Explanation:
The key to area in polar coordinates is the formula for the area of a sector:
a = (1/2)r²θ
Then a differential of area* can be written as ...
da = (1/2)r²·dθ
Filling in the given function for r, we have ...
da = (1/2)(4cos(3θ))²·dθ = 8cos(3θ)²·dθ
The integral will have limits corresponding to the range of values of θ for one loop of the graph: -π/6 to π/6. So, the area is ...

_____
* As with other approaches to finding area (horizontal or vertical slice, for example), we assume that the differential element dθ is sufficiently small that we need not concern ourselves with the fact that r is a function of θ.
Answer:
We conclude that:
∠A = ∠ FEC
Step-by-step explanation:
Given
△ABD ≅ △EFC
To determine:
∠A =
As
△ABD ≅ △EFC
So the triangles △ABD and △EFC are congruent to each other.
- We know that congruent triangles have equal corresponding parts.
Please check the attached graph.
From the graph, it is clear that ∠A is correspondent to ∠E.
∠A = ∠E
From the attached figure, it is clear that:
∠F can also be denoted by ∠ FEC
as
∠A = ∠E
so
∠A = ∠ FEC
Therefore, we conclude that:
∠A = ∠ FEC
I hope my answer help you in your question