + x² - 2x(x - 1)cos 120

(x + 2)² = (x - 1)² + x² - 2x(x - 1)cos (180 - 60)

(x + 2)² = (x - 1)² + x² - 2x(x - 1)(-cos 60)

(x + 2)² = (x - 1)² + x² + 2x(x - 1) cos 60

(x + 2)² = (x - 1)² + x² + 2x(x - 1)(1/2)

(x + 2)² = (x - 1)² + x² + x(x - 1)

x² + 4x + 4 = (x² - 2x + 1) + x² + x² - x

2x² - 7x - 3 = 0

D = (-7)² - 4(2)(-3) = 73

x = (7 ± √73)/4

Answer 2

Answer:

10

Step-by-step explanation:

(x+3)^2=(x/2)^2+(x+2)^2{by using Pythagoras theorem}

on solving this we will get

x=10cm