Question
upstudy study bank question image url

All changes saved 14. In \( \triangle L M N, L M=12, L N=10 \), and \( m \angle L=52^{\circ} \). What is the length, to the nearest tenth of a unit, of \( \overline{M N} \) ? 7.4 units 14.4 units 9.8 units 15.6 units (i)

Ask by Whittaker Ray. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The length of \( \overline{MN} \) is approximately 9.8 units.

Solution

To find the length of \( \overline{MN} \) in \( \triangle LMN \) with given sides \( LM = 12 \), \( LN = 10 \), and angle \( \angle L = 52^\circ \), we can use the **Law of Cosines**. The Law of Cosines states: \[ c^2 = a^2 + b^2 - 2ab \cos(C) \] where: - \( c \) is the side opposite angle \( C \), - \( a \) and \( b \) are the other two sides, - \( C \) is the included angle. In this triangle: - \( c = MN \), - \( a = LM = 12 \), - \( b = LN = 10 \), - \( C = \angle L = 52^\circ \). Plugging in the values: \[ MN^2 = 12^2 + 10^2 - 2 \times 12 \times 10 \times \cos(52^\circ) \] \[ MN^2 = 144 + 100 - 240 \times \cos(52^\circ) \] \[ MN^2 = 244 - 240 \times 0.61566 \quad (\cos(52^\circ) \approx 0.61566) \] \[ MN^2 \approx 244 - 147.7584 \] \[ MN^2 \approx 96.2416 \] \[ MN \approx \sqrt{96.2416} \approx 9.81 \text{ units} \] Rounding to the nearest tenth: \[ MN \approx 9.8 \text{ units} \] **Answer:** 9.8 units

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

Extra Insights

To find the length of \( \overline{M N} \) in triangle \( \triangle L M N \), we can use the Law of Cosines, which states that \( c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \). Here, we identify \( L M = a = 12 \), \( L N = b = 10 \), and \( \angle L = C = 52^{\circ} \). Plugging the values into the formula gives us: \[ MN^2 = 12^2 + 10^2 - 2 \times 12 \times 10 \times \cos(52^\circ) \] Calculating that will yield a number we can then take the square root of to find \( MN \). After calculating, you'll arrive at approximately \( 9.8 \) units for \( \overline{M N} \). This triangle scenario is a fantastic way to explore trigonometry in real life! The Law of Cosines is especially handy in navigation, where understanding distances between points is crucial. For instance, sailors or pilots can rely on this theorem to chart their course accurately when dealing with angled paths, ensuring they reach their destinations efficiently. If you're diving deeper into triangle relationships, don't overlook the world of triangle inequalities! These rules help determine possible triangle conditions and can prevent those "oops" moments where you think a triangle can exist, but it really can't. Remember, the sum of the lengths of any two sides must always be greater than the length of the third side. Happy triangle adventuring!

Related Questions

Exercice 82 Le plan est rapporté à un repère orthonormal direct \( (0, \vec{u}, \vec{v}) \). On appelle \( f \) l'application qui, à tout point \( M \) d'affixe \( z(z \neq-1) \) associe le point \( M^{\prime} \) d'affixe \( z^{\prime} \) telle que : \( z^{\prime}=\frac{-i z-2}{z+1} \). Soient A, B et C les points d'affixes respectives \( a=-1, b=2 i \) et \( c=-i \). 1) Soit \( C^{\prime} \) l'image du point \( C \) par \( f \). Donner l'affixe \( c^{\prime} \) du point \( C^{\prime} \) sous forme algébrique, puis sous forme trigonométrique. 2) Calcule l'affixe \( d \) du point \( D \) ayant pour image par \( f \) le point \( D^{\prime} \) d'affixe \( d^{\prime}=\frac{1}{2} \). 3) Pour tout nombre complexe \( z \) différent de -1 , on note \( p \) le module de \( z+1 \) (c'est-à-dire \( |z+1|=p) \) et \( p^{\prime} \) le module de \( z^{\prime}+i\left(\right. \) c'est-à-dire \( \left.\left|z^{\prime}+i\right|=p^{\prime}\right) \). a) Démontre que pour tout nombre complexe \( z \) différent de -1 , on a : \( p p^{\prime}=\sqrt{5} \). b) Si le point \( M \) appartient au cercle \( (\Gamma) \) de centre A et de rayon 2 , montre qu'alors \( M^{\prime}=f(M) \) appartient à un cercle ( \( \left.\Gamma^{\prime}\right) \) dont on précisera le centre et le rayon. 4) Pour tout nombre complexe \( z \) différent de -1 , on considère le nombre complexe \( \omega=\frac{z-2 t}{z+1} \). a) Interprète géométriquement l'argument du nombre complexe \( \omega \). b) Montre que \( z^{\prime}=-i \omega \). c) Détermine l'ensemble \( (F) \) des points \( M \) d'affixe \( z \) telle que \( z^{\prime} \) soit un réel non nul. d) Vérifie que le point \( D \) appartient aux ensembles \( (\Gamma) \) et \( (F) \). 5) Représente les ensembles \( (\Gamma) \), (F) et \( \left(\Gamma^{\prime}\right) \) en prenant 4 cm pour unité graphique.
Geometry Mali Feb 03, 2025
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy