Question

In an isosceles ∆ABC, if AC = BC and AB2 = 2AC2 , then ∠C = ? (a) 30^0 (b) 45^0 (c) 60^0 (d) 90^0

Answer 1

Answer:

d = 90 deg

Step-by-step explanation:

in ∆ABC  lets draw a perpendicular CD from point C to AB

as AC = BC so CD will bisect AB at D

so AD = BD = AB/2 =

in ∆ACD

Cos∠A = AD/ AC  (base/hypotenuse)

Cos∠A = (AB/2)/AC

Cos∠A = AB/ 2AC

AB^2 = 2 AC^2 given

AB = (√2 * (AC)

Cos∠A = (√2 * (AC) / 2AC

Cos∠A = 1/√2

∠A = 45 deg

∆ABC is isoscelles

∠A = ∠B

so ∠B = 45 ded

∠A + ∠B + ∠C = 180 deg

45 + 45 + ∠C = 180  deg

∠C = 90 deg