#hat A = 102^@, b = 13, c = 10#

We cannot directly use Law of Sines.

#"First let's use Law of Cosines to find the third side " color(brown)(a)#

#a^2 = b^2 + c^2 - (2 b c cos A)#

#a = sqrt(13^2 + 10^2 - (2 * 13 * 10 * cos 102)) = 17.97 " units"#

Only one triangle is possible, having #" " hatA = 102^@# an obtuse angle.

Since we know all the three sides and one angle, we can apply law of sines to find the other two angles.

#sin A / a = sin B / b = sin C / c#

#hat B = sin ^-1 ((b.sin A) / a) = sin^-1 ((13 * sin 102)/17.97) = 45.03^@#

Similarly, #hat C = sin _1 ((10 * sin 102) / 17.97) = 32.97^@#

#"Area of the triangle " A_t = (1/2) b c sin A = (1/2) * 13 * 10 * sin 102 = color(maroon)(63.58 " sq units"#