Question

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

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

eq 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

eq2NM\).

• 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

eq\frac{1}{2}KM\).

Answer:

\(KN = NM\)