If b=14,c=11, and A=60o, find B and C, given thatlog2=.30103, log3=.4771213,Ltan11o44′=9.3174299,and Ltan11o45′=9.3180640.

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Filo · Accepted Answer

tan2B−C​=b+cb−c​cot2A​                      =14+1114−11​cot260o​                      =253​cot30o                      =253​3​                      =4×10033/2​Ltan2B−C​=10+2log2+23​log3−log100                         =10+2×0.30103+23​×0.4771213−2                         =9.3177419Ltan11o45′−Ltan11o44′=9.3180640−9.3174299                                                   =0.0006341and Ltan2B−C​−Ltan11o44′=9.3177419−9.3174299                                                     =0.0003120∴  Difference for 0.0003120=0.000634160×0.0003120​=30"∴  2B−C​=11o44′30"             ---- ( 1 )And 2B+C​=90o−30o            ----- ( 2 )Solving equations ( 1 ) and ( 2 ), we get∠B=71o44′30", ∠C=48o15′30"

Question Text	If $b = 14, c = 11$ , and $A = 6 0^{o}$ , find B and C, given that $l o g 2 = .30103$ , $l o g 3 = .4771213$ , $L tan 1 1^{o} 4 4^{'} = 9.3174299$ , and $L tan 1 1^{o} 4 5^{'} = 9.3180640$ .
Answer Type	Text solution:1
Upvotes	150

If $b = 14, c = 11$ , and $A = 6 0^{o}$ , find B and C, given that
$l o g 2 = .30103$ , $l o g 3 = .4771213$ ,
$L tan 1 1^{o} 4 4^{'} = 9.3174299$ ,
and $L tan 1 1^{o} 4 5^{'} = 9.3180640$ .

Text solutionVerified

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