Answer:
True
Step-by-step explanation:
Given
In JKL, we have:
[tex]\angle J = 27[/tex]
[tex]\angle K = 90[/tex]
In WXY, we have:
[tex]\angle Y = 63[/tex]
[tex]\angle X = 90[/tex]
Required
Is JKL ~ WXY?
In both triangles, we already have one similar angle (90)
Next, is to determine the third angles in both triangles.
In JKL
[tex]\angle J + \angle K + \angle L = 180[/tex]
We have that:
[tex]\angle J = 27[/tex] and [tex]\angle K = 90[/tex]
The expression becomes:
[tex]27 + 90 + \angle L = 180[/tex]
[tex]117 + \angle L = 180[/tex]
[tex]\angle L = 180-117[/tex]
[tex]\angle L = 63[/tex]
In WXY
[tex]\angle W + \angle X + \angle Y = 180[/tex]
We have that:
[tex]\angle Y = 63[/tex] and [tex]\angle X = 90[/tex]
The expression becomes:
[tex]\angle W + 63 + 90 = 180[/tex]
[tex]\angle W + 153 = 180[/tex]
[tex]\angle W = 180-153[/tex]
[tex]\angle W = 27[/tex]
The three angles in JKL are:
[tex]\angle J = 27[/tex] [tex]\angle K = 90[/tex] [tex]\angle L = 63[/tex]
The three angles in WXY are:
[tex]\angle W = 27[/tex] [tex]\angle X = 90[/tex] [tex]\angle Y = 63[/tex]
By comparing the angles, we can conclude that both triangles are similar because of AAA postulate (Angle-Angle-Angle)
