A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{6}$ $(5pi)/6$ . If side C has a length of $3$ $3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ , what are the lengths of sides A and B?

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{6}$ $(5pi)/6$ . If side C has a length of $3$ $3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ , what are the lengths of sides A and B?

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Answer 1 · 2016-10-18T11:06:07

$A = 1.55 (2 d p) u n i t; B = 1.55 (2 d p) u n i t;$ $A=1.55(2dp) unit ; B=1.55(2dp) unit ;$

Explanation:

The angle between sides A and B is $∠ c = (5 \cdot \frac{180}{6}) = 150^{0}$ $/_c = (5*180/6)=150^0$
The angle between sides B and C is $∠ a = (\frac{180}{12}) = 15^{0}$ $/_a = (180/12)=15^0$
The angle between sides C and A is $∠ b = 180 - (150 + 15) = 15^{0}; C = 3$ $/_b = 180-(150+15)=15^0 ;C=3$
Applying sine law; $\frac{A}{sin a} = \frac{B}{sin b} = \frac{C}{sin c}$ $A/sina=B/sinb=C/sinc$ we get $\frac{A}{sin 15} = \frac{3}{sin 150} or A = 3 \cdot (\frac{sin 15}{sin 150}) = 1.55 (2 d p) u n i t$ $A/sin15=3/sin150 or A=3*(sin15/sin150)=1.55(2dp) unit$ and similarly we get $\frac{B}{sin 15} = \frac{3}{sin 150} or B = 3 \cdot (\frac{sin 15}{sin 150}) = 1.55 (2 d p) u n i t$ $B/sin15=3/sin150 or B=3*(sin15/sin150)=1.55(2dp) unit$ {Ans]

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{6}$ $(5pi)/6$ . If side C has a length of $3$ $3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ , what are the lengths of sides A and B?

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A triangle has sides A, B, and C. The angle between sides A and B is 5π6(5pi)/6. If side C has a length of 33 and the angle between sides B and C is π12pi/12, what are the lengths of sides A and B?

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{6}$ $(5pi)/6$ . If side C has a length of $3$ $3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ , what are the lengths of sides A and B?