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Maths ( Problems in 2 dimensions ) ! Urgent !!
P and Q are two points 200 m apart on a beach running east to west ( P being west of Q ). Find the distance of a buoy B from the beach if the true bearing of the buoy from P is 050度 and that from Q is 020度?
最佳解答:
Let PQ extents along the beach, there exist a point C to the east of Q where BC represents the shortest distance of buoy B from the beach. Then BC / CQ = tan (90 - 20) = tan 70 BC = CQ tan 70 -- (1) BC / CP = tan (90 - 50) = tan 40 BC / (CQ + QP) = tan 40 BC / (CQ + 200) = tan 40 BC = (CQ + 200) tan 40 -- (2) Solving (1) & (2): get CQ = ??? -- (3) Put (3) into (1): get BC!! My answer: 87.9m
其他解答:
**********************************buoy B****** ********************************-***-****_****** ****************************-******-*****_****** ************************-*********-******_****** ********************-************-*******_****** ****************-***************-********_****** ************-******************-*********_****** **********P-------200m-------Q*********_R**** 角BPQ =90度 - 50度 =40度 角BQP =90度 + 20度 =110度 看見有個直角三角形沒有？ 設那個直角為R. 角BQR =90度 - 20度 =70度 the distance of a buoy B from the beach = the distance between buoy B and R let the distance between buoy B and R be BR. tangent 角BPQ = BR over PR tan 40 = BR over PR PR = BR over tan 40 tangent 角BQR = BR over QR tan 70 = BR over QR QR = BR over tan 70 因此 不用求PR和QR 就能找到BR PR - QR =200m (BR over tan 40) - (BR over tan 70) = 200m BR = 200m over [ (1 over tan 40) - (1 over tan 70) ] since tan X = 1 over tan (90-X) BR = 200m over [ (1 over tan 40) - (1 over tan 70) ] BR = 200m over [ tan 50 - tan 30 ] BR = 325.52m (corrected to 2 decimal places) 清楚嗎？有問題再給我留言。|||||let the distant be x x/tan20 + x/tan50 = 200 (tan20+tan50)x/tan20 tan50=200 (tan20+tan50)x=86.75256 x=55.76 the distant is 55.76m
Maths ( Problems in 2 dimensions ) ! Urgent !!
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發問:P and Q are two points 200 m apart on a beach running east to west ( P being west of Q ). Find the distance of a buoy B from the beach if the true bearing of the buoy from P is 050度 and that from Q is 020度?
最佳解答:
Let PQ extents along the beach, there exist a point C to the east of Q where BC represents the shortest distance of buoy B from the beach. Then BC / CQ = tan (90 - 20) = tan 70 BC = CQ tan 70 -- (1) BC / CP = tan (90 - 50) = tan 40 BC / (CQ + QP) = tan 40 BC / (CQ + 200) = tan 40 BC = (CQ + 200) tan 40 -- (2) Solving (1) & (2): get CQ = ??? -- (3) Put (3) into (1): get BC!! My answer: 87.9m
其他解答:
**********************************buoy B****** ********************************-***-****_****** ****************************-******-*****_****** ************************-*********-******_****** ********************-************-*******_****** ****************-***************-********_****** ************-******************-*********_****** **********P-------200m-------Q*********_R**** 角BPQ =90度 - 50度 =40度 角BQP =90度 + 20度 =110度 看見有個直角三角形沒有？ 設那個直角為R. 角BQR =90度 - 20度 =70度 the distance of a buoy B from the beach = the distance between buoy B and R let the distance between buoy B and R be BR. tangent 角BPQ = BR over PR tan 40 = BR over PR PR = BR over tan 40 tangent 角BQR = BR over QR tan 70 = BR over QR QR = BR over tan 70 因此 不用求PR和QR 就能找到BR PR - QR =200m (BR over tan 40) - (BR over tan 70) = 200m BR = 200m over [ (1 over tan 40) - (1 over tan 70) ] since tan X = 1 over tan (90-X) BR = 200m over [ (1 over tan 40) - (1 over tan 70) ] BR = 200m over [ tan 50 - tan 30 ] BR = 325.52m (corrected to 2 decimal places) 清楚嗎？有問題再給我留言。|||||let the distant be x x/tan20 + x/tan50 = 200 (tan20+tan50)x/tan20 tan50=200 (tan20+tan50)x=86.75256 x=55.76 the distant is 55.76m
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