Get back to business...V

direct application of $Area=\frac{1}{2}absin\theta$

$\frac{1}{2}8\times 12\times sin150 = 24$

Give Angle ABC = 150, its supplement, Angle ABK, must be 30. That makes Triangle ABK a 30:60:90 and thus side AK must be half the hypotenuse because it is opposite the 30 degree angle and all 30: 60: 90 triangles have sides in proportion a : a(sqrt { 3 }): 2a. As side AK is also the height of Triangle ABC, we have Area = 1/2 4 12 = 24.

Area of a triangle = 0.5 a b*sin x where x is the included angle between the sides a and b.

Easy problem

actually the question should not have included the construction of perpendicular, it would be more elementary

1/2 X 8 X 12 X sin 150 = 4 x 12 x 1/2 = 24

The angle $\measuredangle ABC=150°$ , we can assume that $AK$ is an altitude of the triangule $ABC$ . So $\measuredangle ABK=30°$ and $\measuredangle BAK=60°$ ; so the triangle $AKB$ is half of an equilateral triangle, so $2AK=AB$ so $AK=4$ . So the area is $\boxed { 24 }$

If the angle is 150º, your complement is 30º. Thus, sin 30º= 1/2. Then AK is x, sin 30º = x/8, 1/2 = x/8 --> 2x = 8 --> x = 4

If AK = 4, hence, the area of triangle ABC is height versus base divided by 2. AK is height and BC is base, then area of triangle is: (4 x 12) / 2 --> 48/2 = 24. The answer is 24 cm².