## Thursday, October 30, 2008

### Cosine Law

Today what we did was Exercise 25 # 1-9 , 17 ,18.

1.) a^2 = 15,b^2= 19, < c =" 50" 2=" a" c =" 15^2" 39 =" 586" 39 =" 219.61" a ="15," b =" 10," c =" 12."> A to the nearest degree.

a^2 = b^2 + c^2 - 2bc cos A
a^2 -b^2-c^2 = 2bc cos A
(a^2 -b^2-c^2)/2bc = cos A
A = cos-1 [(a^2 -b^2-c^2)/2bc )]
A = cos-1 [(15^2 - 10^2 - 12^2)/2(10)(12)]
A = 85 degrees

9.a) Find the area of the triangle.

c^2 = a^2 + b^2
c^2 - b^2 = a^2
10^2 - 6^2 = a^2
100 - 36 = a^2
64 = a^2
a = 8
a = h
h=8

A = bh/2
A = (12)(8)/2
A = 48 units^2

9.b) If you know the lengths of all three sides of an isosceles triangle, you have to use the pythagorean theorem to find the height of the triangle. Once you find the height, you can plug in the values to the formula and you can find the area.

17. Does the line 2x + 3y = 39 pass through the point (9, 7)? Justify.

2x + 3y = 39
3y = -2x + 39
y = (-2/3)x + 13

y = (-2/3)(9) + 13
y = -6 + 13
y = 7

18. Draw a diagram to show Triangle ABC if AB = 8, AC = 12, and angle A = 55 degrees. Find the area of the triangle.

A = [bc(sin A)]/2
A = [(12)(8)(sin 55)]/2
A = 39.3 units^2

THE NEXT SCRIBE WILL BE............................
POGIshawn