Dharmendra Singh

Hii Guys, How r your all prepration of pre-boards going. Post if you have any query related with Economics.

Here is the problem: What is the value of sin(A-b) if tan A=9/40 and A is in the first quadrant, while cos B=-4/5 and B is in the third quadrant? I really don't get this homework. What is up with A and B? Do we need to plot the graph as well? =========================== Here's the one I just did with the same angles. ------------ If tan A=9/40, A is in the first quadrant, cos B=-4/5, B is in the third quadrant, find the value of cos(A+B) -------------- Find the cos(A) Tan=y/x Plot the point (40,9) --> hyp = sqrt(40^2 + 9^2) = 41 cos(A) = x/hyp = 9/41 ------------- cos(A + B) = cosA*cosB - sinA*sinB (from Wikipedia) Find sin(A) and sin(B) ------- sin(A) = y/hyp = 40/41 sin(B) = -3/5 in Q3 (it's the 3, 4, 5 triangle) -------------- cos(A + B) = cosA*cosB - sinA*sinB = (9/41)*(-4/5) - (40/41)*(-3/5) = -36/205 + 120/205 = 84/205 ============================= Use some of the values found in the previous problem: sin(A-b) = sin(A)cos(B) - cos(A)sin(B) = (40/41)*(-4/5) - (9/41)*(-3/5) = -200/205 + 27/205 = -173/205 ------------ What is up with A and B? Angles A & B are measured from the x-axis counter-clockwise. All angles on the x-y plane are measured that way. ------ Do we need to plot the graph as well? You don't have to if you don't want to. Not much to graph, just a point for each angle. For A draw a line from the Origin to (40,9) For B it's to (-3,4)

I am a tuition teacher now I want to be a school teacher but I have completed my studies up to B.com in commerce can I go for B.A in arts. Isn't it possible to go for B.A exams?