How do you solve the triangle of $A = 76^{\circ}, a = 34, b = 21$ $A=76^circ, a=34, b=21$ ?

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How do you solve the triangle of $A = 76^{\circ}, a = 34, b = 21$ $A=76^circ, a=34, b=21$ ?

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Answer 1 · 2017-10-23T15:10:09

See the following :)

Explanation:

By the Law of Sines, $\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $(sinA)/a = (sinB)/b = (sinC)/c$
For this question, we know that $A = 76^{\circ}, a = 34 and b = 21$ $A= 76^@, a= 34 and b=21$ , and we can simply plug in the numbers and get B:

$\frac{sin (76^{\circ})}{34} = \frac{sin B}{21}$ $[sin(76^@)]/34 = (sin B)/21$

$B = {36.81980185}^{\circ}$ $B= 36.81980185^@$

$A + B + C = 180^{\circ}$ $A+B+C=180^@$ (angle sum of triangle)
$C = {67.18019815}^{\circ}$ $C=67.18019815^@$

$\frac{sin (76^{\circ})}{34} = \frac{sin C}{c}$ $[sin(76^@)]/34 = (sin C)/c$
$\frac{sin (76^{\circ})}{34} = \frac{{sin 67.18019815}^{\circ}}{c}$ $[sin(76^@)]/34 = (sin67.18019815^@ )/c$
$c = 32.29818587$ $c=32.29818587$

Hope this can help you :)

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How do you solve the triangle of $A = 76^{\circ}, a = 34, b = 21$ $A=76^circ, a=34, b=21$ ?

1 Answer

Explanation:

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How do you solve the triangle of $A = 76^{\circ}, a = 34, b = 21$ $A=76^circ, a=34, b=21$ ?