Question

如圖2-3-13,AB是⊙O的直徑,點P在BA的延長線上,弦CD⊥AB,垂足為E,∠POC =∠PCE.

圖2-3-13

(1)求證:PC是⊙O的切線;

(2)若OE∶EA =1∶2,PA =6,求⊙O的半徑;

(3)在(2)的條件下,求sin∠PCA的值.

Answer 1

思路分析:(1)要證切線PC,仍是先證PC⊥OC.?

(2)要求半徑,可以求OA,先求OE,這可以在Rt△PCO中,利用∠POC=∠PCE,列出有關(guān)方程求解.

(3)求sin∠PCA,先求sin∠ACE =.

(1)證明:在△OCP和△CEP中,?

∵∠POC=∠PCE,∠OPC=∠CPE,?

∴△COP∽△ECP.∴∠OCP=∠CEP.?

∵CD⊥AB,∴∠CEP =90°.?

∴∠OCP =90°.∴PC為⊙O的切線.?

(2)解:設(shè)OE=x,則EA =2x,OA =OC =3x.?

∵∠COP =∠PCE,∴sin∠OPC=sin∠OCE,?

即=,解得x =1.

∴OA =3.?

(3)解:∵∠OCP=90°,∴∠PCA +∠ACO =90°.?

∴sin∠PCA =cos∠ACO.?

又OA =OC,∴∠ACO =∠CAO.?

∴sin∠PCA =cos∠CAO.?

而AE =2,OE =1,OC =3,?

∴ =.?

而cos∠CAO = = =,?

即sin∠PCA =.