1. Trigonometry in Everyday Use

2. PROBLEM # 1: One day, a group of students were playing volleyball. Consider the angle from Sheen Ray’s feet to the highest point of the pole that is elevated 28º and the height of a pole that is 2.94 m. and the angle of depression form the 2 nd pole to the ground is 25º. Find the:

3. Anna Lou Khary Sheen Ray

4. A =? 2.94 M 28º 1 2 D =? C=? B =? E =?

5. Solutions: Find A: sinØ = opp a sin28º = 2.94m a 0.4694715628 = 2.94m a 0.4694715628(a) = 2.94m 0.4694715628 a = 6.262360136m ~ 6.26m A =?

6. Find b: tanØ = opp adj tan28º = 2.94m b 0.5317094317 = 2.94m b 0.5317094317(b) = 2.94m 0.5317094317 0.4694715628 b = 5.52933580m ~ 5.53m b =?

7. Find c: sinØ = adj hyp sin25º = 2.94m c 0.422618261 = 2.94m c 0.422618261(c) = 2.94m 0.422618261 c = 6.956632654m ~ 6.96m C=?

8. Find d: sinØ = adj hyp sin25º = d 6.96m 0.422618261 = d 6.96m 0.422618261(6.96m) = d d = 2.941423102m ~ 2.94m D =?

9. Find e: e = 2.587912698m ~ 2.59m E =? 28º To find e, we have to subtract d from b b – d = e 5.52933580m -2.941423102m = e

10. PROBLEM # 2: While playing tag, Sheila went up a stairwell, to the very top while Derick followed to catch her. But then he notices that Lorraine was still at the bottom. If the distance of Derick from the ground is 1.52m, and the stairwell is elevated 24º from the ground and it is 5.79m long, Find:

11. A.) w= Derick’s distance to Lorraine B.) y= Derick’s distance to Sheila C.) Ø= the angle of elevation from Lorraine to Sheila D.) x= the elevation of Sheila from the ground E.) v= the length from the foot of Lorraine to the ground base of Sheila

12. Sheila Mae Derick Lorraine Faye

13. w=? y=? 24º 1.52m 5.79m v=? x=? Ø=?

14. Find w: (Derick’s distance to Lorraine) 1.52 w=? 24º sinØ = opp hyp sin24º = 1.52m w 0.407 = 1.52m w 0.407(w) = 1.52m 0.407 w =3.73m

15. Find y: (Derick’s distance to Sheila) Let m be the total length of the stairwell y= 2.06m *to find y, subtract w from the total length of the stairwell y = m – w y = 5.79m – 3.73m y=?

16. Find Ø: ( the angle of elevation from Lorraine to Sheila ) cosØ = adj hyp cosØ = 1.52m 3.73m cosØ = 0.408m Ø = cos -1 (0.408) Ø = 66º Ø=?

17. Find x: (Sheila’s elevation from the ground) sinØ = opp hyp sin24º = x 5.79m 0.407 = x 5.79m 0.407(5.79m) = x 0.407 (w) = 1.52m 0.407 X = 2.36m x=?

18. Find v: the length from the foot of Lorraine to the ground base of Sheila v=? tanØ = opp adj tan24º = 2.36m v 0.445 = 2.36m v 0.445(v) = 2.36m 0.445 v = 5.29m