The Ambiguous Case 6.2 The Ambiguous Case
Name: __________________________ Rating: _____________
Course: _________________________ Date: ______________
Instructor: ____________________________
Exercise 6.1
Law of Sines
(Case I)
A. Solve the unknown parts of the following triangles with the given conditions:
1. S = 660 , M = 580 s = 5.8 cm in SMN
2. T = 840 , M = 690 , c = 25.56 , in TMC
B. Solve the following problems. (Show your solutions and draw the figure)
1. One diagonal of a parallelogram is 64 cm. It makes an angle of 470 and 570 respectively, with the sides. Find the sides of the parallelogram.
2. A flagpole 15 ft tall stands on top of a building. From a point in the same horizontal plane with the base of the building, the …show more content…
6.2 Case 2 (The Ambiguous Case) If the given parts of the triangle are two sides and the angle opposite one of the given side, there are many possibilities that may arise, t hat is there maybe no triangle formed, one triangle or two triangles defending upon the given condition, and this is known as the ambiguous case. Following are the conditions to be considered:
1. When a>b, but bb, but m = a sin B, one right triangle is formed
3. When a=b one isosceles triangle is formed.
4. When b>a, two triangles can be formed.
5. When a>b but b >a sin B, two oblique triangles are formed.
Example 1. Solve OMC given that o = 36 cm, c = 48, C = 840
Solution: Let us draw the figure: Solving for O we use the Sine Law M = ? c = 48 cm o = 36 cm sin O = sin O = sin O = ? 840 O C m = ? sin O …show more content…
In symbol we have: a2 = b2 + c2 – 2bc cos A or a = b2 = a2 + c2 – 2ac cos B or b = c2 = a2 + b2 – 2ab cos C or c = The formula above can be used in finding the third side of the triangle in case 3. and simplifying the above formula to solve for the angles of the triangles, we have; cos A = cos B = cos C =
Case 3: Given two sides and included angle: (Law of Cosines)
Example: Solve the triangle given the following conditions:
1. b = 25, A =550, c = 32
Solution:
Course: _________________________ Date: ______________
Instructor: ____________________________
Exercise 6.1
Law of Sines
(Case I)
A. Solve the unknown parts of the following triangles with the given conditions:
1. S = 660 , M = 580 s = 5.8 cm in SMN
2. T = 840 , M = 690 , c = 25.56 , in TMC
B. Solve the following problems. (Show your solutions and draw the figure)
1. One diagonal of a parallelogram is 64 cm. It makes an angle of 470 and 570 respectively, with the sides. Find the sides of the parallelogram.
2. A flagpole 15 ft tall stands on top of a building. From a point in the same horizontal plane with the base of the building, the …show more content…
6.2 Case 2 (The Ambiguous Case) If the given parts of the triangle are two sides and the angle opposite one of the given side, there are many possibilities that may arise, t hat is there maybe no triangle formed, one triangle or two triangles defending upon the given condition, and this is known as the ambiguous case. Following are the conditions to be considered:
1. When a>b, but bb, but m = a sin B, one right triangle is formed
3. When a=b one isosceles triangle is formed.
4. When b>a, two triangles can be formed.
5. When a>b but b >a sin B, two oblique triangles are formed.
Example 1. Solve OMC given that o = 36 cm, c = 48, C = 840
Solution: Let us draw the figure: Solving for O we use the Sine Law M = ? c = 48 cm o = 36 cm sin O = sin O = sin O = ? 840 O C m = ? sin O …show more content…
In symbol we have: a2 = b2 + c2 – 2bc cos A or a = b2 = a2 + c2 – 2ac cos B or b = c2 = a2 + b2 – 2ab cos C or c = The formula above can be used in finding the third side of the triangle in case 3. and simplifying the above formula to solve for the angles of the triangles, we have; cos A = cos B = cos C =
Case 3: Given two sides and included angle: (Law of Cosines)
Example: Solve the triangle given the following conditions:
1. b = 25, A =550, c = 32
Solution: