Triangle with Difference of Two Angles Given

Question

About triangle ABC, we know b = 7 inches, c = 6 inches, and β-γ = 15° [where α = ∠A, β = ∠B, and γ = ∠C]. Calculate α, β, and γ.

Answer 1

Using the law of sines, we have sinβ/7 = sinγ/6.

β = (15 + γ), so

sin (15 + γ) / 7 = sinγ /6

Using the sum of two angle formula, we have (sin15 cos γ + cos15 sin γ) / 7 = sinγ / 6.

Divide both sides by sinγ to get

sin15cotγ / 7 = 1/6.

The only unknown is cot γ, so you can solve for angle γ. (recall the cot γ = 1/ tan γ to use calculator).

Once you have γ, β = 15 + γ.

Since the sum of the angles of a triangle = 180, you can now calculate α.

After rounding, I got α = 61 , β = 67, γ = 52.

Answer 2

You need two formulas here.

The first one is Mollweide's formula (b - c)/ a = [sin(B-A)/2]/cos(A/2)

The second one is the theorem of cosines: a²=b² +c²-2bccosA.

Mollweide's formula yields 1 / a = [ sin 7.5] / cos[A/2]

and the Theorem of cosines: a² =85 -84 cosA

in addition you are going to use Cos[A] = 2cos^2[[A/2] -1.

Now the ball is in your court!!!

Answer 3

Think about how you can use the sine rule to write another equation for β and γ. Now, with two equations and two unknowns, you can find the values of β and γ (this will require some trig identities!). Once you have two angles in a triangle, how do you find the 3rd?