Question

I need help guys Asap

Answer 1

Given:

Two triangles are given in the figure.

To find:

The m∠K and m∠J.

Solution:

In triangle FHK and IGJ,

`m\angle  F=m\angle I` [Given]

`m\angle  H=m\angle G` [Given]

Two corresponding angles are equal. So,

`\Delta FHK\sim \Delta IGJ` [By AA similarity theorem]

All corresponding angles of similar figures are same.

`m\angle  K=m\angle J` ...(i)

`4y^2=6y^2-40`

`4y^2-6y^2=-40`

`-2y^2=-40`

Divide both sides by -2.

`y^2=20` ...(ii)

Taking square root on both sides.

`y=\pm √(20)`

Now,

`m\angle K=(4y^2)^\circ`

`m\angle K=[4(20)]^\circ` [Using (ii)]

`m\angle K=80^\circ`

`m\angle K=m\angle J=80^\circ` [Using (i)]

Therefore, the m∠K is 80° and m∠J is 80°.