Question

1. if cos A = -1/5 and A is between 90 and 180 (degrees) , find sin A/2

2. A plane is flying with an airspeed of 180 miles per hour with a heading of 118 (degrees). The wind currents are a constant 28 miles per hour in the direction due north. Find the true course and ground speed of the plane

Answer 1

Using basic principles of trigonometry you can find the solution of question 1 and using the law of parallelogram of vector addition you can easily find the ground speed and true course of plane. Please find the hand written solution for these questions. I hope it will help you to understand. Please like and comment if you found helpful. Thanks and best wishes.

Answer I. 90<A < 180. Los A = - 12 Cos 20= 1-2 sine Cos2(A) = 1- 2 sin? A 2 sin? A = 1- COSA * - COSA » Sin A = -2 Put the value of cosa - - Sin²A = |- (-Ys) - It is sin A = I3 90 <A <180 45 <A L90 function is the in (Ist quadrant) Ist quadrant, Since, sine Therefore, Sint + 8 = +0.7746 Ang

Answer 2. speed of 180 mph at 118 180 a 108=8 = 180 on diagram, o represents airspeed of 180 mph AB represents wind current speed of 28 mph in of north. Hence, ob represents ground speed of the plane and of LYOB represents true course the plane in Counter clock wise. Now, By using law of parallelogram of vector addition - AR &=208-go ocoa tob resultant vector oc² - BA²+ B²+ 200 OB Cosa - where a = angle between Å and of P = 108-90= 18 OC (180) +(28 ) + 2.8 190 * 28 Cos 18 = 32400+ 784 +9586.65 0c² = 42770.65 oc= 42770.65 = 206.81 mph ground speed of plane = 206.81 mph. or (180) Oc? ground speed of Hence the

from north True and speed vector course = Angle of ground speed vector By using Law of parallelogram, tano = oA sino OB + OA LOSED tano = 180 Sin 18" 28+ 180 Cos18 55.623 199.190 tano 0.27925 o= tan (0.27925) To = 15.6° the course is 15.6° from due north counter clock wire. ground speed of flane is 206-81 mph. Aus Hence and for reference - Law of parallellogram - If two vector acting Simultaneously at a point can be represel agnitude and direction by the adjacent sides of a parallelogram from a point, then the resultant vector is represented both in itude and direction by the diagonal of Parallelogram passine and direction both in magnitude poin drawn from a magnitude and through that point. R² = P²+0² +2PQ Cosa. Q sina a tano = pta sissa