ADDITIONAL SOLVED PROBLEMS
UNIT-03 TRIGONOMETRY _________________________________________________________________________________________________ Prob.1 In the triangle shown AB = 5.0 m, determine AC. Solution: AC/sin1200 = AB/sin340 Therefore AC = [AB/sin340]*sin1200 = 5.0*0.87/0.57 = 7.7 m Prob.2 Two observers A and B simultaneously look at a plane P. Observer A is on the ground while B is sitting on a platform 20 ft. vertically above A. The angles of elevation for the two observers are as shown in the diagram. At the moment the plane was sighted: [a] What was the horizontal distance, H of the plane from A? [b] What was the vertical distance, Y of the plane from the ground? Solution: For observer B: Y – 20 =H*tan450 = H Y = H + 20 …[1] For observer A: Y = H*tan490= 1.15H …[2] From [1] and [2], 1.15H – H = 20 or H = 20/0.15 = 133ft. and Y = 133+20 = 153ft. Prob.3 Consider the right-­‐angle triangle abc with ab = 20.0cm, and bc = 1.0cm.(Note: the diagram is not drawn to scale). [a]Show that tan θ  sin θ
ac = 20 2 + 12 = 20.02cm
tan θ = 1 / 20 = 0.05
sin θ = 1 / 20.02 = 0.0499  0.05
thus tan θ  sin θ [b] Calculate the value of θ in units of radians and show that this value of
θ  tan θ  sin θ . θ = tan −1 (0.05) = 2.86 0 = 0.05rad = tan θ  sin θ Prob.4 In the diagram Δabc and Δcdb are right-­‐angle triangles with dc = 2.0m. What is the length of ac? from Δcdb, dc = bc*cos530 from Δabc bc = ac* cos530 therefore, ac = bc/ cos530 = dc/ cos2 530 = 2/0.36 = 5.55m Prob.5 In the diagram ac = ab = 1.6R, where R = ao is the radius of the circle. What is the value of α? Solution: From o draw a line od perpendicular to ac. Therefore, dc = 1.6R/2 = 0.8R θ = cos −1 (dc / oc) = cos −1 (0.08R / R) = 36.9 0
α = 360 − (∠coa + ∠boa)
where∠coa = ∠boa = 180 0 − 2θ
∴ α =360 0 - (360 0 - 4θ ) = 4θ = 147.6 0