Question

In ΔABC, ∠B = 90, find the measure of the parts of the triangle other than the ones which are given:AB=4, BC=4.

Answer 1

Here we can have 2 ways of approach :

a) By angle sum property and pythagoras theorem b) By applying tan⁻¹ and pythagoras theorem.

Method 1 :

Angles opposite to equal sides of a triangle are equal. So ∠A = ∠C By angle sum property; ∠A + ∠B + ∠C = 180° and 2∠A = 90° and so ∠A = ∠C = 45° By pythagoras theorem; AB² + BC² = AC² , so AC² = 16+16 = 32 and so AC = 4√2 cm

Method 2 :

tan C = AB/BC = 1 and since ΔABC is right angled ∠C < 90° ( Angle sum property). So ∠C = 45° and similarly ∠A = 45° By pythagoras theorem; AB² + BC² = AC² , so AC² = 16+16 = 32 and so AC = 4√2 cm

Pls mark it brainliest if you find it helpful!

Answer 2

In ∆ABC , <B = 90°

AB = 4 , BC = 4 ,

By Phythogarian theorem ,

AC² = AB² + BC²

= 4² + 4²

= 16 + 16

= 32

AC = √32

=> AC = 4√2

And ,

<A = <B = x [ Angles opposite to

equal sides are equal ]

<A + <B + <C = 180°

=> x + x + 90° = 180°

=> 2x = 180° - 90°

=> 2x = 90°

=> x = 90°/2

=> x = 45°

Therefore ,

AC = 4√2 ,

<A = <B = x = 45°

•••••