The two sides of a triangle are 540 and 420 yards long respectively and include an angle of 52o6′. Find the remaining angles, given thatlog2=.30103, Ltan26o3′=9.6891430,Ltan14o20′=9.4074189, and Ltan14o21′=9.4079453.

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Filo · Accepted Answer

Let b=540, c=420 and A=52o6′tan2B−C​=b+cb−c​cot2A​tan2B−C​=540+420540−420​cot26o3′=8tan26o3′1​Ltan2B−C​=log1−(log8+Ltan26.3′)+20                         =0−3×0.30103−9.6891430+20                         =9.4077670Ltan14o21′−Ltan14o20′=9.4079453−9.40791489                                                   =0.0005264And Ltan2B−C​−Ltan14o20′=1.4077670−9.4074189                                                             =0.0003481∴  Difference for 0.0003481=0.00052640.0003481×60​=40"Hence 2B−C​=14o20′40"               ---- ( 1 )And 2B+C​=90o−26o3"=63o57′                  --- ( 2 )Solving ( 1 ) and ( 2 ), we have∠B=78o17′40" and ∠C=49o36′20"

Question Text	The two sides of a triangle are $540$ and $420$ yards long respectively and include an angle of $5 2^{o} 6^{'}$ . Find the remaining angles, given that $l o g 2 = .30103$ , $L tan 2 6^{o} 3^{'} = 9.6891430$ , $L tan 1 4^{o} 2 0^{'} = 9.4074189$ , and $L tan 1 4^{o} 2 1^{'} = 9.4079453$ .
Answer Type	Text solution:1
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The two sides of a triangle are $540$ and $420$ yards long respectively and include an angle of $5 2^{o} 6^{'}$ . Find the remaining angles, given that
$l o g 2 = .30103$ , $L tan 2 6^{o} 3^{'} = 9.6891430$ ,
$L tan 1 4^{o} 2 0^{'} = 9.4074189$ , and $L tan 1 4^{o} 2 1^{'} = 9.4079453$ .

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