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.
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]
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.
