# HW13sol - MAC1114 Homework 13 Due Thursday April 13 •...

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Unformatted text preview: MAC1114 - Homework 13 Due Thursday - April 13 • Refer to the triangle pictured below: • For each problem (a) Solve the triangle (i.e., nd a, b, c, A, B, and C ) (b) Find the area 1. (AAS)∠A = 45◦ , ∠B = 30◦ , b = 6. • C = 180 − 45 − 30 = 105◦ . √ a 6 • sin 45 = sin 30 ⇒ a = 6 2. 6 • sinc c 105 = sin 30 ⇒ √ = 12 sin(105) = 12 sin(60 + 45) = 12 √√ √ 6+ 2 = 3( 6 + 2). 12 4 √√ √ √ 1 • Area= 2 ac sin B = 1 6 23( 6 + 2) sin 30 = 9( 3 + 1). 2 sin 60 cos 45 + cos 60 sin 45 = 2. (ASA) ∠A = 60◦ , ∠B = 45◦ , c = 8. • C = 180 − 45 − 30 = 75◦ . Note that sin 75◦ = sin(105◦ ). • a sin 60 = √ √ 6+ 2 4 • b sin 45 = √ √ 6+ 2 4 √ √ 6+ 2 4 by a similar calculation as we did for √ 16 √ 32√ sin 60 = √ 3 . 6+ 2 ( 3+1) √ 16 2 = (√3+1) √ √ 16 32 6 1√ 3 (8) sin 45 = √3+1 . 2 ( 3+1) 8 ⇒a= 8 ⇒b 1 • Area= 2 ac sin B = 3. (SAS) ∠A = 30◦ , b = 5, c = 7 √ 74 − 35 3. • a2 = b2 + c2 − 2bc cos A = 25 + 49 − 35 cos 30◦ ⇒ a = √ • sin B b = sin A a ⇒ sin B = √ 5 √ 35 3 22 3 2 √ 74−35 3 • C = 150 − B = 150 − arcsin 1 • Area= 2 bc sin A = √ = √ ⇒ B = arcsin √ 5 3 2 √ 74−35 3 √ 5 23 √ 74−35 3 √ 35 3 4. 4. (SSS) a = 3, b = 5, c = 6. • a2 = b2 + c2 − 2bc cos A ⇒ cos A = a2 −b2 −c2 −2bc = 9−25−36 −2(30) = 52 60 ⇒ A = arccos 13 15 • b2 = a2 + c2 − 2ac cos B ⇒ cos B = b2 −a2 −c2 −2ac = 25−9−36 −2(18) = • c2 = a2 + b2 − 3ab cos C ⇒ cos C = c2 −a2 −b2 −2ab = 36−9−25 −2(15) 1 1 = − 15 ⇒ C = arccos(− 15 ) • s = 1 (a+b+c) = 7. From Heron's formula, Area= √2√ 56 = 2 14. 1 20 36 ⇒ B = arccos 5 9 s(s − a)(s − b)(s − c) = 7(7 − 3)(7 − 5)(7 − 6) = ...
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## This note was uploaded on 06/10/2011 for the course MAC 1114 taught by Professor Gentimis during the Spring '11 term at University of Florida.

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