Hard Trig Question

Question

In a triangle, one angle is 45° larger than the other two (combined). The side opposite the smaller angle is 10 cm and the side opposite the larger one is 14 cm. Determine the angles in the triangle using the sine theorem.

Answer 1

Let x and y represent two of the angles of the triangle. The 3rd angle is given to be x + y + 45.

Since the angles of a triangle add to 180 (degrees), that means:

x + y + x + y +45 = 180

2x+2y = 135

x + y = 67.5

Also, x + y + 45 = 112.5 (largest angle is obtuse).

So x and y are acute angles and the 3rd angle must be the x + y + 45 (and that is the angle opposite the longest side (14 cm).

Let x be the angle opposite the side with length 10.:

sin(x)/10 = sin(112.5)/14

sin (x) = 5/7 (sin(112.5)

x = sin^-1 [ 5/7 (sin (112.5)] (at this point we realize that x CANNOT be 180 - sin^-1 [etc...] because x is acute)

Since x + y = 67.5

y = 67.5 - x

and finally the 3rd angle = x + y + 45 (or 180 - (x+y)) [already determined to be 112.5]

FYI, the wording of the problem is a bit confusing, suggesting that 10 is opposite the smallest angle?, but I think the meaning is that 10 is opposite the smaller of x and x+y+45.

Answer 2

Let A be the angle mentioned. Then:

A = B + C + 45° (if B and C are the other two angles)

For any triangle:

A + B + C = 180°

From the first equation:

B + C = A – 45°

Substituting in the second equation:

A + (A-45) = 180°

2A = 225°

A = 112.5°

And

B + C = 112.5° – 45° = 67.5°

Now we can use the law of cosines to determine the length of side C:

a² = b² + c² - 2bc(cos(A))

a² = 10² + 14² - 2(10)(14)(cos112.5°)

a² = 100 + 196 - 280(-0.3827)

a² = 403.15

a = 20.08 ~ 20

Using law of sines:

sin(A)/a = sin(B)/b

sin(B) = (b/a)sin(A)

sin(B) = (10/20)sin112.5°

sin(B) = (1/2)(0.9239)

sin(B) = 0.4620

B = 27.5°

And:

B + C = 67.5°

C = 67.5° - B

C = 67.5° - 27.5°

C = 40°

Answer: A = 112.5°. B = 27.5°, C = 40°