Can you help?

Question

http://i265.photobucket.com/albums/ii233/spare-site-444/mathsquestion.jpg

Answer 1

Sin 67 = x/11.7
.9205 = x/11.7
x = 10.77

Area = 1/2 base X height =
1/2 X 28.3 X 10.77 = 165.55

b² = a² + c² -2ac cos B

b²= 28.3² + 11.7² -2 X 28.3 X 11.7 X cos 67

b² = 800.89 + 136.89 - 2 X 11.7 X 28.3 X .3907 = 808.415

b = 28.433

Answer 2

(a)
Draw a line segment from A down to side BC, meeting BC in a right angle. Let the point of intersection be D. ∆ADB is a right triangle. The length of AD can be obtained from

sin 67° = opposite/hypotenuse = AD/AB = AD/11.7

Solve for AD. Then

Area of ∆ABC = ½(base)(height) = ½ BC*AD

(b)
Use the law of cosines: AC² = AB² + BC² - 2(AB)(BC) cos 67°