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| 1. |
The angle of elevation of a jet plane from a point A on the ground is 60 |
| Answer» Let us suppose that P\xa0be the initial position of the plane and let Q be the final position of the plane respectively and suppose A be the point of observation. Let ABC be the horizontal line through A.According to question, it is given that angles of elevation of the plane in two positions P and Q from a point A are 60° and 30° respectively.{tex} \\therefore \\quad \\angle P A B = 60 ^ { \\circ } , \\angle Q A B = 30 ^ { \\circ }.{/tex}It is also given\xa0that PB = 1500{tex} \\sqrt { 3 }{/tex}\xa0metresNow, In\xa0{tex} \\Delta A B P,{/tex}\xa0we have{tex} \\tan 60 ^ { \\circ } = \\frac { B P } { A B }{/tex}{tex} \\Rightarrow \\quad \\sqrt { 3 } = \\frac { 1500 \\sqrt { 3 } } { A B }{/tex}{tex} \\Rightarrow{/tex}\xa0AB = 1500 mAgain, In\xa0{tex} \\Delta A C Q,{/tex}we have{tex} \\tan 30 ^ { \\circ } = \\frac { C Q } { A C }{/tex}{tex} \\Rightarrow \\quad \\frac { 1 } { \\sqrt { 3 } } = \\frac { 1500 \\sqrt { 3 } } { A C }{/tex}{tex} \\Rightarrow \\quad A C = 1500 \\times 3 = 4500 \\mathrm { m }{/tex}{tex} \\therefore{/tex}\xa0PQ = BC = AC - AB = 4500 - 1500 = 3000 mTherefore, the plane travels 3000 m in 30 seconds.Hence, speed of plane is =\xa0{tex} \\frac { 3000 } { 30 } = 100 \\mathrm { m } / \\mathrm { sec } {/tex} | |