a pyramid OPQR (trianglar base PQO, R at the top,PQ is a horigontal line, P is on the left, O is in the backside of the paper ),the sides OP,OQ,OR are the lengths x,y,z respectively,and the are mutually perpendicular to each other.
i/ express cosPRQ in term of x,y,z
ii/ let s1,s2,s3 & s4 denote the areas of OPR,OPQ,OQR, and PQR respectively,
show that s4^2=s1^2+s2^2+s3^2 (where s1^2 = s1 to the power of 2)
f4 ad mathe1
2007-03-02 8:06 pm
回答 (1)
2007-03-02 8:28 pm
✔ 最佳答案
cosPRQ = (PR^2 + QR^2 - PQ^2) /2PR*QR
= (x^2+z^2)+(y^2+z^2)-(x^2-y^2) /2sqrt(x^2+z^2)sqrt(y^2+z^2)
=z^2/sqrt(x^2+z^2)sqrt(y^2+z^2)
s4^2
=(1/2PR*QR*sinPRQ)^2
=1/4(PR^2*QR^2)(1-cos^2PRQ)
=1/4(x^2+z^2)(y^2+z^2)[1- z^4/(x^2+z^2)(y^2+z^2)]
=1/4[(x^2+z^2)(y^2+z^2)-z^4]
=1/4[(xy)^2+(yz)^2+(zx)^2]
=(xy/2)^2+(yz/2)^2+(zx/2)^2
收錄日期: 2021-04-12 21:24:05
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https://hk.answers.yahoo.com/question/index?qid=20070302000051KK01066