Zestaw XI

Zad 1

sinα=√5/5 √5²+b²=5² 5+b²=25 b²=25-5=20 b=√20=|√4*5|=2√5 tgα=√5/2√5=1/2

Zad 2

tgα=1/3 tgα=sinα/cosα sin²α+cos²α=1 sinα/cosα=1/3 cosα=3sinα

sin²α+(3sinα)²=1 sin²α+9sin²α=1 10sin²α=1|:10 sin²α=1/10

5(2sin²α-1)-5(2(1/10)-1)=5((2/10)-1)=5*(1/5)-5=1-5=-4

Zad 3

tgα=5 tgα=sinα/cosα 5=sinα/cosα sinα=5cosα (sinα-cosα)/sinα+cosα=(5cos-cosα)/(5cos+cosa)=4cosα/6cosα=4/6=2/3

Zad 4

sinα=√3/3 =>cosα=|√1-(√3/3)2|=|√1-(3/9)|=√6/3

tgα=sinα/cosα=√3/3 √6/3=√3/3*3/√6=|√3/6|=|√1/2|=1/√2=√2/2

Jeżeli sinα=√3/3 to tgα≠2/3 więc nie istnieje α dla którego zachodzi podany warunek

Zad 5

sinα+sin(90°-α)=√5/2 z własności f. trygo: sin(90°-α)=cosα więc: (sinα+cosα)²=(√5/2)²

sin²α+cos²α+2sinαcosα=5/4 1+2sinαcosα=5/4 2sinαcosα=5/4-1 2sinαcosα=1/4|:2

sinαcosα=1/8 sin(90°-β)cosα=1/8 z własności f.trygon: sin(90°-β)=cosβ zatem: cosβcosα=1/8

Zad 6

sin(α-β)=sinαcosβ-cosαsinβ

sin(45°-30°)=sin45°*cos30°-cos45°*sin30°=√2/2*√3/2-√2/2*1/2=√6/4-√2/4=(√6-√2)/4

Zad 7

tgα=4/1=h 1²+h²=4² h²=16-1 h=√15

Zad 8

x/(10+x)=tg30° x/(10+x)=√3/3 3x=√3(10+x) 3x=10√3 +√3x 3x-√3x=10√3 ((3-√3)x)=10√3|:(3-√3) x=(10√3)/(3-√3) x=(10√3)/(3√3)*(3+√3)/3+√3) x=(10√3(3+√3))/9-3 x=(30√3+30)/6 x=(30√3+30)/6 x=(30(√3+1)/6 x=5(√3+1)

Zad 9

[ cos²a-cosa]/[ sin³a-sina]=[cosa*( cos²a-1)]/[sina*(sin²a-1)]=[cosa*(-sin²a)]/[ sina*(-cos² a)] =- sina /(- cosa) = tga

Zad 10

Dane podłoża = 30m α = 55° H/x = tg 55° = 1,4281 podłoże = H/1,4281 = 30/1,4281=21m Lina=√H²+x²=|√30²+ 21²|= |√900 + 441| = √1341 = 36,62 m

Zad 11

	1		5
tgα +		=
	tgα		sinα

sinα		cosα		5
	+		=		/*sinαcosα
cosα		sinα		sinα

sin²α + cos²α = 5cosα 5cosα = 1 /:5

	1
cosα =
	5

sin²α = 1 − cos²α

	1
sin^2α = 1−
	25

	24
sin^2α =
	25

	√4*6		2√6
sinα =		=
	5		5

	sinα		2√6   5
tgα =		=		= 2√6
	cosα		1   5