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Setler [38]
3 years ago
7

What additional information is needed to prove triangle LMN is congruent to triangle KMN using the HL theorem?

Mathematics
2 answers:
Alexxx [7]3 years ago
6 0
<h3>Answer</h3>

leg MN of both triangle is equal.

<h3>Step-by-step explanation</h3>

HL stands for "Hypotenuse, Leg" (the longest side of a right-angled triangle is called the "hypotenuse", the other two sides are called 'legs', or 'base' and 'height'.

It means we have two right-angled triangles with

  1. the same length of hypotenuse
  2. the same length for one of the other two legs

Since ∠ LMN and ∠KMN are right angle , the hypotenuse LN and and one leg LN of one right-angled triangle LMN are equal to the corresponding hypotenuse KN and leg MN of another right-angled triangle MKN, hence the two triangles are congruent.

IgorC [24]3 years ago
3 0
<h2>Answer:</h2>

∠LMN is a right angle

<h2>Step-by-step explanation:</h2>

If we want to prove that two right triangles are congruent by knowing that the corresponding hypotenuses and one leg are congruent, we begin as follows:

  • Since two legs are congruent and we know this by the hash marks, then the triangle ΔLKN is isosceles.
  • By definition LN ≅ NK
  • If ∠LMN is a right angle, then MN is the altitude of triangle ΔLKN
  • Also MN is the bisector of LK, so KM ≅ ML
  • So we have two right triangles ΔLMN and ΔKM having the same lengths of corresponding sides
  • In conclusion, ΔLMN ≅ ΔKMN
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Step-by-step explanation:

I'm going to use Physics here for this concept of vectors. Here are some stipulations I have set for the problem (aka rules I set and then followed throughout the problem):

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Now let's begin.

We need to find the x and y components of both of these vectors. I am going to call the first vector A and the second B, while the resultant vector will be C. Starting with the x components of A and B:

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A_y=50sin(89.75) so

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B_y=0 and we need to add those results together.

C_y=50

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C_{mag}=\sqrt{(200)^2+(50)^2}  and that gives us a final magnitude of

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Now for the angle:

Since both the x and y components of the resultant vector are in quadrant 1, we don't need to add anything to the angle to get it right, so

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