1) If <A = 30º, and c = 20, find a, b, and <B
2) If a = 10 and b = 14, find c, <A, and <B
3) If a = 6 and <B = 53º, find b, c, and <A
4) If c = 13 and b = 12, find a, <A, and <B
5) If <A = 20º and b = 10, find <B, a and c
6) If c = 30 and <A = 40º, find <B, a, and b
Unless these are right triangles, more information is needed to answer your questions. Is there a figure that is supposed to go with these questions?
on the paper all the info is given is : Given Trianlge ABC, and i think all of em are right triangles...
To solve these questions, we need to use the trigonometric relationships between the sides and angles of a triangle. Specifically, we can use the sine, cosine, and tangent functions.
1) Given <A = 30º and c = 20, we need to find a, b, and <B.
a) To find <B, we know that the sum of the angles in a triangle is 180º. So, <B = 180º - <A - <C = 180º - 30º - 90º (since <C is a right angle) = 60º.
b) To find a, we can use the sine function: sin(A) = a/c. Rearranging the formula, we get a = c * sin(A) = 20 * sin(30º) = 10.
c) To find b, we can use the Pythagorean theorem: b² = c² - a². Substituting the given values, we have b² = 20² - 10² = 300, so b ≈ √300 ≈ 17.32.
2) Given a = 10 and b = 14, we need to find c, <A, and <B.
a) To find c, we can use the Pythagorean theorem: c² = a² + b². Substituting the given values, we have c² = 10² + 14² = 296, so c ≈ √296 ≈ 17.20.
b) To find <A, we can use the cosine function: cos(A) = a/c. Rearranging the formula, we get <A = cos⁻¹(a/c) = cos⁻¹(10/17.20) ≈ 55.43º.
c) To find <B, we know that the sum of the angles in a triangle is 180º. So, <B = 180º - <A - <C = 180º - 55.43º - 90º = 34.57º.
3) Given a = 6 and <B = 53º, we need to find b, c, and <A.
a) To find b, we can use the sine function: sin(B) = b/c. Rearranging the formula, we get b = c * sin(B) = c * sin(53º).
b) To find c, we can use the cosine function: cos(B) = a/c. Rearranging the formula, we get c = a / cos(B) = 6 / cos(53º).
c) To find <A, we know that the sum of the angles in a triangle is 180º. So, <A = 180º - <B - <C = 180º - 53º - 90º = 37º.
4) Given c = 13 and b = 12, we need to find a, <A, and <B.
a) To find a, we can use the Pythagorean theorem: a² = c² - b². Substituting the given values, we have a² = 13² - 12² = 25, so a = √25 = 5.
b) To find <A, we can use the sine function: sin(A) = a/c. Rearranging the formula, we get <A = sin⁻¹(a/c) = sin⁻¹(5/13) ≈ 23.58º.
c) To find <B, we know that the sum of the angles in a triangle is 180º. So, <B = 180º - <A - <C = 180º - 23.58º - 90º = 66.42º.
5) Given <A = 20º and b = 10, we need to find <B, a, and c.
a) To find <B, we know that the sum of the angles in a triangle is 180º. So, <B = 180º - <A - <C = 180º - 20º - 90º = 70º.
b) To find a, we can use the sine function: sin(A) = a/c. Rearranging the formula, we get a = c * sin(A) = c * sin(20º).
c) To find c, we can use the cosine function: cos(A) = b/c. Rearranging the formula, we get c = b / cos(A) = 10 / cos(20º).
6) Given c = 30 and <A = 40º, we need to find <B, a, and b.
a) To find <B, we know that the sum of the angles in a triangle is 180º. So, <B = 180º - <A - <C = 180º - 40º - 90º = 50º.
b) To find a, we can use the sine function: sin(A) = a/c. Rearranging the formula, we get a = c * sin(A) = 30 * sin(40º).
c) To find b, we can use the cosine function: cos(B) = b/c. Rearranging the formula, we get b = c * cos(B) = 30 * cos(50º).
By using these trigonometric relationships, you can solve various problems involving triangles and find the missing sides and angles.