jpjzhf9的部落格

標題:

angle in rectilinear figures

發問:

find the unknowns 圖片參考：http://imgcld.yimg.com/8/n/HA00101881/o/701205220034013873407931.jpg

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2. angle BAC + angle ABC = angle ACD (exterior angle of triangle) (x - 20?) + (x - 50?) = x + 30? 2x - 70? = x + 30? 2x = x + 100? x = 100? angle ACD + angle ABC = 180? (adjacent angles on straight line) (x + 30?) + y = 180? (100? + 30? + y) = 180? y = 50? 3. angle HIJ + angle JIK = 180? (adjacent angles on straight line) 108? + angle JIK = 180? angle JIK = 72? In triangle IJK: angle JIK + angle IJK = angle JKL (exterior angle on triangle) 72? + (2n + 10?) = 142? 2n + 82? = 142? 2n = 60? n = 30? 5. AC = CD = AD (given) So triangle ACD is equilateral. angle ACD = angle CDA = angle DAC = 60? (property of equilateral triangle) angle ABC + angle ACD = 180? (adjacent angles on straight line) (9k - 6?) + 60? = 180? 9k + 54? = 180? 9k = 126? k = 14? 6. UT = ST = US (given) So triangle UST is equilateral. angle UST = angle STU = angle TUS = 60? (property of equilateral triangle) RS = RT (given) So triangle RST is isosceles. angle RST = angle RTS (base angles, isosceles triangle) In triangle RST: angle RST + angle RTS + angle SRT = 180? (angle sum of triangle) angle RST + angle RTS + 42? = 180? angle RST + angle RTS = 138? So angle RST = angle RTS = 69? angle UTS + angle RTU = angle RTS 60? + p = 69? p = 9? 1. RU is perpendicular to PT (given) So angle PUQ = angle TUQ = 90? In triangle PUQ: angle UPQ + angle PUQ = angle UQS (exterior angle of triangle) 46? + 90? = angle UQS angle UQS = 136? In triangle SQR: angle QSR + angle SRQ = angle UQS (exterior angle of triangle) p + 32? = 136? p = 104? 4. s + 130? = 180? (interior angles, IH // FG) s = 50? IH = IK (given) So triangle IHK is isosceles. angle IKH = angle IHK (base angles, isosceles triangles) angle IHK = t In triangle IHK: angle IKH + angle IHK = angle HIG (exterior angle of triangle) t + t = 50? 2t = 50? t = 25?

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