Question

Solve. Trigonometry using cosine, tangent and sine

Answer 1

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) ${BD \over 20} = sin 54.$, 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 ${{16.18} \over CD} = tan 28$, 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 ${{JL} \over 14} = tan 51$. Follows that JL = 14.81. Now, similarly, trinagle JLK is a right triangle, with hypotenuse JL. From the definition of sine, as before,${KL \over 14.81} = sin 72.$. 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