5.1 Right Triangle Trig Review

PreCalc 5.1 Right Triangle Trig Review Name: Key Date: 1. Find the missing side of each triangle below. Use can use the Pythagorean Theorem, the rules for Special Right Triangles, or SOН САН ТОА. 1. 12 X 2. x ^ 2 - 63 x ^ 2 + 9 ^ 2 = 12 ^ 2 x ^ 2 + ft =f44 9 x = 3sqrt(7) 45 X 4. 5. 45 deg 12404 (2sqrt(3)) ^ 2 + 8 ^ 2 = lambda ^ 2 3 3. 42 X 6. 8 19. y x = 2sqrt(19) Tan 55 lambda =19-mod 5 x = 27.1 7. 21 8. 貸 underline 7 (3 =x) 609 2 5 4 0 Csc 02 4 3 Sec 0-53 EX. II. Given the following triangles, the 6 trig SOH CAH y 15 35° X 9. 2 459 x = 15 * 6r ^ 3 * Y boxed x =12.3| x/(Let * 35) = x/15 y = underline r5.m 35 m * 35 = 4/15 Tan 0 = 413 10. 0 Csc 0 43 257 8 X:√28=257 6 ^ 2 + x ^ 2 = r ^ 2 6 11. 33 0 9 (3sqrt(6)) ^ 2 + x ^ 2 - 9 ^ 2 x ^ 2 = 27 sin theta = 4/5 cos theta = 3/5 cot theta = 3/4 Sec*theta = boxed sqrt[4] 5 4 sin theta= boxed (sqrt(omega))/3 Ta - 67 = 45/x X_{e} = 45 Tm67 x = t ^ 9, t ] y = 12/60517 155 = 19/T 3sqrt(6) Cot*theta = boxed sqrt L 2 y = 14sqrt(3) y = s * sqrt(6) Csc * 0 = boxed sqrt b 3 4sqrt(3) = x 4sqrt(6) y = root(5, 4) cot 0= boxed (sqrt(7))/3 6/(2sqrt(2)) = 3/(sqrt(7)) Tan theta= boxed (3sqrt(7))/2 3/(sqrt(6)) = (3sqrt(4))/(4e) (3sqrt(3))/7 * cos theta = boxed sqrt 5 3 2sqrt(3) 3/(sqrt(3)) = 50/3 1/1 * sqrt(2) sin theta = 6/8 > boxed 3 4 (j * sqrt(u))/(p * sqrt(u)) = sqrt(u) * ran * 0 = boxed sqrt 2 t 16sqrt(5) = x 8sqrt(6) (21sqrt(3))/5 7sqrt(4) * sqrt(3) -> 16sqrt(3) y = 34sqrt(3) (3sqrt(6))/8 cos theta= 257 8 z boxed 47/47 * Soc*theta = 44/67 = boxed 4 sqrt 7 7 30 deg 8sqrt(6) * sqrt(2) plus/minus sqrt(5) 67 deg (sqrt(2))/2 55 deg - III. Use SOH CAH TOA to find the missing angle measures in cach triangle below. 12. 600 13. 850 14. 19 28 13 13 Te 24 33 y = 45.1 deg y/(cos y) = (a * 0 ^ 2)/850 442.7 y/(Sx - y) = 19/(Sigma*y) IV. To "Solve a Triangle" means to find all of the missing sides/angles in any given triangle. Solve the right triangles below by any means. You should start by writing in the sides/angles that you already know in the table below. Remember, side a is opposite angle A. This holds true for all sides and angles. 15. B 9 16. B 40 C 12 A C 17 B C = 15 225 = e ^ 7 A A=Tan(%) A = 36.5 deg 13=\$ 0 - 36.9 b A 19.8 17 C x cos x = 19/(2r) x = 47.3 ^ 2 S max = (Gcb)/(8gamma ^ 2) x =44.9^ * c = 17/(S*mu*q) Q = 17/(t * 45) x = 57 x Ta=x^ * = 70/(1 ^ 3) Tcm * 49 = 17/a c = 90/6 aligned A= 36.4^ B = 5.18/(60 deg) * b = 12/15 \\ c= underline 40^ &c= underline 15 aligned T cm A= 1 r^ 2 9 ^ 2 + 12 ^ 2 = c ^ 2 sin 41 = 17/c c = - 22.5 A = (4t)/(110 deg)