Question

Triangle J K L is shown. Angle J K L is 120 degrees and angle K L J is 40 degrees. The length of K L is 2 and the length of J L is k. Law of sines: StartFraction sine (uppercase A) Over a EndFraction = StartFraction sine (uppercase B) Over b EndFraction = StartFraction sine (uppercase C) Over c EndFraction What is the approximate value of k? Use the law of sines to find the answer. 2.9 units 3.8 units 5.1 units 6.2 units

Answer 1

$\textbf{Given:}$

$\text{In \triangle JKL, \angle{JKL}=120^{\circ} , \angle{KLJ}=40^{\circ} , JL=k and KL= 2 units}$

$\textbf{To find:}$

$\text{The approximate value of k}$

$\textbf{Solution:}$

$\textbf{Concept used:}$

$\textsf{Sine formula:}$

$\textsf{In triangle ABC,}$

$\mathsf{\displaystyle\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}}$

$\text{Applying Sine formula in \triangle JKL, we get}$

$\dfrac{KL}{sin\angle{KJL}}=\dfrac{JL}{sin\angle{JKL}}=\dfrac{KJ}{sin\angle{KLJ}}$

$\dfrac{2}{sin\,20^{\circ}}=\dfrac{k}{sin\,120^{\circ}}=\dfrac{KJ}{sin\,40^{\circ}}$

$\implies\dfrac{2}{sin\,20^{\circ}}=\dfrac{k}{sin\,120^{\circ}}$

$\implies\dfrac{2}{sin\,20^{\circ}}=\dfrac{k}{cos\,30^{\circ}}$

$\implies\dfrac{2}{sin\,20^{\circ}}=\dfrac{k}{\frac{\sqrt{3}}{2}}$

$\implies\dfrac{2}{sin\,20^{\circ}}=\dfrac{2k}{\sqrt{3}}$

$\implies\dfrac{1}{sin\,20^{\circ}}=\dfrac{k}{\sqrt{3}}$

$\implies\,k=\dfrac{\sqrt{3}}{sin\,20^{\circ}}$

$\implies\,k=\dfrac{1.732}{0.3420}$

$\implies\,k=5.0643$

$\implies\bf\,k=5.1\,\text{units}$

$\therefore\textbf{The approximate value of k is 5.1 units}$

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