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.

$y=\pm \sqrt{20}$

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°.