Let's think about each of the given tasks one at a time, starting with finding all of the supplementary .
are any two angles whose sum is 180^(∘). Since ∠AFE is 180^(∘), any division of it will create supplementary angles. Using FB as the divider, we get the following scenario.
This means that ∠AFB and ∠BFE are supplementary angles. We can also use FC as a divider.
This means that ∠AFC and ∠EFC are also supplementary angles. Finally, we can divide ∠AFE using FD.
Therefore, ∠AFD and ∠DFE is our last pair of supplementary angles.
Complementary Angles
are any two angles whose sum is 90^(∘). Since ∠AFE is a and ∠CFE is a , ∠AFC is also a right angle measuring 90^(∘).
Any division of either of these will create complementary angles.
Looking at the left-hand side, we can see that ∠AFB and ∠BFC are complementary angles. Similarly, on the right-hand side, ∠CFD and ∠DFE are also complementary angles.
m∠DFE
From the diagram, we know that m∠CFD is 27^(∘).
Since ∠CFD and ∠DFE are complementary angles, the sum of their measures is 90^(∘).
m∠CFD + m∠DFE = 90^(∘)
27^(∘) + m∠DFE = 90^(∘)
m∠DFE = 63^(∘)
Therefore, m∠DFE is 63^(∘).
m∠BFC
We are given that m∠AFB is 39^(∘).
Since ∠AFB and ∠BFC are complementary angles, their measures add up to 90^(∘).
m∠AFB + m∠BFC = 90^(∘)
39^(∘) + m∠BFC = 90^(∘)
m∠BFC = 51^(∘)
Hence, m∠BFC is 51^(∘).
m∠BFE
Finally, we know that ∠AFB and ∠BFE are supplementary angles.
That means that their measures sum to be 180^(∘).
m∠AFB + m∠BFE = 180^(∘)
39^(∘) + m∠BFE = 180^(∘)
m∠BFE = 141^(∘)
Our last angle, m∠BFE, is 141^(∘).
