How do you solve the triangle given $△ A B C, a = 22, b = 18, m ∠ C = 130$ $triangleABC, a=22, b=18, mangleC=130$ ?

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Answer 1 · 2017-01-04T16:42:30

Solution is $a = 22$ $a=22$ , $b = 18$ $b=18$ , $c = 36.29$ $c=36.29$ ,

$m ∠ A = {27.67}^{\circ}$ $m/_A=27.67^@$ , $m ∠ B = {22.33}^{\circ}$ $m/_B=22.33^@$ and $m ∠ C = 130^{\circ}$ $m/_C=130^@$

Solving a triangle means finding all sides and angles of the triangle. For this we use sine formulas i.e.

$\frac{a}{sin A} = \frac{B}{sin B} = \frac{c}{sin C}$ $a/sinA=B/sinB=c/sinC$ and

cosine formulas $a^2=b^2+c^2-2bccosA$ , $b^2=c^2+a^2-2cacosB$ and $c^2=a^2+b^2-2abcosC$ .

Here we are given $a=22$ , $b=18$ and $m/_C=130^@$ , hence

$c^2=22^2+18^2-2xx22xx18xxcos130^@$

$=484+324-792xx(-0.6428)=808+509.0976=1317.0976$

i.e. $c=36.29$ and using sine formula, we get

$22/sinA=18/sinB=36.29/(sin130^@)=36.29/0.766=47.376$

Hence, $sinA=22/47.376=0.4644$ and $A=27.67^@$

and $sinB=18/47.376=0.3799$ and $B=22.33^@$

Solution is $a=22$ , $b=18$ , $c=36.29$ ,

$m/_A=27.67^@$ , $m/_B=22.33^@$ and $m/_C=130^@$

How do you solve the triangle given triangleABC, a=22, b=18, mangleC=130△ABC,a=22,b=18,m∠C=130triangleABC, a=22, b=18, mangleC=130?