標題:
F.2 Maths trigonometric ratios
The base angles of an isosceles triangle are both 52°. If its area is25cm^2, find its base.
最佳解答:
let h = height of the isosceles triangle, and b = base of the isosceles triangle then A= area of the isosceles triangle = 25 sq. cm since isosceles triangle and h perpendicular with base b in right angle we have: A/2 = 1/2 x 1/2 x b x h So A = 2 x 1/2 x 1/2 x b x h = (b x h ) /2 .............(1) h/(b/2) = tan 52 degree So h = 0.64 x b.......................................................(2) Subst (2) into (1), we have 25 = (b x 0.64 x b)/2 b = sqrt (50/0.64) = (25 x sqrt(2))/4 = 8.84 Please try it yourself agin in different isosceles angle (e.g. 15, 30, 45 & 60 etc.)! Hope this will help you.
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F.2 Maths trigonometric ratios
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發問:The base angles of an isosceles triangle are both 52°. If its area is25cm^2, find its base.
最佳解答:
let h = height of the isosceles triangle, and b = base of the isosceles triangle then A= area of the isosceles triangle = 25 sq. cm since isosceles triangle and h perpendicular with base b in right angle we have: A/2 = 1/2 x 1/2 x b x h So A = 2 x 1/2 x 1/2 x b x h = (b x h ) /2 .............(1) h/(b/2) = tan 52 degree So h = 0.64 x b.......................................................(2) Subst (2) into (1), we have 25 = (b x 0.64 x b)/2 b = sqrt (50/0.64) = (25 x sqrt(2))/4 = 8.84 Please try it yourself agin in different isosceles angle (e.g. 15, 30, 45 & 60 etc.)! Hope this will help you.
其他解答:
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