How do you solve the triangle given $B=10^circ35', a=40, c=30$ ?

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Answer 1 · 2017-01-23T17:37:06

b= 11.87, $C= 27.69^o$ , $A= 141.71^o$

Use cosine formula $cos B=(a^2 +c^2 -b^2)/(2ac)$

Thus $cos 10^o 35' = (40^2 +30^2 -b^2)/(2(40)(30))$
(for calculation purpose consider $10^o 35'=10.6^o)$

$0.9829=(2500 -b^2)/2400$

$b^2= 2500- 0.9829*2400 = 2500-2359=141$

$b=sqrt 141$ = 11.87

Next using sine formula $sin B/ b=sin C/c$

$sinC=30/sqrt148sin 10^o 35'=0.4647$

C= $27.69^o$

$A= 180-10^o 35' - 27.69^O = 141.71^o$

How do you solve the triangle given B=10^circ35', a=40, c=30B=10^circ35', a=40, c=30?