Question

In angle ABC, AB=BC and AC 2√2 ∆ ABC 90° find AB​

Answer 1

Step-by-step explanation:

here is your ANSWER

in above picture

Answer 2

Answer:

Given :

In ∆ABC , <B = 90°

AB = BC

AC² = AB² + BC²

=> AC² = AB² + AB² [ from ( 1 ) ]

[ By Phythogarian theorem ]

=> AC² = 2AB²

=> AC = √2 (AB) ---( 2 )

Now ,

i ) CosecC = ( Hypotenuse )/(opp.side to <C )

= AC /AB

= √2 ( AB )/AB [ from ( 2 ) ]

= √2

ii ) CotC = ( Adjecent side to<C )/( opp.side )

= BC/AB

= BC/AB [ from ( 1 ) ]

= 1

Therefore ,

CosecC = √2 ,

cot C = 1

Step-by-step explanation:

Concept:

In an isoceles triangle, the angles opposite to equal sides are equal.

since AB=BC, triangle ABC is isoceles triangle.

since triangles ABC is isoceles,

we have∠A = ∠C = 45°

Now,

cosecC

=cosec45°

= 1/tan45

=1/1

=1