In ΔABC, m∠B = 90, find the measure of the parts of the triangle other than the ones which are given below:

(1) m∠C = 45, AB = 5

(2) m∠A = 30, AC = 10

(3) AC = 6√2, BC = 3√6

(4) AB = 4, BC = 4

(1) If ∠C = 45 and ∠B = 90

⇒ ∠A = 45 (angle sum property of a triangle)

∠A = ∠C

⇒ AB = BC (sides opposite to equal angles)

and since, AB = 5

⇒ BC = 5

and by Pythagoras theorem

AB² + BC² = AC²

⇒ 5² + 5² = AC

⇒ AC = 5√2

(2) if ∠A = 30 and ∠B = 90

⇒ ∠C = 60 (Angle sum property of triangle)

put the known values

also,

(3)

⇒ ∠C = 30

⇒ ∠A = 60 (by angle sum property)

also, by Pythagoras theorem

AB² = AC² – BC²

⇒ AB = 3√2

(4) Since ∠B = 90

⇒ AB² + BC² = AC²

substituting AC = 4 and BC = 4

we get

AC = 4√2

Also since, AB = BC

⇒ ∠A = ∠C (angles opposite to equal sides of the same triangle)

∵ ∠B = 0

⇒ ∠A = ∠C = 45 (Angle sum property)