《换面法》制图答案

1.求作直线AB的真长和倾角、。a’b’OXabVHO1X1H1Va1b1AB实长O1X1V1Hb1’a1’2.求作点A与直线BC间的真实距离。c’b’a’acbOXVHa1O1X1VH1O1b1(c1)(d1)真实距离d’d3.求作吸气罩的相邻壁面之间的夹角的真实大小。OXVHb’a’f’c’e’d’abcfedO1X1H1Va1c1e1d1O2X2H1V2a2’e2’c2’d2’夹角4.求作两平行直线AB、CD所确定的平面的倾角和。b’a’c’d’abcdXO1’12’2O1X1HV1c1’b1’a1’b1’X1O1H1Vd1b1c1a16.求作∠ABC的真实大小。a’b’c’abcOXVH1’1O1X1HV1c1’a1’b1’X2O2V1H2c2b2a2夹角8.补全矩形ABCD的水平投影。a’b’c’d’abOXVHO1X1HV1a1’b1’d1’dc1’c9.求作ΔABC与正面V的倾角的，并过点D作ΔABC的垂线DE，作出垂足E，注明点D与ΔABC之间的真实距离。OXa’b’c’d’abcdm’mO1X1H1Vb1c1m1a1d1e1e’eDE实长10.已知∠ABC=45°，点C在直线AB的前方，补全∠ABC的水平投影。OXb’a’c’baO1X1VH145°b1a1c1c1、已知直线DE平行于ΔABC平面，与ΔABC平面的距离为5mm，求作DE的水平投影。两解任求一解。d’e’a’b’c’abcOXVHX1VH1O1c1a1(f1)b1截取c1n1=5mmn1d1e1ff’den’2.已知等腰三角形ABC的底边BC，ΔABC对V面的倾角＝45°，三角形的高为20mm，补全ΔABC的两面投影。OXb’c’bcO1X1VH1b1c145°截取20mma1a’a3、已知点D与ΔABC平面的距离为12mm，BC为水平线，补全ΔABC的正面投影。两解任求一解。d’b’c’bcadOXVHd1’圆半径=12mmm1’a1’a1’m1’a’a’HV1X1O14.已知点A与直线BC的距离为10mm，求作点A的水平投影，两解任求一解。OXb’c’bca’O1X1VH1b1c1O2X2H1V2b2’c2’圆半径=10mma1a2’a5.作两交叉直线AB、CD的公垂线，并注明它们之间的最短距离。OZa’a’’c’c’’b’b’’d’d’’O1X1WV1a1’c1’b1’d1’O2X2V1W2a2’’b2’’c2’’d2’’e2’’f2’’公垂线EF实长e1’f1’f’’e’’f’e’6.已知两交叉直线AB、CD的距离为10mm，补全CD的正面投影，两解任求一解。OXa’b’c’acbdO1X1V1Hb1’a1’c1’O2X2H2V1c2b2a2d2d1’d’7、求作直线AB上与ΔCDE、ΔCDF平面等距的点G。a’b’d’c’e’f’dcefOXVHO1X1HV1c1’d1’e1’f1’aba1’b1’O2X2V1H2d2(c2)f2e2a2b2g2g1’gg’g2在∠e2d2f2角平分线上8.已知正方形ABCD的对角线BD在直线PQ上，补全正方形ABCD的两面投影。a’p’q’aqpOXO1X1HV1a1’q1’p1’O2X2V1H2q2P2a2b2d2c2c1’b1’d1’bdcb’d’c’9.已知等边三角形ABC的一边AB，平面的倾角＝30°，顶点C在直线AB的前上方，补全ΔABC的两面投影。a’b’abOXO1X1VH1a1b130°O2X2H1V2a2’b2’c2’60°60°c1c’c10、求作直线AB与ΔCDE平面的交角θ。ea’b’abc’d’e’cdXOPABKLθδVHX1O1H1V1’1c1d1e1b1a1m1m’H1V2O2X2a2’(m2’)b2’O3X3V2H3m3a3b3∠b3a3m3＝δ已知正方形ABCD的一边BC在直线MT上，用换面法补全正方形ABCD的两面投影。d'd2d1'a2b2HVOXa'm't'atmO1X1HV1t1'm1'a1'b1'bb'X2O2H2V1b2c2m2t2a2c1'cdc'