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Questions on trigonometry
Please help me answer these questions: http://www.flickr.com/photos/81115255@N00/3589330674/ thanks a lot! 更新: 第三題 tan入面個個係12度
最佳解答:
1.. a).tan 45*=h/DB DB=h -----------------(1) CD=h/tan 60*------------(2) b). note that DB=DC+100 DC=DB-100 by (1) DC=h-100 -----(3) put (3) into (2) h-100 = h/tan 60* h tan 60*-100 tan 60* = h h(tan 60*-1)=tan 60* (100) h = (100 tan 60*)/(tan 60*-1) = 236.60 4..LHS=(1/tan^2 Q)- cos ^2 Q =(cos^2 Q/sin^2 Q) - cos^2 Q =(cos^2 Q/sin^2 Q) - (cos^2 Q sin^2 Q)/sin^2 Q =(cos^2 Q(1-sin^2 Q))/sin ^2 Q =(cos^2 Q cos^2 Q)/sin^2 Q =(cos^2 Q/sin^2 Q)(cos^2 Q) =1/tan^2 Q cos ^2 Q =RHS 2009-06-02 22:50:32 補充： 第二 if 係 = 9/5 就識做 =3= 第三 tan入面個個係12定係12度...係2個唔同既ans=.=
其他解答:7638E748CCC65837
Questions on trigonometry
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發問:Please help me answer these questions: http://www.flickr.com/photos/81115255@N00/3589330674/ thanks a lot! 更新: 第三題 tan入面個個係12度
最佳解答:
1.. a).tan 45*=h/DB DB=h -----------------(1) CD=h/tan 60*------------(2) b). note that DB=DC+100 DC=DB-100 by (1) DC=h-100 -----(3) put (3) into (2) h-100 = h/tan 60* h tan 60*-100 tan 60* = h h(tan 60*-1)=tan 60* (100) h = (100 tan 60*)/(tan 60*-1) = 236.60 4..LHS=(1/tan^2 Q)- cos ^2 Q =(cos^2 Q/sin^2 Q) - cos^2 Q =(cos^2 Q/sin^2 Q) - (cos^2 Q sin^2 Q)/sin^2 Q =(cos^2 Q(1-sin^2 Q))/sin ^2 Q =(cos^2 Q cos^2 Q)/sin^2 Q =(cos^2 Q/sin^2 Q)(cos^2 Q) =1/tan^2 Q cos ^2 Q =RHS 2009-06-02 22:50:32 補充： 第二 if 係 = 9/5 就識做 =3= 第三 tan入面個個係12定係12度...係2個唔同既ans=.=
其他解答:7638E748CCC65837
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