Question

what is the measure of ∠n to the nearest whole degree? 35° 45° 55° 65° law of cosines: ( a^2 = b^2 + c^2 - 2bccos(a) )

Explanation:

Step1: Identify sides for ∠N

In right triangle \( \triangle MLN \), for \( \angle N \), opposite side \( ML = 6.74 \), hypotenuse \( MN = 7 \), adjacent side \( LN = 4 \). We can use sine or cosine. Let's use sine: \( \sin(N)=\frac{ML}{MN} \).

Step2: Calculate \( \sin(N) \)

\( \sin(N)=\frac{6.74}{7}\approx0.9629 \)

Step3: Find \( \angle N \)

\( N=\arcsin(0.9629)\approx74^\circ \)? Wait, no, wait. Wait, \( ML = 6.74 \), \( LN = 4 \), \( MN = 7 \). Wait, maybe use cosine: \( \cos(N)=\frac{LN}{MN}=\frac{4}{7}\approx0.5714 \). Then \( N=\arccos(0.5714)\approx55^\circ \)? Wait, no, wait, \( LN = 4 \), \( ML = 6.74 \), \( MN = 7 \). Let's check with law of cosines. Let's label the triangle: \( L \) is right angle, so \( LN = 4 \) (side \( a \)? No, law of cosines: \( a^2 = b^2 + c^2 - 2bc\cos(A) \). Let's take \( \angle N \), so side opposite \( \angle M \) is \( LN = 4 \), side opposite \( \angle N \) is \( ML = 6.74 \), hypotenuse \( MN = 7 \). Wait, maybe better to use sine: \( \sin(N)=\frac{ML}{MN}=\frac{6.74}{7}\approx0.9629 \), so \( N=\arcsin(0.9629)\approx74^\circ \)? No, that's not matching. Wait, maybe I made a mistake. Wait, \( ML = 6.74 \), \( LN = 4 \), so \( \tan(N)=\frac{ML}{LN}=\frac{6.74}{4}=1.685 \), so \( N=\arctan(1.685)\approx59^\circ \)? No, the options are 35,45,55,65. Wait, maybe the correct approach: \( \cos(N)=\frac{LN}{MN}=\frac{4}{7}\approx0.571 \), so \( N=\arccos(0.571)\approx55^\circ \) (since \( \cos(55^\circ)\approx0.5736 \), which is close to 0.5714). So \( \angle N \approx 55^\circ \).

Answer:

\( 55^\circ \) (the option with \( 55^\circ \))