Zad. 1
Z wlasnosci trojkata o katach 90, 60, 30 wynika, ze:
[tex]|AD|=\frac{|DC|}2\\5=\frac{|DC|}2 /*2\\10=|DC|\\|AC|=\frac{|DC|\sqrt3}2\\|AC|=\frac{10\sqrt3}2\\\\\text{Odcinek DC ma }10\sqrt3cm \text{ dlugosci} - falsz\\\\|BC|=2|AC|\\|BC|=2*\frac{10\sqrt3}2cm=10\sqrt3cm\\|AB|=|AC|\sqrt3=\frac{10\sqrt3}2*\sqrt3cm=5*3=15cm\\\\|AB|=|AD|+|DB|\\15cm=5cm+|DB| /-5cm\\10cm=|DB|\\\\Ob_{CDB}=|CD|+|DB|+|BC|\\OB_{CDB}=10+10+10\sqrt3=20+10\sqrt3cm\\\\\text{Obwod trojkata CBD jest rowny } 20+10\sqrt3cm - prawda[/tex]
Zad. 2
Cosinus kata ostrego to stosunek dlugosci przyprostokatnej lezacej na przeciwko kata ostrego do przeciwprostokatnej.
[tex]3^2+4^2=c^2\\9+16=c^2\\25=c^2\\5=c\\\\cos\alpha=\frac35\\\\Odp. C[/tex]
Zad. 3
[tex]b=2,5a\\a^2+b^2=c^2\\a^2+(2,5a)^2=c^2\\a^2+6.25a^2=c^2\\7.25a^2=c^2\\c=\sqrt{7.25a^2}=\sqrt{7\frac14a^2}=\sqrt{\frac{29}4a^2}=\frac{\sqrt{29}}2a[/tex]
[tex]sin\alpha=\frac{a}c=\frac{a}{\frac{\sqrt{29}}2a}=1:\frac{\sqrt{29}}2=1*\frac2{\sqrt{29}}*\frac{\sqrt{29}}{\sqrt{29}}=\frac{2\sqrt{29}}{29}[/tex]
[tex]cos\alpha=\frac{b}c=\frac{2.25a}{\frac{\sqrt{29}}2a}=2\frac14:\frac{\sqrt{29}}2=\frac94*\frac2{\sqrt{29}}=\frac{9}{2\sqrt{29}}*\frac{\sqrt{29}}{\sqrt{29}}=\frac{9\sqrt{29}}{2*29}=\frac{9\sqrt{29}}{58}[/tex]
[tex]tg\alpha=\frac{a}b=\frac{a}{2.25a}=1:2.25=1:2\frac14=1:\frac94=1*\frac49=\frac49\\ctg\alpha=\frac{b}a=\frac{2.25a}a=2.25[/tex]
[tex]sin\beta=\frac{b}c=cos\alpha=\frac{9\sqrt{29}}{58}\\cos\beta=\frac{a}c=sin\alpha=\frac{2\sqrt{29}}{29}\\tg\beta=\frac{b}a=ctg\alpha=2.25\\ctg\beta=\frac{a}b=tg\alpha=\frac49[/tex]
