In triangle ABC angle a = x Angle b = 3x c = 2x AC= 4 units then find AB , BC and x ( SOLVE BY PYTHAGORAS THEROEM )
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We are given that in △ABC,∠=90
o
tyhe sides of △ABC is given by
AB=x cm,BC=(4x+4)cm &
AC=(4x+5) cm
From the pythagoras theorem,
AC
2
=AB
2
+BC
2
So,
(4x+5)
2
−(x)
2
+(4x+4)
2
16x
2
+40x+25=x
2
+16x
2
+22z−16
∴ 16x
2
+40x+25=17x
2
+32x+16
∴ x
2
−8x−9=0
∴ (x−9)(x+1)=0
∴ x=9 or x=−1
but distance cannot be negative
So, x=9
Now, sides are AB=x=9 cm
BC=4(9)+4=40 cm
AC=4(9)+5=41 cm
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