Angle I = 20 degrees, side HI = 5, angle L = 20 degrees, and side KL = 5. What additional information would you need to prove that ΔHIJ ≅ ΔKLM by SAS? Angle H is congruent to angle K. Angle J is congruent to angle M. Side IJ is congruent to side LM. Side HJ is congruent to side KM.
Got a picture also?
no
ok
Im drawing this
Its not that serious, u dont have to draw it but, I think its the length of angle IJ and LM
\(\color{#0cbb34}{\text{Originally Posted by}}\) @xxDeppressionxx ➡Who be tired? ⬅ \(\color{#0cbb34}{\text{End of Quote}}\) bro if ur not tryna help then leave
Yeah i don't have to draw this, anyways we know one side and 1 angle so we need to know the side on the other side of angle I, / angle L so Side IJ is congruent to side LM. is correct
\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight Yeah i don't have to draw this, anyways we know one side and 1 angle so we need to know the side on the other side of angle I, / angle L so Side IJ is congruent to side LM. is correct \(\color{#0cbb34}{\text{End of Quote}}\) thanxxxxxxxxxxxx
why does it say "Please remember to upload any diagrams associated with the problem!"
there was no image.
HI = 5 and KL = 5, so HI = KL. That's one S of SAS angle I = angle L = 20 which forms the "A" in SAS so we need another pair of sides, and you have the correct answer. We need IJ = LM to be true These two pairs of sides must sandwich the angle in between |dw:1606865462523:dw| x = y must be the case for SAS to apply
|dw:1606865584672:dw|
Thank you, genius
No problem
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