In equilateral ΔABC, D and E are points on the sides AB and AC, respectively, such that AD = CE. BE and CD intersect at F. The measure (in degrees) of ∠CFB is:
120°
Given:
D and E are points located on the sides AB and AC respectively.
AD = CE
BE and CD intersect at F.
Concept Used:
Concept of triangle congruence,
The exterior angle is always equal to the sum of the opposite interior angles.
Calculation:

ΔCBE ≅ ΔACD [SAS Congruence]
Thus, all three angles of these two triangles are equal,
Let ∠EBC be θ, then ∠ACD is also θ.
Now,
∠BEC = 180° - (60° + θ)
⇒ 120° - θ
Now, in ΔECF
The exterior angle ∠CFB = (120° - θ) + θ
⇒ 120°
∴ ∠CFB is 120° .
Consider the following statements:
1. If ABC is a right-angled triangle, right-angled at A, and if sin \(\rm B = \frac 1 3,\) then cosec C = 3.
2. If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles.
Which of the above statements is/are correct?
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