what is the distance from a to b?

Question

Points A and B are on opposite sides of a lake. Point C is 81 meters from A. The measure of angle BAC is determined to be 102°, and the measure of angle ACB is determined to be 49°. What is the distance from A to B, rounded to the nearest whole meter?

Answer 1

The lake is irrelevant to this problem. First, draw triangle ABC with the given values. Now, let's recognize that we have two angles in this triangle, so we can find the third: B = 180 - A - C = 180 - 102 - 49 = 29. Now we can use the law of sines, which says c/sinC = b/sinB. We can rearrange that to get c = b*sinC/sinB. Since you know the length of side b and the measures of angles B and C, just plug in those values and calculate.

Answer 2

angels:

A=γ=102°

B=β=49º

C=α=29°

point C (α) is an end of side "b"=81m

side "a"=a/sinα=b/sinβ=c/sinγ

c/.978=81/.754=107.326; side "c"=104.98m

a^2+b^2=c^2

104.98^2-81^2=4460^.5=66.783m side "a"

Answer 3

Please see video. Really needed a figure to explain this one...