命题若△A1B1C1的三个内角的余弦值分别等于△A2B2C2的三个内角的正弦值,则△A1B1C1是锐角三角形,△A2B2C2是钝角三角形.证明由于△A1B1C1的三个内角的余弦值分别为△A2B2C2的三个内角的正弦值,故△A1B1C1的三个内角的余弦值均为正,因此,△A1B1C1是锐角三角形.下面证明△A2B2C2 If the cosine of the three internal angles of △A1B1C1 is equal to the sine of the three internal angles of △A2B2C2, then △A1B1C1 is an acute triangle and △A2B2C2 is an obtuse triangle. Prove that the cosine of the three internal angles of △A1B1C1 is △ The sine of the three inner angles of A2B2C2, so the cosine of the three inner angles of △A1B1C1 are all positive, therefore, △A1B1C1 is an acute triangle. The following proves that △A2B2C2