{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Pre-Calc Exam Notes 39

Pre-Calc Exam Notes 39 - for h in equation(2.5 and...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
The Law of Sines Section 2.1 39 Another way of stating the Law of Sines is: The sides of a triangle are proportional to the sines of their opposite angles. To prove the Law of Sines, let ABC be an oblique triangle. Then ABC can be acute, as in Figure 2.1.1(a), or it can be obtuse, as in Figure 2.1.1(b). In each case, draw the altitude 1 from the vertex at C to the side AB . In Figure 2.1.1(a) the altitude lies inside the triangle, while in Figure 2.1.1(b) the altitude lies outside the triangle. h b a c A B C (a) Acute triangle h b a c A B C 180 B (b) Obtuse triangle Figure 2.1.1 Proof of the Law of Sines for an oblique triangle ABC Let h be the height of the altitude. For each triangle in Figure 2.1.1, we see that h b = sin A (2.4) and h a = sin B (2.5) (in Figure 2.1.1(b), h a = sin (180 B ) = sin B by formula (1.19) in Section 1.5). Thus, solving
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: for h in equation (2.5) and substituting that into equation (2.4) gives a sin B b = sin A , (2.6) and so putting a and A on the left side and b and B on the right side, we get a sin A = b sin B . (2.7) By a similar argument, drawing the altitude from A to BC gives b sin B = c sin C , (2.8) so putting the last two equations together proves the theorem. QED Note that we did not prove the Law of Sines for right triangles, since it turns out (see Exercise 12) to be trivially true for that case. 1 Recall from geometry that an altitude of a triangle is a perpendicular line segment from any vertex to the line containing the side opposite the vertex....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online