A triangle ABC has side AB being 10cm and side BC being 8cm. The angle BAC is 30°

Question

a.) use cosine rule to find 2 possible values of side AC

b.) find the differences between the 2 possible measurements of angle ABC

c.) find 2 possible distances from point C to side AB

Answer 1

Here is a start for you.

8² = x² + 10² - 2(10)x (cos30); This is SSA, which does not guarantee one triangle.8 is the side opposite the 30 degree angle. Cos 30 = √3/2.

Solving the above quadratic equation using quadratic formula will give the two possible values for side AC.

x² - x10√3 +36 = 0; use the quadratic formula with a = 1, b = -10√2, c = 36.

Hopefully that is sufficient to get you on your way to answering the asked questions. If you need some more help, post a comment asking for additional clarification.

Answer 2

Lets put this in trig figures: α=30°

β=?

γ=?

Sides: a=8

b=10

c?

a/sinα=b/sinβ=c/sinγ

8/sin30=15.736=10/.635asin=39.455=β; γ=180-39.455-30=110.545

c/sin110.545=15.763 so c/.936;c=14.735

The imfo you gave is just for 1 triangle (sas).