Suppose a=8 b=8√3

Question

a. If γ = 90°, solve the triangle.

b. If c = 9(1 +√3), how many triangles can be formed? If there are none, explain why.

Otherwise, solve the triangle/s.

c. If α = 30° what are the possible values for β?

SHOW COMPLETE SOLUTION

Answer 1

a. We assume that γ = 90° is the angle between a and b.

Thus, we have a right triangle and c² = a²+ b², and thus c² = 8²+ 3*8² = 4*8², and thus c = 16.

What if γ = 90° is the angle between a and c? It says to solve the triangle, but may be worth considering this possibility as well: in this case: 8²+ c² = 8²*3, giving us : c = 8√2

b.We can start by using the law of cosines: c² = a²+ b²-2*a*b* cos γ

We plug in the values we are given to try to solve for cos γ.

81*(4+2√3) = 64+64*3 - 128√3 cos γ

324+ 162√3 = 256 - 128√3 cos γ

Thus:

128√3 cos γ = -68 - 162√3

cos y = -68/(128√3) - 162/128 < -1

This would be impossible! Any angle y would give -1 <= cos y < =1.

So no such triangle exists.

c. We denote α with A use the law of sines:

sin A/ a = sin B/ b = sin C/ c

sin A = 1/2

and thus sin B = b/a * sin A = √3/2, which gives B = π/3 or 60°, or B = 2π/3 or 120°

Answer 2

a)

a=8, b=8sqr3

y=90 degrees

3rd side, opposite the right angle = 16

angle A, opposite 8 = 30 degrees

angle B, opposite 8sqr3, = 60 degrees

It's a familiar right triangle . If you don't recognize it then

use the Pythagorean Theorem to solve for the 3rd side

then use Law of Sines to solve for one other angle

then 90-A= B

or 90-B=A

b)

Use the Law of Cosines, to find the the Cosine of angle C, angle opposite side c

It's less than -1, which is impossible, as all Cosines are greater than or equal to -1

so no triangle is possible

c)

if A = 30 degrees what are possible values for B?

0<B<180-30=150

but narrow it down further with the Law of Sines

2 possibilities, one in Quadrant I, 2nd in Quadrant II

60 and 120 degrees