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7. SOLVING OBLIQUE TRIANGLES: THE LAW OF SINES An oblique triangle is one without an angle of measure 90 o . When either two angles and a side are known (AAS) in the triangle Δ ABC or two sides and the angle opposite one of them (SSA) is given, then the law of sines may be applied to solve the triangle. Theorem 7.1: The Law of Sines In the general triangle Δ ABC sin α sin β sin γ = = a b c To prove the law of sines for the oblique triangle shown in Figure 7.1, we will sin α sin β = . first show that a b . C γ b A . a α β . B c Figure 7.1 For this, we drop a perpendicular C D from vertex C to the straight line containing the vertices A and B. Because both angles α and β are acute in the figure, D lies between A and B. In the right triangles Δ ADC and Δ BDC, we have sin α = CD b and sin β = CD a ‘ so b sin α = C D and a sin β = C D . Consequently, b sin α = a sin β . Dividing both sides of the last equation by ab , we obtain sin α sin β = . a b 47 E . C h γ′ a b l . β B . α D Figure 7.2 A sin α sin γ = , we observe that angle γ is obtuse. Let B E be a c the perpendicular from vertex B to the line l containing side AC . Let B E = h. Now to verify that Then h sin γ ′ = = sin (180 o − γ ) = sin γ a so sin α = and h c a sin γ = c sin α . Thus, sin α sin γ = . a c The proof of the law of sines in the case of an acute triangle Δ ABC is similar. Notice that the law of sines (Theorem 7.1) can be written in the alternative form: a b c = = sin α sin β sin γ . Solving a Triangle (AAS) If two angles of the triangle Δ ABC are given, then the third angle can be found by using the relationship: α + β + γ = 180 o ; 48 hence, the three denominators sin α , sin β , and sin γ can be found using a calculator. Now, if any one of the sides a, b, or c is also given, then the equations a b c = = sin α sin β sin γ can be solved for the remaining two sides. The following example indicates the procedure for solving a triangle when two angles and one side are given (or can be determined from the information provided.) Unless otherwise indicated, we shall round off all angles to the nearest hudredth of a degree, and all side lengths to four significant digits. Example 7.1 ---------------------------- -----------------------------------------------------------In Δ ABC , suppose that α = 41o , β = 77 o , and a = 74. Solve for γ , b, and c. B β = 77 o c α = 41o A Solution Figure 7.3 a = 74 γ C b γ = 180 o − 41o − 77 o = 62 o a b c = = By the law of sines, ; o o sin 41 sin 77 sin 62 o hence, since a = 74, b = a sin 77 o sin 41o = 74 sin 77 o sin 41o = 109.9 and c = a sin 62 o sin 41o = 74 sin 62 o sin 41o = 99.59 . ___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ Note that in this case there is always a unique solution by the Angle-Side-Angle criteria for congruent triangles. Solving a Triangle (SSA) Because there are several possibilities, the situation in which you are given the lengths of two sides of a triangle and the angle opposite one of them is called the ambiguous case. For instance, suppose you are given side a, side b, and angle α in Δ ABC . You might try to construct Δ ABC from this information by drawing a line segment A C of length b and a ray l that starts at A and makes an angle α with A C (Figure 7.4). To find the 49 remaining vertex B , you could use a compass to draw an arc of a circle of radius a with center C. If the arc intersects the ray l at point B, then Δ ABC is the desired triangle. C γ a b Figure 7.4 α A β l B c I As figure 7.5 illustrates, there are actually four possibilities if you try to construct Δ ABC by the above method in case α is acute: ( i ) The circle does not intersect the ray l at all and there is no triangle Δ ABC (Figure 7.5a). ( ii ) The circle intersects the ray l in exactly one point B and there is just one right triangle Δ ABC (Figure 7.5b) ( iii ) The circle intersects the ray l in two points B1 and B2 and there are two triangles Δ A B1 C and Δ A B2 C (Figure 7.5c). ( iv ) The circle intersects the ray l in exactly one point B and there is just one acute triangle Δ ABC (Figure 7.5d). Figure 7.5 C b α C a b a α l A A ( a ) a < b sin α ⇒ no triangle possible l B ( b ) a = b sin α ⇒ one right triangle 50 C C b a b α α l A a B1 l A B2 ( c ) b sin α < a < b ⇒ two triangles ( d ) a > b ⇒ one acute triangle II In case α is obtuse, then there are only two possibilities as shown in Figure 7.6. Figure 7.6 C a b a C α b l A α l A ( a ) a > b ⇒ one triangle ( b ) a < b ⇒ no triangle possible In the ambiguous case, you can always use a calculator to solve the triangle Δ ABC. Just use the law of sines, sin α sin β = a b to evaluate sin β : sin β = b sin α , 0 < β < 180 o . a 51 Recall that the sine of an angle is never greater than 1; hence, if b sin α > 1, then this a trigonometric equation has no solution, and no triangle satisfies the given conditions. If sin α sin α b = 1, then the equation has only one solution, β = 90 o . If b < 1, then the a a trigonometric equation has two solutions. Namely, ⎛ sin α ⎞ ⎟ β 1 = arcsin ⎜ b and β 2 = 180 o − β 1 . ⎝ a ⎠ Once you have determined β ( or β 1 and β 2 ), you know two angles and two sides of the triangle (or triangles), and you can solve the triangle by using the methods previously explained. Even if there are two solutions β 1 and β 2 of the trigonometric equation for β , it is possible that only one of these solutions corresponds to an actual triangle satisfying the given conditions (see Figure 7.5d and Figure 7.6a). Example 7.2 ---------------------------- -----------------------------------------------------------Solve the triangle in each case. 1. α = 30 o , a = 8, b = 5. Since α is acute and a > b there is only one acute triangle constructible. sin β sin 30o = 5 8 5 ⎛ 5 ⎞⎛ 1 ⎞ ⇒ sin β = ⎜ ⎟⎜ ⎟ = ⎝ 8 ⎠⎝ 2 ⎠ 16 ⎛ 5⎞ ⇒ β = arcsin ⎜ ⎟ = 18.21o . ⎝ 16 ⎠ Then γ = Finally, C γ 5 8 β 30 o A B c 180 o − 30 o − 18.21o = 131.79 o . ⎛ sin 131.79o ⎞ sin 131.79o sin 30o ⎟ = 11.93. = ⇒ c = 8⎜⎜ o ⎟ 8 c sin 30 ⎠ ⎝ ___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ 2. α = 30 o , a = 5, b = 8. C 8 <a<b 2 ⇒ there are two triangles constructible. Here α is acute and b sin α = sin 30 5 o ⇒ sin β = γ2 8 sin β 8 5 4 8 1 = ⎛⎜ ⎞⎟⎛⎜ ⎞⎟ = . ⎝ 5 ⎠⎝ 2 ⎠ 5 A 52 30 o β2 B’ β1 γ1 5 β1 B ⎛4⎞ Thus, β 1 =arcsin ⎜ ⎟ = 53.13 o . Then γ 1 = 180 o − 30 o − 53.13 o = 96.87 o and ⎝5⎠ ⎛ sin 96.87o ⎞ sin 30o sin 96.87o ⎟ = 9.928 = ⇒ c = 5⎜⎜ o ⎟ 5 c ⎝ sin 30 ⎠ To find the measure of β 2 observe that sin β 2 = sin (π − β 1 ) = sin π cos β 1 - cos π sin β 1 = sin β 1 . Thus, β 2 = 180 o − β 1 = 180 o − 53.13 o = 126.87 o and γ 2 = 180 o − 30 o − 126.87 o = 23.13 o . Then ⎛ sin 23.13o ⎞ sin 30o sin 23.13o ⎟ = 3.928 . = ⇒ c = 5⎜⎜ o ⎟ 5 c sin 30 ⎠ ⎝ ___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ Section 7 Problems--------------- ------- -----------------------------------------------------------In problems 1 to 10 use the law of sines to solve each triangle Δ ABC. Round off angles to the nearest hundredth of a degree and side lengths to four significant digits. 1. a = 32, β = 38 o , γ = 21o . 2. a = 50, b = 30, α = 45 o . 3. a = 40, b = 70, α = 30 o . 4. b = 4.5, c = 9, γ = 60 o . 5. a = 31, b = 33, α = 60 o . 6. α = β = 14 o , c = 30. 7. α = 30 o , a = 8, b = 5. 8. α = 30 o , a = 5, b = 8 9. β = 60 o , a = 11, b = 12. 10. β = 60 o , a = 12, b = 11. 53