Question

In ΔABC, ∠B = 90, find the measure of the parts of the triangle other than the ones which are given:m∠C=45, AB=5.

Answer 1

SOLUTION :

Given :

In ∆ ABC ,∠B = 90°, ∠C = 45°, AB= 5

∠A + ∠B + ∠C = 180°

∠A + 90° + 45° = 180°

135° + ∠A = 180°

∠A = 180° - 135°

∠C = 45°

∠A = ∠C = 45°

BC = AB

[Sides opposite to equal angles are equal]

For sinC , base = BC , Perpendicular = AB, hypotenuse = AC

sinC = Perpendicular/ hypotenuse = AB/AC

sin 45 = AB/AC

1/√2 = 5/AC

AC = 5√2

Hence, ∠C = 45° , AC = 5√2

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

It is given that ,

In ∆ABC , <B = 90° ,

<C = 45° , AB = 5 ,

i ) Sin 45° = AB/AC

1/√2 = 5/AC

AC = 5√2

ii ) Cos 45 ° = BC/AC

1/√2 = BC/5√2

( 5√2/√2 ) = BC

5 = BC

BC = 5

Therefore b,

In ∆ABC ,

AB = 5 ,

BC = 5 ,

AC = 5√2

Or

In ∆ABC , <B = 90° , <C = 45° ,

<A = 45°

Therefore , ∆ABC is right angled

isosceles triangle.

AB = BC = 5

[ Sides opposite to equal angles are

equal ]

AC² = AB² + BC²

AC² = 5² + 5²

AC² = 2 × 25

AC = √ 2 × 25

AC = 5√2

Therefore ,

AB = BC = 5 ,

AC = 5√2

I hope this helps you.

: )