{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Unit 4 Section 10 - 4.10 SOLUTION OF OBLIQUE TRIANGLES...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
4.10 SOLUTION OF OBLIQUE TRIANGLES Again, to solve a triangle means to find the lengths of all its sides and the measures of all its angles. Solving Oblique Triangles Definition 4.10.1 A triangle with no right interior angle is called oblique . Such a triangle contains either three acute angles or two acute angles and one obtuse angle. Figure 1 Triangle Inequality : The sum of the lengths any two sides of any triangle is longer than the length of the third side; or the absolute value of the sum of two numbers is never larger than the sum of their absolute values. Any triangle, right or oblique, can be solved if at least one side and any other two measures are known . The five possible situations are illustrated below. 1. AAS : Two angles of a triangle and a side opposite one of them are known. Figure 2 2. ASA : Two angles of a triangle and the included side are known. Figure 3 3. SSA : Two sides of a triangle and an angle opposite one of them are known. (In this case, there may be no solution, one solution, or two solutions. The latter is known as the ambiguous case.)
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Figure 4 4. SAS : Two sides of a triangle and the included angle are known. Figure 5 5. SSS : All three sides of the triangle are known. Figure 6 The list above does not include the situation in which only the three angle measures are given. The reason for this lies in the fact that the angle measures determine only the shape of the triangle and not the size , as shown with the following triangles. Thus we cannot solve a triangle when only the three angle measures are given. Figure 7
Image of page 2
In order to solve oblique triangles, we need to derive the law of sines and the law of cosines . The law of sines applies to the first three situations listed above. The law of cosines applies to the last two situations. The Law of Sines In any triangle (see the figure below), the ratio of a side and the sine of the opposite angle is a constant; that is, or . Figure 8 Proof of the Law of Sines: Consider any oblique triangle. It may or may not have an obtuse angle. Although we look at only the acute-triangle case, the derivation of the obtuse-triangle case is essentially the same. In an acute at the left, we have drawn an altitude from vertex . It has length . From , we have or . From , we have or . With and , we now have . There is no danger of dividing by 0 here because we are dealing with triangles whose angles are never 0 o or 180 o . Thus the sine value will never be 0. If we were to consider altitudes from vertex and vertex in the triangle shown above, the same argument would give us or .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Solving Oblique Triangles (AAS and ASA) When two angles and a side of any triangle are known, the law of sines can be used to solve the triangle. Example 4.10.1 In , , and . Solve the triangle. Solution : We first make a drawing. We know three of the six measures.
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern