Since your question doesn't specify which angle is 40 degrees, we will take two cases:
i) When ∠B = 40°
ii) When ∠A = 40°
Solving part (i):
If ∠B = 40°, we can draw a diagram,
Using sine, we can find AB and using tangent, we can find BC.
The ratio and steps to solve are shown below:
[tex]\begin{gathered} \sin 40=\frac{12}{AB} \\ AB\sin 40=12 \\ AB=\frac{12}{\sin 40} \\ AB=18.7 \end{gathered}[/tex]and,
[tex]\begin{gathered} \tan 40=\frac{12}{BC} \\ BC\tan 40=12 \\ BC=\frac{12}{\tan 40} \\ BC=14.3 \end{gathered}[/tex]AnswerWhen ∠B = 40°,AB = 18.7BC = 14.3----------------------------------------------------------------------------------Solving part (ii):If ∠A = 40°, we can draw a diagram,
Using cosine, we can find AB and using tangent, we can find BC.
The ratio and steps to solve are shown below:
[tex]\begin{gathered} \cos 40=\frac{12}{AB} \\ AB\cos 40=12 \\ AB=\frac{12}{\cos 40} \\ AB=15.7 \end{gathered}[/tex]and,
[tex]\begin{gathered} \tan 40=\frac{BC}{12} \\ 12\tan 40=BC \\ BC=10.1 \end{gathered}[/tex]AnswerWhen ∠A = 40°,AB = 15.7BC = 10.1
