Use the Law of Sines to solve the triangle? 6.) A=60 degrees, a=9, c=10.

1 Answer
Jan 1, 2017

Check for the Ambiguous Case and, if appropriate, use the Law of Sines to solve the triangle(s).

Explanation:

Here is a reference for The Ambiguous Case

angle A is acute. Compute value of h:

h = (c)sin(A)

h = (10)sin(60^@)

h ~~ 8.66

h < a < c, therefore, two possible triangles exist, one triangle has angle C_("acute") and the other triangle has angle C_("obtuse")

Use The Law of Sines to compute angle C_("acute")

sin(C_("acute"))/c = sin(A)/a

sin(C_("acute")) = sin(A)c/a

C_("acute") = sin^-1(sin(A)c/a)

C_("acute") = sin^-1(sin(60^@)10/9)

C_("acute") ~~ 74.2^@

Find the measure for angle B by subtracting the other angles from 180^@:

angle B = 180^@ - 60^@ - 74.2^@

angle B = 45.8^@

Use the Law of Sines to compute the length of side b:

side b = asin(B)/sin(A)

b = 9sin(45.8^@)/sin(60^@)

b ~~ 7.45

For the first triangle:

a = 9,b ~~ 7.45, c = 10, A = 60^@, B ~~ 45.8^@, and C ~~ 74.2^@

Onward to the second triangle:

angle C_("obtuse") ~~ 180^@ - C_("acute")

C_("obtuse") ~~ 180^@ - 74.2^@ ~~ 105.8^@

Find the measure for angle B by subtracting the other angles from 180^@:

angle B = 180^@ - 60^@ - 105.8^@ ~~ 14.2^@

Use the Law of Sines to compute the length of side b:

b = 9sin(14.2^@)/sin(60^@)

b ~~ 2.55

For the second triangle:

a = 9,b ~~ 2.55, c = 10, A = 60^@, B ~~ 14.2^@, and C ~~ 105.8^@