University:
Jackson College
Course:
MAT-039 | Beginning Algebra
Academic year:

2021
Views:

366

Pages:

8
Author:

Aspen Harper

Trigonometry Lesson #8: Practice Test Use the following information to answer the first three questions. right triangle DEF with sides DF = 15, DE = 8, and EF = 17 1. The value of cos F is A. 8/15 B. 8/17 (correct) C. 17/8 D. 62 2. The value of x° is A. 15 B. 8 C. 28° (correct) D. 62 3. Which statement is false? A. cos E = sin F (correct) B. sin E = cos F C. tan E ≠ tan F D. tan F = sin F / cos F Recall: Ensure you Calculator is in degree mode 4. The value of a is: A. 14 tan 51° B. 14 cos 51° C. 14 / tan 51° D. 14 / cos 51° 5. The value of x, to the nearest tenth of a degree, is: A. 35.0° B. 61.9° C. 72.3° D. 29.3° 6. A guy wire to a pole makes an angle of 71.4° with the level ground and is 13.8 feet from the pole at ground level. How far above the ground is the guy wire attached to the pole? A. 4.6 feet B. 13.1 feet C. 14.6 feet D. 41.0 feet 7. The value of length AC, to the nearest tenth, is: A. 0.1 B. 2.1 C. 7.2 D. 7.5 8. A corner flag in a World Cup soccer match is 5 feet high. At game time, the flag casts a shadow which is 3.2 feet long. To the nearest tenth of a degree, the angle of elevation of the sun is: A. 57.4° B. 50.2° C. 39.8° D. 32.6° Use the following information to answer the next three questions. The diagram shows how a horse breeder has used fencing to divide a parcel of land into two triangular paddocks for his horses. Each of the line segments represents fencing. E G 55° 0.85 km (hyp) 43° D F Problem: Not enough information is provided for ADEF in order to solve for DF. We must solve for a shared side. 9. The length of the fencing represented by DF is: A. 0.89 km B. 0.79 km C. 0.62 km D. 0.58 km Sr. Shared side EF Cos 43° = EF/0.85 EF = 0.62... Use side EF to solve for DF tan 55° = DF/0.62... DF = 0.89... 10. The value of cos G is: A. 0.68 B. 0.72 C. 0.80 D. impossible to determine from the given information. Since we know two of the three angles in ΔEFG, we can apply the rule that a triangle's angles add to 180°. < EGF = 180° - 90° - 43° = 47° cos(47°) = 0.68... Numerical Response 1. The total area of the two triangular paddocks, to the nearest 0.1 km², is: (Record your answer in the numerical response box from left to right) 0.5 Recall: Area = ½ bh Total area = ½(DF)(EF) + ½(DF)(EG) = ½(0.89...)(0.62...) + ½(0.62...)(0.57...) = 0.455... km²