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2009-11-28T20:22:48+01:00
Tg α = 3, α - miara kąta ostrego
tg α = y/x = 3 ---> y = 3x
r² = x² +y² = x² +(3x)² = x² +9x² = 10x²
r =√(10x²) = x*√10
sin α = y/r = 3x/(x*√10) = 3/√10 =(3√10)/10
cos α = x/r = x/(x*√10) = 1/√10 =√10/10
[8 cosα - 7 sin α]/[5 cosα + 2 sin α] =
=[8√10/10 - 7*3√10/10]/[5√10/10 + 2*3√10/10] =
= [- 13√10/10]/[ 11√10/10] = [-13√10/10]*[10/(11√10)] =
= - 13/11
2009-11-28T20:32:59+01:00
α ∈ (0°; 90°)
tg α=3
tg α=sinα/cosα
sinα/cosα=3
sinα=3cosα
(8cosα - 7sinα)/(5cosα + 2sinα)= (8cosα-7*3cosα)/(5cosα+2*3cosα)=
=(8cosα-21cosα)/(5cosα+6cosα)=
-13cosα/11cosα=-13/11
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