In right triangle \(\triangle KMF\), \(\tan (\angle MKF)=\frac{MF}{KM}\). \(MF=KM\cdot \tan (\angle MKF)=47\cdot \tan (67^{\circ })\). In \(\triangle LFM\), by the law of sines, \(\frac{LF}{\sin (\angle MKF)}=\frac{KM}{\sin (\angle LFM)}\). \(LF=\frac{KM\cdot \sin (\angle MKF)}{\sin (\angle LFM)}=\frac{47\cdot \sin (67^{\circ })}{\sin (46^{\circ })}\). \(LF\approx \frac{47\cdot 0.92}{0.72}\approx 60.03\).

1 answer

To clarify and confirm the calculations in your problem regarding triangles \( \triangle KMF \) and \( \triangle LFM \), we will break it down step by step.

Step 1: Find \( MF \)

Given that: \[ MF = KM \cdot \tan(\angle MKF) \] with \( KM = 47 \) and \( \angle MKF = 67^\circ \).

We apply the tangent function: \[ MF = 47 \cdot \tan(67^\circ) \] Using a calculator, we find: \[ \tan(67^\circ) \approx 2.35585 \] Thus, \[ MF \approx 47 \cdot 2.35585 \approx 110.71 \]

Step 2: Use the Law of Sines in Triangle \( LFM \)

For triangle \( LFM \), we apply the Law of Sines: \[ \frac{LF}{\sin(\angle MKF)} = \frac{KM}{\sin(\angle LFM)} \]

We solve for \( LF \): \[ LF = \frac{KM \cdot \sin (\angle MKF)}{\sin(\angle LFM)} \] Where:

  • \( \angle MKF = 67^\circ \)
  • \( KM = 47 \)
  • \( \angle LFM = 46^\circ \)

Calculating \( LF \): First, we find \( \sin(67^\circ) \) and \( \sin(46^\circ) \): \[ \sin(67^\circ) \approx 0.92050, \quad \sin(46^\circ) \approx 0.71934 \]

Plugging in the values: \[ LF = \frac{47 \cdot \sin(67^\circ)}{\sin(46^\circ)} = \frac{47 \cdot 0.92050}{0.71934} \] Calculating the numerator: \[ 47 \cdot 0.92050 \approx 43.24 \] Now, calculating \( LF \): \[ LF \approx \frac{43.24}{0.71934} \approx 60.18 \]

Summary

  • We first calculated \( MF \approx 110.71 \).
  • Then, applying the Law of Sines, we concluded that \( LF \approx 60.18 \).

Your approximation of \( LF \approx 60.03 \) seems quite close, but with precise calculations, the value is approximately \( 60.18 \). If we round appropriately based on the angles and the operations: \[ \text{Final Answer: } LF \approx 60.18 \]