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(1) Question: Find the equation of the straight line which passes through (-1, 2) and is perpendicular to the line x - y + 2 = 0. Solution: The slope of x - y + 2 = 0 is (-1)/(-1) = 1 Since the required line ⊥ x - y + 2 = 0, the slope of the required line is (-1)/1 = -1. By point-slope form, the equation is: -1 = (y - 2)/(x + 1) -x - 1 = y - 2 x + y - 1 = 0 (2) 圖片參考：http://i127.photobucket.com/albums/p138/hawkele/PICT0002.jpg Question: MN is parallel to BC, AM=6 , MB=2 and AN=4. Find the length of NC and ∠ACB and ∠ABC. Solution: ∠AMN + ∠ANM + ∠MAN = 180° (∠ sum of △) ∠AMN + 20° + 10° = 180° ∠AMN = 150° So ∠ABC = ∠AMN = 150° (corr. ∠s, MN // BC) and ∠ACB = ∠ANM = 20° (corr. ∠s, MN // BC) Also ∠BAC = ∠MAN = 10° (common ∠) So △BAC ～ △MAN (A. A. A.)【OR we write (equiangular)】 So AB/AM = AC/AN (corr. sides of ～△s) (AM+MB)/AM = (AN+NC)/AN (6+2)/6 = (4+NC)/4 4/3 = (4+NC)/4 16/3 = 4+NC NC = 4/3 (3) Question: Find the values in surd form of cos 120? and tan (-π/6) without using a calculator. Solution: cos 120° = cos (180° - 60°) = - cos 60°【cos (180° - x) = - cos x】 = - 1/2 tan (-π/6) = - tan (π/6)【tan (- x) = - tan x】 = -√3 /3 (4) Question: A boy of 1m tall has a shadow of length 2m long . Find the angle of elevation of the sun correct to the nearest degree. Solution: Let θ be the angle of elevation. tan θ = 1/2 θ = tan-1 1/2 θ = 26.5651° θ = 27° (corr. to the nearest degree) Hope it helps! ^^ Also wish a Merry Christmas and Happy New Year.
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1.Find the equation of the straight line which passes through (-1, 2) and is perpendicular to the line x-y+2=0. 2.MN is parallel to BC, AM=6 , MB=2 and AN=4 . link ....http://i127.photobucket.com/albums/p138/hawkele/PICT0002.jpg Find length of NC,, 更新: quote q2 find 角ACB and 角ABC 3.Find the values in surd form of cos120? and tan (-π/6) without using a calculator. 4.A boy of 1m tall has a shadow of length 2m long . Find the angle of elevation of the sun correct to the nearest degree.最佳解答:
(1) Question: Find the equation of the straight line which passes through (-1, 2) and is perpendicular to the line x - y + 2 = 0. Solution: The slope of x - y + 2 = 0 is (-1)/(-1) = 1 Since the required line ⊥ x - y + 2 = 0, the slope of the required line is (-1)/1 = -1. By point-slope form, the equation is: -1 = (y - 2)/(x + 1) -x - 1 = y - 2 x + y - 1 = 0 (2) 圖片參考：http://i127.photobucket.com/albums/p138/hawkele/PICT0002.jpg Question: MN is parallel to BC, AM=6 , MB=2 and AN=4. Find the length of NC and ∠ACB and ∠ABC. Solution: ∠AMN + ∠ANM + ∠MAN = 180° (∠ sum of △) ∠AMN + 20° + 10° = 180° ∠AMN = 150° So ∠ABC = ∠AMN = 150° (corr. ∠s, MN // BC) and ∠ACB = ∠ANM = 20° (corr. ∠s, MN // BC) Also ∠BAC = ∠MAN = 10° (common ∠) So △BAC ～ △MAN (A. A. A.)【OR we write (equiangular)】 So AB/AM = AC/AN (corr. sides of ～△s) (AM+MB)/AM = (AN+NC)/AN (6+2)/6 = (4+NC)/4 4/3 = (4+NC)/4 16/3 = 4+NC NC = 4/3 (3) Question: Find the values in surd form of cos 120? and tan (-π/6) without using a calculator. Solution: cos 120° = cos (180° - 60°) = - cos 60°【cos (180° - x) = - cos x】 = - 1/2 tan (-π/6) = - tan (π/6)【tan (- x) = - tan x】 = -√3 /3 (4) Question: A boy of 1m tall has a shadow of length 2m long . Find the angle of elevation of the sun correct to the nearest degree. Solution: Let θ be the angle of elevation. tan θ = 1/2 θ = tan-1 1/2 θ = 26.5651° θ = 27° (corr. to the nearest degree) Hope it helps! ^^ Also wish a Merry Christmas and Happy New Year.
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