Question

i) In AABC, ABC = 120°, AB = BC, seg BD I side AC, A-D-C. Prove that AC =V3AB​

Answer 1

Answer:

AC = 3AB = √3BC

Step-by-step explanation:

given that in a triangle ∠ABC = 120°

and AB = BC.

when two sides are equal it is a isosceles triangle

which states two angles are also equal

=> now let AC = y and AB = BC = x

side of Δle is given by the formula

c = √(a²+b²-2abcosФ)

where c = AB, a = BC, b = AC and Ф is the angle between the side BC and AC

now AC = √(AB²+BC²-2.AB.BC.cosФ)

=> AC = √(x²+x²-2.x.x.cos120°)

=>AC = √(2x²+2sin30°)

=>AC = √(3x²)

=> AC = √3BC = √3 AB