angles A B C are acute angles such that sin A = 0.1 cos B = 0.4 sin C = 0.7
without finding the angles A B C use the addition formula to calculate to 2 decimal places
sin(A+B) cos(B-C)
the answers are 0.95 and 0.93 please show workings
without finding the angles A B C use the addition formula to calculate to 2 decimal places
sin(A+B) cos(B-C)
the answers are 0.95 and 0.93 please show workings
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First, we find cosA, sinB, sinC. In each case draw a right-angled triangle. In the 1st, mark angle A and mark Opp = 1, Hyp = 10, so by the Theorem of Pythagoras ADJ = sqrt (100 - 1) = sqrt99 so cos A = (sqrt99)/10 = 3rt11/10. Do exactly the same with a second triangle having an acute angle B and mark Adj = 4, hyo = 10 so sinB = (2rt21/10); do the same again for sinC (rt51/10). Now use Sin(A + B) = sinAcosB + cosAsinB ; put in the values from above to get sin(A + B) = (1/10)*(4/10) + (3rt11/10)(2rt21/10) - at this point, use your calculator, and you should get the required answer. Cos(B - C) = cosCcosB + ssinCsinB. Do exactly the same here. Rounf off in both cases.
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sin A = 0.1 cos A = sqrt 0.99 cos B = 0.4 sin B = sqrt 0.84 sin C = 0.7 cos C = sqrt 0.51
sin(A + B) = sin Acos B + cosAsin B = 0.1 x 0.4 + sqrt 0.99 x sqrt 0.84 = 0.04+ 0.912 = 0.952 = 0.95
cos(B-C) = cosBcosC + sinBsinC = 0.4sqrt0.51 + 0.7sqrt0.84 = 0.2857 +0.6416=0.9273 =0.93
sin(A + B) = sin Acos B + cosAsin B = 0.1 x 0.4 + sqrt 0.99 x sqrt 0.84 = 0.04+ 0.912 = 0.952 = 0.95
cos(B-C) = cosBcosC + sinBsinC = 0.4sqrt0.51 + 0.7sqrt0.84 = 0.2857 +0.6416=0.9273 =0.93
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Do it urself noob