How do you solve the triangle given $△ A B C, a = 8, m ∠ B = 35, m ∠ C = 55$ $triangleABC, a=8, mangleB=35, mangleC=55$ ?

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Answer 1 · 2018-06-25T12:21:48

See below

We know
$a = 8$ $a=8$
$ˆ B = 35$ $hatB=35$
$ˆ C = 55$ $hatC=55$

Then however $ˆ A + ˆ B + ˆ C = 180$ $hatA+hatB+hatC=180$ then $ˆ A = 90$ $hatA=90$

ABC is a right triangle

By $sin$ $sin$ theorem $\frac{a}{sin A} = \frac{b}{sin B}$ $a/sinA=b/sinB$ or $\frac{8}{1} = \frac{b}{sin 35}$ $8/1=b/sin35$ then

$b = 8 \cdot sin 35 = 4.5886$ $b=8·sin35=4.5886$

And again with leg c: $\frac{c}{sin 55} = \frac{8}{sin 90}$ $c/sin55=8/sin90$ and $c = 8 \cdot sin 55 = 6.5532$ $c=8·sin55=6.5532$

Or by Pitagorean theorem $8^{2} = {4.5886}^{2} + c^{2}$ $8^2=4.5886^2+c^2$

How do you solve the triangle given triangleABC, a=8, mangleB=35, mangleC=55△ABC,a=8,m∠B=35,m∠C=55triangleABC, a=8, mangleB=35, mangleC=55?