The document contains 25 solved trigonometric problems involving the law of sines, law of cosines, and graphs of sine and cosine functions. It provides step-by-step solutions to problems finding missing sides and angles of triangles using trigonometric identities and relationships. Several examples calculate lengths, angles, and areas using information about one or two sides or angles of a triangle.

1. 25 solved Trigonometric Problems Law of Sine Law of Cosines Law of Tangents Graphs of Sine and Cosine Graphs of Other Functions Pampanga Agricultural College Magalang, Pampanga, Philippines

2. Solve ∆ABC if A=39°, a=21inches and b=11inches. Find the remaining side and angles. Solution Using the Sine Law, a = b sinA sinB 21 = 11 sin39 ° sinB sinB = 11sin39 21 ≈ 0.329644 B ≈ 19,25° We can now say that, C ≈ 180-(39+19.25) C ≈121.75 To solve for the remaining side, c = a sinC sinA c = a sin121.75 ° sin19.25 ° c= 21sin121.75° sin19.25° c≈54.16inches

3. Given a ∆ABC, A=48°, b=16, c=32. Find a. Solution: a2 = b2 + c2 - 2·b ·c·Cos A a2 = 102 + 202 - 2·10 ·20·Cos 30 a = ( 102 + 202 - 2·10 ·20·Cos 30 )0.5 a = 12.393137

4. Solve ∆ABC shown at the right with A=65°, B=48°, and c+76ft. Find the remaining sides and angle. Solution A+B+C=180 ° C=180°-(B+C) C=180°-(65+48) C=180°-113 C=67° To get the length of two sides, we use the Law of Sines. sinA = sinC a c sin65 ° = sin67° a 76 76sin65 °=asin67° 72sin65 °=a sin67° 75=a sinB = sinC b c sin48° = sin67° b 76 76sin48°=bsin67° 76sin48° =b sin67° 67°=b

5. Two sides of ∆ ABC measures 6cm and 8cm and their included angle 40°. Find the third side. Solution: c ²=a²+b²-2abcosC c²=6²+8²-2(6)(8)cos40° c=5.144cm

6. Given a triangle whose angles are A=40 °, B=95° and side b=30cm. Find the length of the bisector of angle C. Solution: Angle C=180°-40°-95°=45° In triangle ADC: Ө =180 °-40°-22.5°=117.5° c= x = 30 sin40 ° sin117.5° x=21.74cm

7. Given ∆ ABC : C=100 °, b=20. Find c. Solution: Using the Cosine Law: c ²=a²+b²-2abcosC c²=15²+20²-2(15)(20)cos100° c=27

8. Given ∆ABC, if A=40°°, B=20°, a=2, find C and b. A+B+C=180 ° C=180-(A+B) C=180-(40+20) C=180-60 C=120° sinA = sinB a b sin40 = sin20 2 b bsin40=2sin20 b= 2sin20 sin40 b=1.0642

9. Given a ∆ABC, A=12°, C=45°, a=5m. Find c. Solution: a = b = c sinA sinB sinC c= a(sinC) = 5(sin45) sinA sin12 c=17.005

10. A small electric component is in the shape of a triangle with sides 6.23, 8.146 and 11.392 millimeter. Find the largest angle Solution: Let a = 6.23, b = 8.146 and c = 11.392 Angle opposite to the largest side is the largest angle. cos C = a2 + b2 - c2 = (6.23)2 + (8.146)2 -(11.392)2 = -0.242

11. Given a ∆ABC, A=36°, B=23°,b=11mm. Find a. Solution: a = b = c sinA sinB sinC b= b(sinA) = 11(sin35) sinB sin23 a=16.1475mm

12. Given angle A=32 °, angle B=70°, and side c=27 units. Solve for side a of the ∆. Solution Sine Law A+B+C=180 ° C=180°-32°-70° C=78° a = b sinA sinB a = 27 sin32 ° sin78 ° a=14.63units

13. Solve for b in ∆ABC if a=4, c=7 and B=130° Solution b ²=a²+c²-2accosB 4²+7²-2(4)(7)cos130° 16+49-56cos130° 65+56(.6428) 100.9961 b=10.05

14. Given a ∆ABC, B=36°, C=24°, c=12cm. Find b. Solution: a = b = c sinA sinB sinC b= c(sinB) = 12(sin36) sinC sin24 b = 17.3415cm

15. In a ∆ABC, A=45° and angle C=70°. The side opposite angle C is 40m long. What is the side of opposite angle A? Solution a = b sinA sinB a = b sin45° sin70 ° a=30.1m

16. Two sides of a ∆ABC are 50m and 60m long. The angle included between these sides is 30°. What is the interior angle opposite the longest side? Solution: Solving for the third side by cosine law: c²=a²+b²-2abcosC c=30.064 Solving for the angle opposite the 60m side by sine law: 60 = 30.064 sin Ө sin30 ° sin Ө =0.9978 Ө =86.26° and 93.74°

17. In ∆ABC, find the side c if angle C=100°, side b=20, and side a=15. Solution By Cosine Law: c ²=a²+b²-2abcosC c ²=(15) ²+(20) ²-2(15)(20)cos100° C=27

18. Find the difference between the largest and smallest angles of a triangle if the lengths of sides are 10, 19 and 23. Solution:

19. The area of the triangle whose angles are 61 °9’32”, 34°14’46”, and 84°35’46” is 680.60. The length of the longest side is: Solution: A= ½ ac sinB=680.6 a = c sin34°14’46” sin84°35’46” A=0.5653c1 680.6= ½ (0.5653 c) (c) sin61°9’32” c=52.43units

20. Given a ∆ABC, A=30°, b=10m, c=20m. Find a. Solution: a2 = b2 + c2 - 2·b ·c·Cos A a2 = 102 + 202 - 2·10 ·20·Cos 30 a = ( 102 + 202 - 2·10 ·20·Cos 30 )0.5 a = 12.393137m

21. Graphs of Sine, Cosine and Tangents

22. Some Vocabulary Amplitude: height Frequency: number of curves in a normal period Period: how long it takes to complete one curve Domain: x values (the angle in radians) Range: y values (the trig value)

23. Find the Amplitude and Period of each function: y=-2cos 5/4 x Amplitude=-2 Period=2 π/5/4 = 8π/5 y=2sin2x Amplitude=2 Period=2π/2 = π

24. Y= ¼ cos4x Amplitude= ¼ Period= 2π/4 = π/2 Y=3sin ¼ x Amplitude=3 Period=2π/ ¼ = 8π Y= ½ cos ¼x Amplitude= ½ Period=2 π/ ¼ = 8π

29. Prepared by: Paula P. Alfonso Agricultural Science Curriculum 3 rd year High School [email_address]