How do you solve the triangle given A= 48 deg C=97 deg a=12?

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How do you solve the triangle given A= 48 deg C=97 deg a=12?

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Answer 1 · 2018-03-16T09:34:08

Angle $ˆ A = 48^{\circ}, ˆ B = 35^{\circ}, ˆ C = 97^{\circ}$ $hat A = 48^@, hat B = 35^@, hat C = 97^@$

Side $a = 12, b = 9.26, c = 16.03$ $a = 12, b = 9.26, c = 16.03$ units

Perimeter $P = 37.29$ $P = 37.29$ Units

Area $A_{t} = 55.15$ $A_t = 55.15$ sq units

Explanation:

$a = 12, ˆ A = 48^{\circ}, ˆ C = 97^{\circ}$ $a = 12, hatA = 48^@, hat C = 97^@$

$ˆ B = 180 - 48 - 97 = 35^{\circ}$ $hat B = 180 - 48 - 97 = 35^@$

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$\frac{12}{{sin 48}^{\circ} =} \frac{b}{{sin 35}^{\circ} =} \frac{c}{{sin 97}^{\circ}}$ $12 / sin 48^@ = b / sin 35^@ = c / sin 97^@$

$b = \frac{12 \cdot sin 35}{sin 48} \approx 9.26$ $b = (12 * sin 35) / sin 48 ~~ 9.26$

$c = \frac{12 \cdot sin 97}{sin 48} \approx 16.03$ $c = (12 * sin 97) / sin 48 ~~ 16.03$

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Perimeter of the triangle $P = 12 + 9.26 + 16.03 = 37.29$ $P = 12 + 9.26 + 16.03 = 37.29$ units

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Area of the triangle $A_{t} = (\frac{1}{2}) \cdot a \cdot b \cdot sin C = (\frac{1}{2}) \cdot 12 \cdot 9.26 \cdot sin 97 = 55.15$ $A_t = (1/2) * a * b * sin C = (1/2) * 12 * 9.26 * sin 97 = 55.15$ sq units

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How do you solve the triangle given A= 48 deg C=97 deg a=12?

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