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發問:
http://chingyihouse.linknethost.com/P1020025.JPG PQR is a straight line . RSQ is a straight line prependicular to the x-axis . OP =2.5cm , OS = 6cm and RS = 8cm . If OQ is the terminal side of theta , find sin theta
最佳解答:
在O的┴線和PQ交點作符號A RO^2=RS^2+OS^2 畢氏定理 RO=10 ∠ROS=∠OPQ (Corr.∠s, OS//PQ) ∠OAP=∠RSO=90 degree ∠POA=∠PRQ (Corr.∠s, OS//PQ) ∴△OPA～△ROS (aaa) OP/RO=OA/RS OA=2 SQ=2 SQ^2+SO^2=OQ^2 畢氏定理 OQ=√40 SIN∠XOQ=SQ/OQ SIN∠XOQ=2/√40 SIN(360-∠XOQ)= -2/√40 SINθ= -√10/10
其他解答:
I don't know eat zone go
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maths!!!!!!!!F.4發問:
http://chingyihouse.linknethost.com/P1020025.JPG PQR is a straight line . RSQ is a straight line prependicular to the x-axis . OP =2.5cm , OS = 6cm and RS = 8cm . If OQ is the terminal side of theta , find sin theta
最佳解答:
在O的┴線和PQ交點作符號A RO^2=RS^2+OS^2 畢氏定理 RO=10 ∠ROS=∠OPQ (Corr.∠s, OS//PQ) ∠OAP=∠RSO=90 degree ∠POA=∠PRQ (Corr.∠s, OS//PQ) ∴△OPA～△ROS (aaa) OP/RO=OA/RS OA=2 SQ=2 SQ^2+SO^2=OQ^2 畢氏定理 OQ=√40 SIN∠XOQ=SQ/OQ SIN∠XOQ=2/√40 SIN(360-∠XOQ)= -2/√40 SINθ= -√10/10
其他解答:
I don't know eat zone go
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