these are the problems that are in my extra exercise books,please show me the stesp clearly that you slove the problems!
1. if tanA =4/3 and A+B=90。(degrees),find cosB.
2. if tanA=2ab /(a square - b square),find sinA.
maths problems
2008-05-23 3:31 am
回答 (3)
2008-05-23 4:11 am
✔ 最佳答案
As follows~圖片參考：http://i248.photobucket.com/albums/gg191/ncy_0916/trigo121.jpg?t=1211458259
2008-05-22 20:15:00 補充：
2)By cot^2A +1 = csc^2A,
then csc^2A﹣1 = cot^2A
As tanA > 0, sinA must be > 0
2008-05-22 20:15:50 補充：
1)AC^2 = AB^2 + BC^2
AC^2 = 3^2 + 4^2
AC^2 = 25
AC = 5
2008-05-23 4:17 am
1.tanA =4/3
用計數機..a=53.13....
b=90。-a
大約
計數機..
cos36.8699....=4/5
2.畫個直角三角形...
搵其中一個角=a...但不是90度個隻...
tanA=對邊/鄰邊...
....
用畢氏定理搵出斜邊....
(2ab)^2+(a^2-b^2)^2=斜邊^2
即斜邊=開方(2ab)^2+(a^2-b^2)^2
最後...sinA=對邊/斜邊..
=2ab /上面求出的斜邊..
=2ab/開方(2ab)^2+(a^2-b^2)^2
^2= square
2008-05-22 20:20:07 補充：
開方(2ab)^2+(a^2-b^2)^2
應該有得再化簡...
用計數機..a=53.13....
b=90。-a
大約
計數機..
cos36.8699....=4/5
2.畫個直角三角形...
搵其中一個角=a...但不是90度個隻...
tanA=對邊/鄰邊...
....
用畢氏定理搵出斜邊....
(2ab)^2+(a^2-b^2)^2=斜邊^2
即斜邊=開方(2ab)^2+(a^2-b^2)^2
最後...sinA=對邊/斜邊..
=2ab /上面求出的斜邊..
=2ab/開方(2ab)^2+(a^2-b^2)^2
^2= square
2008-05-22 20:20:07 補充：
開方(2ab)^2+(a^2-b^2)^2
應該有得再化簡...
參考: myself
2008-05-23 4:11 am
1. Since A + B = 90, therefore, A = 90 - B,
tanA=tan(90 - B)= 1/tanB. That is 4/3 = 1/tanB,
tanB=3/4, therefore, cosB = 4/sqrt(3^2 + 4^2) = 4/5.
2. tanA = 2ab/(a^2 - b^2), therefore, sinA = 2ab/sqrt[(2ab)^2 + (a^2 - b^2)^2]
= 2ab/sqrt[4a^2b^2 + a^4 -2a^2b^2 + b^4] = 2ab/sqrt[a^4 + 2a^2b^2 + b^4]
=2ab/sqrt[(a^2 + b^2)^2] = 2ab/(a^2 + b^2).
tanA=tan(90 - B)= 1/tanB. That is 4/3 = 1/tanB,
tanB=3/4, therefore, cosB = 4/sqrt(3^2 + 4^2) = 4/5.
2. tanA = 2ab/(a^2 - b^2), therefore, sinA = 2ab/sqrt[(2ab)^2 + (a^2 - b^2)^2]
= 2ab/sqrt[4a^2b^2 + a^4 -2a^2b^2 + b^4] = 2ab/sqrt[a^4 + 2a^2b^2 + b^4]
=2ab/sqrt[(a^2 + b^2)^2] = 2ab/(a^2 + b^2).
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