Amy's Reflection #12!

here are some things i understand:

to solve a right triangle you use SOHCAHTOA

Area of a right angle = 1/2 bh

Laws of Sines: sin A/a = sin B/b = sin C/c
*used when pairs are known in non-right angles
*set up a proportion

Area of a non-right triangle = 1/2 (leg) (leg) sin (angle b/w)

Examples:

1. B = 90 degrees C = 50 degrees b = 10

Angle A
180 - 90 - 50 = 40

Side a
cos 50 = a/10
a = 10 cos 50
a = 6.428

Side c

sin 50 = c/10
c = 10 sin 50
c = 7.660

area = 1/2 (6.428) (7.660) = 24.620

2. two sides of a triangle have the lengths 7cm & 4cm. the angle between the sides measures 73 degrees. Find the area of the triangle.

1/2 (7) (4) sin (73) = 13.388 cm^2

Hints (word problems):
if it mentions anything about anything that has height or something on the ground that means you're gonna draw a right angle and solve...and the other problems you're gonna draw an isosceles triangle and remember to draw down the middle to make it two separate triangles...
(i hope that makes sense....)

the worksheets are due tuesday, huh?