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Triangle J K L is shown. Angle J K L is 67 degrees and angle K L J is 74 degrees. The length of K L is 2.3
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 perimeter of the triangle? Use the law of sines to find the answer.

4.6 units
5.7 units
6.9 units
9.2 units
could someone explain why that answer is what it is

Sagot :

sqdancefan

Answer:

  (d)  9.2 units

Step-by-step explanation:

Given ∆JKL with K=67°, L=74°, and KL=2.3, you want the perimeter of the triangle.

Law of sines

The law of sines tells you the relationship between the sides and angles:

  [tex]\dfrac{JK}{\sin(L)}=\dfrac{KL}{\sin(J)}=\dfrac{LJ}{\sin(K)}[/tex]

To use this relation, we need to know the angle opposite the given side.

  ∠J +∠K +∠L = 180°

  ∠J +67° +74° = 180°

  ∠J = 39° . . . . . . . . . . . . subtract 141° from both sides

Now, we can write ...

  [tex]\dfrac{JK}{\sin(74^\circ)}=\dfrac{2.3}{\sin(39^\circ)}=\dfrac{LJ}{\sin(67^\circ)}[/tex]

Side lengths

Solving this equation for JK and LJ, we have ...

  [tex]JK=2.3\cdot\dfrac{\sin(74^\circ)}{\sin(39^\circ)}\\\\\\LJ=2.3\cdot\dfrac{\sin(67^\circ)}{\sin(39^\circ)}\\\\\\P=KL +JK +LJ\\\\\\P=2.3+2.3\cdot\dfrac{\sin(74^\circ)}{\sin(39^\circ)}+2.3\cdot\dfrac{\sin(67^\circ)}{\sin(39^\circ)}\\\\\\P=2.3\left(1+\dfrac{\sin(74^\circ)+\sin(67^\circ)}{\sin(39^\circ)}\right)\approx 9.177[/tex]

The perimeter of the triangle is about 9.2 units, choice D.

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