How do you solve the triangle given $B=75^circ20', a=6.2, c=9.5$ ?

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Answer 1 · 2017-05-27T14:12:49

$angleB=14°40',angleA=75°20',b=1.623$

angle B in the question should be angle A

$angleA=75°20',a=6.2,c=9.5$

$:.$ Ajacent/opposite $=b/a= Cot 75°20'$

$:.b/6.2= Cot 75°20'$

multiply both sides by 6.2

$:.b=Cot 75°20' xx 6.2$

$:.b=0.261723366 xx 6.2$

$:.b=1.622684873$

$:.color(blue)(b=1.623$ to the nearest 3 decimal places.

$:.180°-(90°+75°20')=14°40'$

$:.color(blue)(angleB=14°40'$

How do you solve the triangle given B=75^circ20', a=6.2, c=9.5B=75^circ20', a=6.2, c=9.5?