MCQ on Practical Application of Trigonometry

MCQ on Practical Application of Trigonometry Prepared by: RAMIL T. FRANCISCO IsmaelMathay Sr. High School

Question 1: The angle of elevation of the top a tower 40 meters high is 37 degrees when seen from a point on level ground. Find the distance of the point from the foot of the tower A. 53.08 meters C. 63. 08 meters B. 45.08 meters D. 35.08 meters

Question 2: From the window of a building 45 meters above the ground level, the angle of depression of a car on level ground is 42 degrees, How far is the car from the building? A. 53.08 meters C. 63. 98 meters B. 49.98 meters D. 94.08 meters

Question 3: A lighthouse is 72 meters high. Find the angle of elevation of its top from a ship 125 meters away on ground level A. 53.08 degrees C. 29.90 degrees B. 49.98 degrees D. 94.08 degrees

Question 4: A tower stands at the top of a cliff. At a distance of 60 m from the foot of the cliff which is at ground level, the angle of elevation of the top of the tower as well as the cliff are 65 and 53 degrees respectively, Find the height of the tower. A. 53.08 meters C. 29.90 meters B. 49.98 meters D. 49.05 meters

Question 5: From the top of a cliff 52 m high, the angles of depression of two ships due east of it are 36 and 24 degrees respectively. Find the distance between the ships. A. 54.68 meters C. 45.22 meters B. 49.22 meters D. 71.57 meters

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C Solution for Question 1 The diagram shows the tower from a point A 40 m Let AB = x m 37 degrees B A x m Tan 37 = BC / AB = 40 / x x = 40 / tan 37 = 53.08 to 4 significant digit

Solution for Question 2 C D 42 degrees In the diagram, A represents the car, BC represents the height of the building where C is the point of observation and angle ACD is the angle of depression 45 m Let AB = x m /_ACB = 90-42 = 48 degrees B A Tan 48 = x /45 x = 45 tan 48 = 49.98 meters X m

Solution for Question 3 C 72 m B 125 m A The diagram shows the lighthouse BC of height 72 m at a distance of 125 m away on level ground Let the angle of elevation be X Tan X = 72 / 125 = 0.576 X= 29.9 degrees

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Solution for Question 4 R In the diagram, PQ represents the cliff and QR the tower, S is 60 m from P In SPQ, tan 53 = PQ /SP = PQ /60 PQ = 60 tan 53 = 79.62 Q In SPR, tan 65 = PR /SP = PR /60 PR = 60 tan 65 = 128.67 60 53 degrees S P 60 m QR = PR – PQ = 49.05

Question 5: From the top of a cliff 52 m high, the angles of depression of two ships due east of it are 36 and 24 degrees respectively. Find the distance between the ships. A. 54.68 meters C. 70.68 meters D. 45.22 meters B. 77.56 meters

Solution for Question 5 Q 24 degrees 36 degrees 52 m 36 degrees P A 24 degrees B In the figure PQ represents the cliff, A and B the ships PBQ = 24 and PAQ = 36 PAQ = tan 36 = PQ/PA = 71.57 PBQ = tan 24 = PQ/PB = 116.79 The distance between the 2 ships is 116.79 – 71.57 = 45.22 m

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MCQ on Practical Application of Trigonometry