Question

Plz give correct answer in step by step explanation plz...correct answer will be marked BRAINLIEST and be followed....✌

Answer 1

Given, AM=BM=CM

therefore, ∆AMC and ∆MCB are isosceles.

since, angle CAM = Angle MCA,

angle MBC = MCB

Now in (I) case,

we have, angle CAM = 60°

angle CAM = angle MCA = 60° (given)

In ∆CMA,

angle( CAM + AMC + ACM ) = 180°

120° + angle CMA = 180°

angle CMA = 60°

angle( CMA + CMB ) = 180°. (linear pair)

angle CMB = 120°

In ∆MCB,

angle( MCB + MBC + CMB ) = 180°

2 angle MCB = 180 - 120

angle MCB = 30°

angle ACB = angle( ACM + MCB)

angle ACB = 60 + 30

angle ACB = 90°

Second case,

when, AMC = 36°

In∆AMC,

angle( AMC + CAM + MCA ) = 180°

2 angle MCA = 180

angle MCA = 90°

angle( AMC + CMB) = 180°. { linear pair}

angle CMB = 144°

In∆CMB,

angle( MCB + MBC + CMB ) = 180°

2 angle MCB = 180 - 144

2 angle MCB = 36

angle MCB = 18°

angle ACB = angle( MCA + MCB)

angle ACB = 90 + 18

angle ACB = 108°

third case,

angle BMC = 100°

In ∆ MBC,

angle( MBC + MCB+ CMB) = 180

2 MCB = 180 - 100

Angle MCB = 40°

angle CMA = 180 - 100 { linear pair }

angle CMA = 80°

In ∆ CMA,

angle ( MCA + CMA + CAM) = 180°

2 MCA = 180 - 80

angle MCA = 50°

angle ACB = angle ( ACM + BCM}

angle ACB = 50 + 40

angle ACB = 90°

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