Sinuslar va kosinuslar teoremasi

Bu nima va nega kerak

Pifagor teoremasi faqat to'g'ri burchakli uchburchak tomonlarini bog'laydi. Ammo amaliyotda uchraydigan uchburchaklarning aksariyati to'g'ri burchakli emas: yer uchastkalari, qurilish fermalari va navigatsiya yo'llari ixtiyoriy burchakli uchburchaklardan iborat.

Ixtiyoriy uchburchakda tomonlar va burchaklar orasidagi bog'liqlikni qanday ifodalash mumkin?

Buning uchun ikkita muhim qonun mavjud: Kosinuslar teoremasi va Sinuslar teoremasi. Kosinuslar teoremasi Pifagor teoremasining ixtiyoriy burchak uchun umumlashtirilgan shaklidir. Sinuslar teoremasi esa tomonlar va burchaklar nisbatini tashqi chizilgan aylana radiusi bilan bog'laydi.

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Asosiy g'oya va teoremalar

Ixtiyoriy $ABC$ uchburchakda burchaklar $\alpha, \beta, \gamma$, qarshisidagi tomonlar $a, b, c$ bo'lsin.

          B
         / \
      c /   \ a
       /     \
      /_______\
     A    b    C

1. Kosinuslar teoremasi

[!IMPORTANT] Kosinuslar teoremasi. Uchburchak ixtiyoriy tomonining kvadrati qolgan ikki tomon kvadratlari yig'indisidan shu tomonlar va ular orasidagi burchak kosinusi ko'paytmasining ikkilangani ayirmasiga teng: $a^2 = b^2 + c^2 - 2bc \cos \alpha$ $b^2 = a^2 + c^2 - 2ac \cos \beta$ $c^2 = a^2 + b^2 - 2ab \cos \gamma$

$\alpha = 90^\circ$ bo'lganda $\cos 90^\circ = 0$ bo'lib, $a^2 = b^2 + c^2$ (Pifagor teoremasi) hosil bo'ladi.

Muhim natijalar:

1. Burchak kosinusi: $\cos \alpha = \dfrac{b^2 + c^2 - a^2}{2bc}$
2. Burchak turini aniqlash:
   • $b^2 + c^2 > a^2 \implies \alpha < 90^\circ$ (o'tkir);
   • $b^2 + c^2 = a^2 \implies \alpha = 90^\circ$ (to'g'ri);
   • $b^2 + c^2 < a^2 \implies \alpha > 90^\circ$ (o'tmas).
3. Parallelogramm diagonallari xossasi: $d_1^2 + d_2^2 = 2(a^2 + b^2)$

2. Sinuslar teoremasi

[!IMPORTANT] Sinuslar teoremasi. Uchburchak tomonlari qarshisidagi burchaklar sinuslariga proporsional bo'lib, bu nisbat tashqi chizilgan aylananing diametriga teng: $\dfrac{a}{\sin \alpha} = \dfrac{b}{\sin \beta} = \dfrac{c}{\sin \gamma} = 2R$ Bu yerda $R$ — tashqi chizilgan aylana radiusi.

Bundan $R = \dfrac{a}{2\sin \alpha}$ kelib chiqadi.

3. Trigonometrik yuz formulalari

$S = \dfrac{1}{2}ab \sin \gamma = \dfrac{1}{2}bc \sin \alpha = \dfrac{1}{2}ac \sin \beta$ $S = \dfrac{abc}{4R}$

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Yechilgan misollar

1-misol. Uchinchi tomonni topish

Uchburchakning ikki tomoni $5\text{ sm}$ va $8\text{ sm}$, ular orasidagi burchak $60^\circ$. Uchinchi tomonni toping.

• Kosinuslar teoremasiga ko'ra: $a^2 = 5^2 + 8^2 - 2 \cdot 5 \cdot 8 \cdot \cos 60^\circ$
• $\cos 60^\circ = 0{,}5$: $a^2 = 25 + 64 - 80 \cdot 0{,}5 = 89 - 40 = 49$ $a = \sqrt{49} = 7\text{ sm}$

Javob: $7\text{ sm}$.

2-misol. Burchak turini va kattaligini aniqlash

Tomonlari $a = 3\text{ sm}$, $b = 5\text{ sm}$, $c = 7\text{ sm}$ bo'lgan uchburchakning eng katta burchagini toping.

• Eng katta burchak katta tomon ($c = 7\text{ sm}$) qarshisida yotadi: $\cos \gamma = \dfrac{3^2 + 5^2 - 7^2}{2 \cdot 3 \cdot 5} = \dfrac{9 + 25 - 49}{30} = \dfrac{-15}{30} = -0{,}5$
• $\cos \gamma = -0{,}5 \implies \gamma = 180^\circ - 60^\circ = 120^\circ$.

Javob: $120^\circ$.

3-misol. Tashqi aylana radiusini hisoblash

Uchburchakda $a = 12\text{ sm}$ tomon qarshisida $\alpha = 30^\circ$ burchak yotadi. Tashqi chizilgan aylana radiusini toping.

• Sinuslar teoremasidan: $2R = \dfrac{a}{\sin \alpha} = \dfrac{12}{\sin 30^\circ} = \dfrac{12}{0{,}5} = 24\text{ sm}$
• $R = 24 : 2 = 12\text{ sm}$.

Javob: $12\text{ sm}$.

4-misol. Sinuslar teoremasi orqali tomonni topish

$ABC$ uchburchakda $c = 6\sqrt{2}\text{ sm}$, $\angle C = 45^\circ$, $\angle A = 60^\circ$. $a = BC$ tomonni toping.

• Sinuslar teoremasi: $\dfrac{a}{\sin 60^\circ} = \dfrac{6\sqrt{2}}{\sin 45^\circ} \implies a = \dfrac{6\sqrt{2} \cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = 6\sqrt{3}\text{ sm}$

Javob: $6\sqrt{3}\text{ sm}$.

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Mashq

1. Uchburchakning ikki tomoni $4\text{ sm}$ va $6\text{ sm}$, orasidagi burchak $60^\circ$. Uchinchi tomonini toping.
2. Tomonlari $7\text{ sm}, 8\text{ sm}, 9\text{ sm}$ bo'lgan uchburchakning kichik burchagi kosinusini hisoblang.
3. Uchburchakda $\alpha = 45^\circ$ qarshisidagi tomon $8\text{ sm}$. Tashqi aylana radiusini toping.
4. Parallelogramm tomonlari $7\text{ sm}$ va $9\text{ sm}$, bitta diagonali $8\text{ sm}$. Ikkinchi diagonalini toping.
5. $ABC$ uchburchakda $a = 10\text{ sm}$, $\angle A = 30^\circ$, $\angle B = 45^\circ$. $b$ tomonni toping.
6. Tomonlari $6\text{ sm}$ va $10\text{ sm}$, orasidagi burchak $30^\circ$ bo'lgan uchburchakning yuzini toping.
7. Tomonlari $3\text{ sm}, 5\text{ sm}, 7\text{ sm}$ bo'lgan uchburchakning eng katta burchagini toping.
8. Teng yonli uchburchakning yon tomoni $10\text{ sm}$, uchidagi burchagi $120^\circ$. Asosini hisoblang.
9. Tomonlari $5\text{ sm}, 6\text{ sm}, 7\text{ sm}$ bo'lgan uchburchak burchaklariga ko'ra qanday uchburchak?
10. Tomonlari $13\text{ sm}, 14\text{ sm}, 15\text{ sm}$ bo'lgan uchburchakka tashqi chizilgan aylana radiusini toping.

Javoblar va yechim yo'llari

1. $c^2 = 16 + 36 - 2(4)(6)(0{,}5) = 28 \implies c = \sqrt{28} = 2\sqrt{7}\text{ sm}$. Javob: $2\sqrt{7}\text{ sm}$.
2. Kichik tomon $7\text{ sm}$ qarshisida: $\cos \alpha = \dfrac{64 + 81 - 49}{2(8)(9)} = \dfrac{96}{144} = \dfrac{2}{3}$. Javob: $\dfrac{2}{3}$.
3. $2R = \dfrac{8}{\sin 45^\circ} = 8\sqrt{2} \implies R = 4\sqrt{2}\text{ sm}$. Javob: $4\sqrt{2}\text{ sm}$.
4. $d_2^2 = 2(49 + 81) - 64 = 260 - 64 = 196 \implies d_2 = 14\text{ sm}$. Javob: $14\text{ sm}$.
5. $b = \dfrac{10 \sin 45^\circ}{\sin 30^\circ} = \dfrac{10(\sqrt{2}/2)}{0{,}5} = 10\sqrt{2}\text{ sm}$. Javob: $10\sqrt{2}\text{ sm}$.
6. $S = \dfrac{1}{2} \cdot 6 \cdot 10 \cdot \sin 30^\circ = 30 \cdot 0{,}5 = 15\text{ sm}^2$. Javob: $15\text{ sm}^2$.
7. $\cos \gamma = \dfrac{9 + 25 - 49}{30} = -0{,}5 \implies \gamma = 120^\circ$. Javob: $120^\circ$.
8. $a^2 = 100 + 100 - 200(-0{,}5) = 300 \implies a = 10\sqrt{3}\text{ sm}$. Javob: $10\sqrt{3}\text{ sm}$.
9. Katta tomon $7$: $5^2 + 6^2 = 61 > 7^2 = 49$. Burchak o'tkir, demak o'tkir burchakli. Javob: O'tkir burchakli.
10. Geron formulasi: $p = 21$, $S = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84\text{ sm}^2$. $R = \dfrac{13 \cdot 14 \cdot 15}{4 \cdot 84} = 8{,}125\text{ sm}$ ($\dfrac{65}{8}\text{ sm}$). Javob: $8{,}125\text{ sm}$.

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Keng tarqalgan xatolar

1. Kosinuslar teoremasida ishorani adashtirish. ✗ $a^2 = b^2 + c^2 + 2bc \cos \alpha$ deb yozish. ✓ Formulada ikkilangan ko'paytma ayriladi: $-2bc \cos \alpha$.

2. O'tmas burchakda kosinus manfiyligini inobatga olmaslik. ✗ O'tmas burchak kosinusi manfiy bo'lsa ham formuladagi minusni o'zgartirmaslik. ✓ Agar $\alpha > 90^\circ$ bo'lsa, $\cos \alpha < 0$ bo'lib, minus bilan minus plyus beradi: $b^2 + c^2 + 2bc|\cos \alpha|$.

3. Sinuslar teoremasidagi diametrni radius deb hisoblash. ✗ $\dfrac{a}{\sin \alpha} = R$ deb yozish. ✓ Nisbat radiusga emas, balki diametrga ($2R$) tengdir.

4. Mos tomon va burchakni almashtirib qo'yish. ✗ $\dfrac{a}{\sin \beta} = \dfrac{b}{\sin \alpha}$ deb olish. ✓ Har doim tomon o'zining qarshisidagi burchak sinusiga nisbatlanadi.

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DTM uslubidagi test

Ushbu mavzuni mustahkamlash uchun berilgan 5 ta test savolini ishlab, javoblaringizni tekshiring.

Keyingi qadam

Sinuslar va kosinuslar teoremalari orqali uchburchak tomonlari va burchaklarini hisoblashni o'rgandik. Keyingi darsda bu bilimlarni birlashtirib, berilgan 3 ta element bo'yicha butun uchburchakni to'liq tiklash — Uchburchakni yechish mavzusini o'rganamiz.