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| 1. |
A triangle ABC at b 90 and c tan =^3 |
| Answer» Let ABC be a triangle right angled at C such that tan {tex}A = \\frac{1}{{\\sqrt 3 }}{/tex} and tan B ={tex}\\sqrt 3 {/tex}tan {tex}A = \\frac{1}{{\\sqrt 3 }}{/tex}{tex}\\Rightarrow {/tex}{tex}\\frac{{BC}}{{AC}} = \\frac{1}{{\\sqrt 3 }}{/tex}{tex}\\Rightarrow {/tex}{tex}\\frac{{BC}}{1} = \\frac{{AC}}{{\\sqrt 3 }} = k{/tex}(say)where k is a positive number{tex} \\Rightarrow \\left. \\begin{gathered} BC = k \\hfill \\\\ AC = \\sqrt 3 k \\hfill \\\\ \\end{gathered} \\right\\}{/tex}....(1){tex} \\Rightarrow \\frac{{AC}}{{BC}} = \\sqrt 3 {/tex}{tex} \\Rightarrow \\frac{{AC}}{{\\sqrt 3 }} = \\frac{{BC}}{1} = k(say){/tex}where k is a positive number{tex}\\left. \\begin{gathered} AC = \\sqrt 3 k \\hfill \\\\ BC = k \\hfill \\\\ \\end{gathered} \\right\\}{/tex} ...(2)From (1) and (2),AC ={tex}\\sqrt 3 k{/tex}BC = KIn {tex}\\vartriangle {/tex} ABC,{tex}\\because {/tex}{tex}\\angle{/tex} C={tex}{90^ \\circ }{/tex}{tex}\\therefore {/tex} AB2\xa0= AC2\xa0+ BC2 ...By Pythagoras theorem{tex}\\Rightarrow {/tex} AB2 = ({tex}\\sqrt 3 k{/tex})2\xa0+ (k)2{tex}\\Rightarrow {/tex} AB2 = 3k2\xa0+ k2{tex}\\Rightarrow {/tex} AB2\xa0= 4k2{tex}\\Rightarrow {/tex} AB =\xa0{tex}\\sqrt {4{k^2}} {/tex}{tex}\\Rightarrow {/tex} AB = 2kTherefore,sin A =\xa0{tex}\\frac{{BC}}{{AB}} = \\frac{k}{{2k}} = \\frac{1}{2}{/tex}cos B =\xa0{tex}\\frac{{BC}}{{AB}} = \\frac{k}{{2k}} = \\frac{1}{2}{/tex}cos A =\xa0{tex}\\frac{{AC}}{{AB}} = \\frac{{\\sqrt 3k }}{{2k}} = \\frac{{\\sqrt 3 }}{2}{/tex}sin B ={tex}\\frac{{AC}}{{AB}} = \\frac{{\\sqrt 3 k}}{{2k}} = \\frac{{\\sqrt 3 }}{2}{/tex}Now, sin A cos B + cos A sin B{tex}\\frac{1}{2}.\\frac{1}{2} + \\frac{{\\sqrt 3 }}{2}.\\frac{{\\sqrt 3 }}{2} = \\frac{1}{4} + \\frac{3}{4} = 1{/tex} | |