To'g'ri burchakli uchburchak: matematika darsi

Katetlari $6$ va $8$ bo'lgan uchburchakning gipotenuzasi $10$ , balandligi $4{,}8$ , ichki aylana radiusi $2$ , tashqi aylana radiusi $5$ . Bularning barchasi bitta rasmdan chiqadi. To'g'ri burchakli uchburchak planimetriyaning eng boy qismi: deyarli har bir geometriya masalasi unga keltiriladi.

Pifagor teoremasi

To'g'ri burchak qarshisidagi tomon gipotenuza, qolgan ikkitasi katetlar. Katetlar kvadratlari yig'indisi gipotenuza kvadratiga teng.

\angle C=90^\circ

a

b

katetlar,

c

gipotenuza

a^2+b^2=c^2

c=\sqrt{a^2+b^2}

a=\sqrt{c^2-b^2}

Gipotenuza doim eng katta tomon

Teskari teorema: $a^2+b^2=c^2$ bajarilsa, uchburchak to'g'ri burchakli va $c$ qarshisidagi burchak to'g'ri. $7$ , $24$ , $25$ uchburchagi shunday: $49+576=625$ .

Pifagor uchligi	Karralilari
$3,\ 4,\ 5$	$6,\ 8,\ 10$ ; $9,\ 12,\ 15$ ; $15,\ 20,\ 25$
$5,\ 12,\ 13$	$10,\ 24,\ 26$ ; $15,\ 36,\ 39$
$8,\ 15,\ 17$	$16,\ 30,\ 34$
$7,\ 24,\ 25$	$14,\ 48,\ 50$
$20,\ 21,\ 29$ ; $9,\ 40,\ 41$

Katetlar nisbati berilsa, $a=3k$ , $b=4k$ deb olamiz. Gipotenuza $35$ bo'lsa, $5k=35$ , $k=7$ , kichik katet $21$ . Gipotenuza katetdan aniq songa katta bo'lsa, $c=b+9$ deb Pifagorga qo'yamiz, kvadratlar qisqaradi va chiziqli tenglama qoladi.

Trigonometrik nisbatlar

$\alpha$ o'tkir burchak, $a$ uning qarshisidagi katet, $b$ yopishgan katet, $c$ gipotenuza bo'lsin:

\sin\alpha=\frac{a}{c}

\cos\alpha=\frac{b}{c}

Qarshi katet va yopishgan katet gipotenuzaga nisbati

\mathrm{tg}\,\alpha=\frac{a}{b}

\mathrm{ctg}\,\alpha=\frac{b}{a}

Qarshi katet yopishganga va aksincha

Bundan katetni topish qoidalari: $a=c\sin\alpha=b\,\mathrm{tg}\,\alpha$ , $b=c\cos\alpha=a\,\mathrm{ctg}\,\alpha$ . Ikkala o'tkir burchak yig'indisi $90^\circ$ bo'lgani uchun $\sin\beta=\cos\alpha$ , $\cos\beta=\sin\alpha$ , $\mathrm{tg}\,\beta=\mathrm{ctg}\,\alpha$ . Pifagordan $\sin^2\alpha+\cos^2\alpha=1$ kelib chiqadi.

$\alpha$	$\sin\alpha$	$\cos\alpha$	$\mathrm{tg}\,\alpha$	$\mathrm{ctg}\,\alpha$
$\alpha$ $30^\circ$	$\sin\alpha$ $\dfrac{1}{2}$	$\cos\alpha$ $\dfrac{\sqrt3}{2}$	$\mathrm{tg}\,\alpha$ $\dfrac{\sqrt3}{3}$	$\mathrm{ctg}\,\alpha$ $\sqrt3$
$\alpha$ $45^\circ$	$\sin\alpha$ $\dfrac{\sqrt2}{2}$	$\cos\alpha$ $\dfrac{\sqrt2}{2}$	$\mathrm{tg}\,\alpha$ $1$	$\mathrm{ctg}\,\alpha$ $1$
$\alpha$ $60^\circ$	$\sin\alpha$ $\dfrac{\sqrt3}{2}$	$\cos\alpha$ $\dfrac{1}{2}$	$\mathrm{tg}\,\alpha$ $\sqrt3$	$\mathrm{ctg}\,\alpha$ $\dfrac{\sqrt3}{3}$

Gipotenuzasi $10$ , o'tkir burchaklari sinuslari yig'indisi $\dfrac{7}{5}$ bo'lsa:

\frac{a}{10}+\frac{b}{10}=\frac{7}{5}

a+b=14

Sinuslar katetlarni beradi

Maxsus uchburchaklar

30^\circ

qarshisidagi katet gipotenuzaning yarmi. Tomonlar nisbati

1:\sqrt3:2

Teng yonli to'g'ri burchakli uchburchak. Tomonlar nisbati

{1:1:\sqrt2}

Teskari fakt ham o'rinli: katet gipotenuzaning yarmiga teng bo'lsa, uning qarshisidagi burchak $30^\circ$ . Burchaklar $1:1:2$ nisbatda bo'lsa, ular $45^\circ$ , $45^\circ$ , $90^\circ$ ; $1:2:3$ nisbatda bo'lsa, $30^\circ$ , $60^\circ$ , $90^\circ$ .

Bir burchagi

60^\circ

bo'lgan shakldan perpendikulyar tushirilsa,

30^\circ

li uchburchak hosil bo'ladi:

AD=5

AE=10

ED=5\sqrt3

Gipotenuzaga balandlik

CH=h

BH=p

AH=q

katetlarning gipotenuzadagi proyeksiyalari:

a

ning proyeksiyasi

p

b

ning proyeksiyasi

q

p+q=c

Balandlik uchburchakni o'zaro va berilganga o'xshash ikki uchburchakka ajratadi. Barcha formulalar shu o'xshashlikdan chiqadi:

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