Law of Sines/Cosines

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lual0209

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Please could you verify that I have the following 3 questions correct:

1. Given the triangle ABC, with A = 58 degrees, C = 40 degrees, and b = 18, find side a (to 2 decimal places).

A + B + C = 180 degrees. 58 degrees + B + 40 degrees = 180 degrees. B = 82 degrees.

a = a/sinA = b/sinB = a/sin58 degrees = 18/sin82 degrees = 18 sin58 degrees/sin82 degrees = 15.41488231 = 15.41.

2. In triangle ABC, B = 38 degrees, a = 9.2, and c = 15.3. Find side b (to 2 decimal places).

b squared = a squared + c squared - 2 ac cosB = 9.2 squared + 15.3 squared - 2 (9.2) (15.3) cos38 degrees = 96.88921264.

b = square root of 96.88921264 = 9.843231819 = 9.84.

3. In triangle ABC, a = 22, b = 18, and c = 11. Find B (to the nearest degree).

b squared = a squared + c squared - 2 ac cosB, cosB = a squared + c squared - b squared/2 ac = 22 squared + 11 squared - 18 squared/2 (22) (11) = 281/484 = 0.580578512.

B = inverse cos (0.580578512) = 54.50875762 degrees = 55 degrees.

Thank you.
 
lual0209 said:
Please could you verify that I have the following 3 questions correct:

1. Given the triangle ABC, with A = 58 degrees, C = 40 degrees, and b = 18, find side a (to 2 decimal places).

A + B + C = 180 degrees. 58 degrees + B + 40 degrees = 180 degrees. B = 82 degrees.

a = a/sinA = b/sinB = a/sin58 degrees = 18/sin82 degrees = 18 sin58 degrees/sin82 degrees = 15.41488231 = 15.41.

2. In triangle ABC, B = 38 degrees, a = 9.2, and c = 15.3. Find side b (to 2 decimal places).

b squared = a squared + c squared - 2 ac cosB = 9.2 squared + 15.3 squared - 2 (9.2) (15.3) cos38 degrees = 96.88921264.

b = square root of 96.88921264 = 9.843231819 = 9.84.

3. In triangle ABC, a = 22, b = 18, and c = 11. Find B (to the nearest degree).

b squared = a squared + c squared - 2 ac cosB, cosB = a squared + c squared - b squared/2 ac = 22 squared + 11 squared - 18 squared/2 (22) (11) = 281/484 = 0.580578512.

B = inverse cos (0.580578512) = 54.50875762 degrees = 55 degrees.

Thank you.
Click to expand...

Looks good to me....
 
lual0209 said:
a = a/sinA = b/sinB = a/sin58 degrees = 18/sin82 degrees = 18 sin58 degrees/sin82 degrees = 15.41488231 = 15.41.
Click to expand...
Keereck! But quite "messy" :shock: Keep the process CLEAR:
a/SIN(A) = b/SIN(B)
a/SIN(58) = 18/SIN(82)
a = 18SIN(58) / SIN(82)
a = 15.414 ; a = 15.41
 
lual0209 said:
2. In triangle ABC, B = 38 degrees, a = 9.2, and c = 15.3. Find side b (to 2 decimal places).

b squared = a squared + c squared - 2 ac cosB = 9.2 squared + 15.3 squared - 2 (9.2) (15.3) cos38 degrees = 96.88921264.
Click to expand...
b^2 = a^2 + c^2 - 2acCOS(B)

OK?
 
lual0209 said:
Please could you verify that I have the following 3 questions correct:

1. Given the triangle ABC, with A = 58 degrees, C = 40 degrees, and b = 18, find side a (to 2 decimal places).

A + B + C = 180 degrees. 58 degrees + B + 40 degrees = 180 degrees. B = 82 degrees.

a = a/sinA = b/sinB = a/sin58 degrees = 18/sin82 degrees = 18 sin58 degrees/sin82 degrees = 15.41488231 = 15.41.

2. In triangle ABC, B = 38 degrees, a = 9.2, and c = 15.3. Find side b (to 2 decimal places).

b squared = a squared + c squared - 2 ac cosB = 9.2 squared + 15.3 squared - 2 (9.2) (15.3) cos38 degrees = 96.88921264.

b = square root of 96.88921264 = 9.843231819 = 9.84.

3. In triangle ABC, a = 22, b = 18, and c = 11. Find B (to the nearest degree).

b squared = a squared + c squared - 2 ac cosB, cosB = a squared + c squared - b squared/2 ac = 22 squared + 11 squared - 18 squared/2 (22) (11) = 281/484 = 0.580578512.

B = inverse cos (0.580578512) = 54.50875762 degrees = 55 degrees.

Thank you.
Click to expand...

P.S. Thank you to all who responded. Further to this, please can anyone let me know if in any of these cases (Q. 1, Q. 2, or Q. 3), there is no triangle that exists, and why not.
 
lual0209 said:
lual0209 said:
Please could you verify that I have the following 3 questions correct:

1. Given the triangle ABC, with A = 58 degrees, C = 40 degrees, and b = 18, find side a (to 2 decimal places).

A + B + C = 180 degrees. 58 degrees + B + 40 degrees = 180 degrees. B = 82 degrees.

a = a/sinA = b/sinB = a/sin58 degrees = 18/sin82 degrees = 18 sin58 degrees/sin82 degrees = 15.41488231 = 15.41.

2. In triangle ABC, B = 38 degrees, a = 9.2, and c = 15.3. Find side b (to 2 decimal places).

b squared = a squared + c squared - 2 ac cosB = 9.2 squared + 15.3 squared - 2 (9.2) (15.3) cos38 degrees = 96.88921264.

b = square root of 96.88921264 = 9.843231819 = 9.84.

3. In triangle ABC, a = 22, b = 18, and c = 11. Find B (to the nearest degree).

b squared = a squared + c squared - 2 ac cosB, cosB = a squared + c squared - b squared/2 ac = 22 squared + 11 squared - 18 squared/2 (22) (11) = 281/484 = 0.580578512.

B = inverse cos (0.580578512) = 54.50875762 degrees = 55 degrees.

Thank you.
Click to expand...

P.S. Thank you to all who responded. Further to this, please can anyone let me know if in any of these cases (Q. 1, Q. 2, or Q. 3), there is no triangle that exists, and why not.
Click to expand...

Check with triangle inequality (a + b >c, b + c > a and c + a> b) and A + B + C = 180°

for a triangle in euclidean space - all these four conditions must be met.
 
Thanks Subhotosh. I think that Q. 1 and Q. 3 are valid triangles, and that Q. 2 is actually two triangles. Therefore, a triangle exists in all three cases. Is this right?
 
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