Answer:
CD = 30.43; LK = 14.08
Step-by-step explanation:
Problem 12.
Let's first find BD. Triangle ADB is a right triangle with hypotenuse AB. By definition of sine ( or by theorem, they're equivalent) [tex]{BD \over 20} = sin 54.[/tex], hence BD= 16.18. Jump on the other side now. Triangle BDC is a right triangle with sides BD and CD. By definition of tangent [tex] {{16.18} \over CD} = tan 28 [/tex], thus CD= 30.43.
Problem 13
Assuming the 72° is angle JLK
As usual, triangle MJL is a right triangle with sides MJ and LJ. By definition of tangent [tex] {{JL} \over 14} = tan 51 [/tex]. Follows that JL = 14.81. Now, similarly, trinagle JLK is a right triangle, with hypotenuse JL. From the definition of sine, as before,[tex]{KL \over 14.81} = sin 72.[/tex]. Solving for the only unknown value, you find LK = 14.08.
Please double check the calculations and feel free to add more digits if required
