Kelly was asked to determine whether △KLN is congruent to △MNL. She noted that KL ≅ MN, LM ≅ KN, and that the three pairs of cor
responding angles are congruent. She said that this is only five pairs of congruent corresponding parts, so it is not possible to conclude that △KLN is congruent to △MNL. Do you agree with Kelly's reasoning? Complete the explanation.
1 answer:
Answer:
<h2>Kelly is wrong, with this congruent parts, we can conclude that triangles are congruent.</h2>Step-by-step explanation:
To demonstrate congruent triangles, we need to use the proper postulates. There are at least 5 postulates we can use.
- Angle-Angle-Side Theorem (AAS theorem).
- Hypotenuse-Leg Theorem (HL theorem).
- Side-Side-Side Postulate (SSS postulate).
- Angle-Side-Angle Postulate (ASA postulate).
- Side-Angle-Side Postulate (SAS postulate).
In this case, Kelly SAS postulate, because the corresponding sides-angles-sides are congruent, i.e., KL ≅ MN and LM ≅ KN, also, all corresponding angles are congruent.
So, as you can see, only using SAS postulate, the congruency can be demonstrated. (Refer to the image attached to see an example of SAS postulate)
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You line up the numbers shown in the photo, then add each column
Given that 
, then 
The slope of a tangent line in the polar coordinate is given by:
Thus, we have:
Part A:
For horizontal tangent lines, m = 0.
Thus, we have:
Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:
</span><span>θ = 0
</span>θ = <span>2.02875783811043
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θ = <span>4.91318043943488
Part B:
For vertical tangent lines,
Thus, we have:
</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:
</span>θ = <span>4.91718592528713</span>
The slope of a tangent line in the polar coordinate is given by:
Thus, we have:
Part A:
For horizontal tangent lines, m = 0.
Thus, we have:
Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are horizontal are:
</span><span>θ = 0
</span>θ = <span>2.02875783811043
</span>
θ = <span>4.91318043943488
Part B:
For vertical tangent lines,
Thus, we have:
</span>Therefore, the <span>values of θ on the polar curve r = θ, with 0 ≤ θ ≤ 2π, such that the tangent lines are vertical are:
</span>θ = <span>4.91718592528713</span>