form 6 maths > triangle
2017/06/21 00:14
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form 6 maths > triangle
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cos (angleQPR) = [ 1 - sin2(angleQPR)]0.5 = [ 1 - 4/9 ]0.5 = sqrt(5)/3 QR^2 = (PQ^2 + PR^2) / [2(PQ)(PR)cos(angleQPR)] = (49 +1225) /[2(7)(35)sqrt(5)/3] = 39sqrt(5)/25
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form 6 maths > triangle
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A triangle PQR in which PQ = 7 and PR = 35. Given that sin (angleQPR) = 2/3 and that angleQPR is acute, Find the exact value of cos (angleQPR) in its simplest form, (2) Show that QR = 2x6^1/2, (4) Find... 顯示更多 A triangle PQR in which PQ = 7 and PR = 35. Given that sin (angleQPR) = 2/3 and that angleQPR is acute, Find the exact value of cos (angleQPR) in its simplest form, (2) Show that QR = 2x6^1/2, (4) Find (anglePQR) in degrees to 1 decimal place. (3) 更新: which formula did you used to get the second line?? '' [ 1 - sin2(angleQPR)]0.5 ''最佳解答:
cos (angleQPR) = [ 1 - sin2(angleQPR)]0.5 = [ 1 - 4/9 ]0.5 = sqrt(5)/3 QR^2 = (PQ^2 + PR^2) / [2(PQ)(PR)cos(angleQPR)] = (49 +1225) /[2(7)(35)sqrt(5)/3] = 39sqrt(5)/25
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