Question

In ΔABC, ∠B = 90, find the measure of the parts of the triangle other than the ones which are given:AC=6√ 2, BC=3√ 6.

Answer 1

Let AB = c, AC = b and BC = a (general notations)

By pythagoras theorem, a² + c² = b² or (AB)² + (BC)² = (AC)²,

c² = 36*2 - 9*6 = 72 - 54 = 18

c = √18 = 3√2

Also ∠C can be found using tanC

tan C = AB/BC = 3√2 / 3√6 = 1/√3

so ∠C = 30°

By angle sum property, ∠B = 180-90-30 = 60°

Answer 2

In ∆ABC ,

<B = 90° ,

AC = 6√2 , BC = 3√6 ,

By Phythogarian theorem ,

AC² = BC² + AB²

=> AB² = AC² - BC²

=> AB² = ( 6√2 )² - ( 3√6 )²

=> AB² = 72 - 54

=> AB² = 18

=> AB = √18

=> AB = 3√2

In ∆ABC ,

ii ) sin C = AB/AC

= 3√2/6√2

= 1/2

SinC = 1/2

=> SinC = sin 30°

C = 30°

iii ) CosA = AB/AC

= 3√2/6√2

= 1/2

=> CosA = Cos 60°

=> A = 60°

Therefore ,

AB = 3√2 ,

<C = 30° , <A = 60°

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