Given: BFCE, AB perpendicular to BE, DE perpendicular to BE, angle BFD equal to angle ECA
Prove triangle ABC is similar to triangle DEF

Step-by-step explanation:

From the given picture,

∠ABE = ∠DEF = 90° [Since, AB and DE are perpendicular to DE]

m∠ECA = m∠BFD [Given]

m∠ECA + m∠ACB = 180° [Liner pair of angles]

m∠BFD + m∠DFE = 180° [Liner pair of angles]

m∠ACB + m∠ECA = m∠BFD + m∠DFE [Transitive property]

m∠ACB = m∠DEF [Since, m∠ECA = m∠BFD]

Therefore, ΔABC ≅ ΔDEF [By AA property of similarity]

7. 153.86 ft

8. 12.56 ft

9. 78.5 m

10. 379.94 in

11. 132.665 m

12. 12.56 yd

13. 226.865 ft

14. 113.04 ft

15. 63.585 ft

16. 254.34 m

17. 50.24 cm

step-by-step explanation:

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.