Question

POTRZEBUJE ROZWIĄZANIE. PILNE NA JUŻ. PROSZĘ O POMOC.

DAM MAX PTK ,

Answer 1

Ctgα=4/3⇒tgα=3/4

sin²α=sin²α/1=sin²α/(sin²α+cos²α)=tg²α/(1+tg²α)=(9/16)/(25/16)=9/25
cos²α=1-sin²α=16/25
wiec
sinα=3/5  cosα=4/5
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(1-cos²α)/sinα*cosα/sinα=sinα·cosα/sinα=cosα
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DANE
α=40  b=4
β=90-40=50
cosα=b/c⇒c=b/cosα=4/cos40=5,2216
tgα=a/b⇒a=b·tgα=4·tg40=3,3564