triangle knm is shown. what is true about the sides of knm? kn = nm kn + nm = km km = 2(nm) kn =…

triangle knm is shown. what is true about the sides of knm? kn = nm kn + nm = km km = 2(nm) kn = \\frac{1}{2}km

triangle knm is shown. what is true about the sides of knm? kn = nm kn + nm = km km = 2(nm) kn = \\frac{1}{2}km

Answer

Explanation:

Step1: Recall the property of isosceles triangle

In a triangle, if two angles are equal, then the sides opposite to them are equal.

Step2: Identify equal angles in (\triangle KNM)

In (\triangle KNM), (\angle K = 25^{\circ}) and (\angle M=25^{\circ}).

Step3: Determine the sides opposite to equal angles

The side opposite to (\angle K) is (NM) and the side opposite to (\angle M) is (KN). Since (\angle K=\angle M), by the isosceles - triangle property, (KN = NM).

Now, check the other options:

  • For the option (KN + NM=KM): By the triangle - inequality theorem, in a triangle (a + b>c) (where (a,b,c) are the sides of the triangle). Here, if (a = KN), (b = NM), (c = KM), then (KN+NM>KM), so (KN + NM\neq KM).
  • For the option (KM = 2(NM)): There is no information (such as a right - triangle with a (30 - 60-90) angle relationship or other special - triangle properties) to support this. Using the law of sines (\frac{KN}{\sin\angle M}=\frac{NM}{\sin\angle K}=\frac{KM}{\sin\angle N}). (\sin\angle N=\sin130^{\circ}=\sin(180 - 50)^{\circ}=\sin50^{\circ}), (\sin\angle K=\sin\angle M=\sin25^{\circ}). (\frac{KM}{\sin130^{\circ}}=\frac{NM}{\sin25^{\circ}}), (KM=\frac{\sin130^{\circ}}{\sin25^{\circ}}NM\approx\frac{0.766}{0.423}NM\approx1.81NM\neq2NM).
  • For the option (KN=\frac{1}{2}KM): From the law of sines (\frac{KN}{\sin\angle M}=\frac{KM}{\sin\angle N}), (KN=\frac{\sin\angle M}{\sin\angle N}KM=\frac{\sin25^{\circ}}{\sin130^{\circ}}KM\approx\frac{0.423}{0.766}KM\approx0.55KM\neq\frac{1}{2}KM).

Answer:

(KN = NM)