Uchburchakni yechish

Bu nima va nega kerak

Xaritagrafiya, geodeziya va navigatsiyada uzoq masofalarni o'lchash uchun triangulyatsiya usuli qo'llaniladi. Bu usul yerni uchburchaklar to'riga ajratib, har bir uchburchakning noma'lum tomon va burchaklarini hisoblashga asoslanadi.

Uchburchakni yechish — uchburchakning berilgan ba'zi elementlari orqali qolgan barcha noma'lum tomon va burchaklarini topish jarayonidir. Uchburchakning 6 ta asosiy elementi bor: 3 ta tomon ($a, b, c$) va 3 ta burchak ($\alpha, \beta, \gamma$). Uchburchakni yechish uchun kamida 3 ta element (ulardan kamida bittasi tomon) ma'lum bo'lishi shart.

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Asosiy g'oya va 4 ta asosiy masala turi

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

Uchburchakni yechish 4 asosiy holatga bo'linadi:

1-holat. Bir tomon va ikkita burchak berilganda ($a, \beta, \gamma$)

1. Uchinchi burchak: $\alpha = 180^\circ - (\beta + \gamma)$;
2. Sinuslar teoremasidan qolgan tomonlar topiladi: $b = \dfrac{a \sin \beta}{\sin \alpha}, \quad c = \dfrac{a \sin \gamma}{\sin \alpha}$

2-holat. Ikki tomon va ular orasidagi burchak berilganda ($a, b, \gamma$)

1. Kosinuslar teoremasi bilan uchinchi tomon topiladi: $c = \sqrt{a^2 + b^2 - 2ab \cos \gamma}$
2. Kosinuslar teoremasi bilan burchak topiladi: $\cos \alpha = \dfrac{b^2 + c^2 - a^2}{2bc}$
3. Uchinchi burchak: $\beta = 180^\circ - (\alpha + \gamma)$.

3-holat. Uchta tomon berilganda ($a, b, c$)

Uchburchak tengsizligi tekshiriladi ($a+b>c, a+c>b, b+c>a$).

1. Kosinuslar teoremasidan burchaklar kosinusi topiladi: $\cos \alpha = \dfrac{b^2 + c^2 - a^2}{2bc}, \quad \cos \beta = \dfrac{a^2 + c^2 - b^2}{2ac}$
2. Uchinchi burchak: $\gamma = 180^\circ - (\alpha + \beta)$.

4-holat. Ikki tomon va birining qarshisidagi burchak berilganda ($a, b, \alpha$)

Sinuslar teoremasi orqali: $\sin \beta = \dfrac{b \sin \alpha}{a}$.

• $\sin \beta > 1 \implies$ yechim yo'q;
• $\sin \beta = 1 \implies$ bitta to'g'ri burchakli uchburchak ($\beta = 90^\circ$);
• $\sin \beta < 1 \implies$ 1 ta yoki 2 ta yechim.

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Yechilgan misollar

1-misol. Bir tomon va ikki burchak bo'yicha yechish

$ABC$ uchburchakda $a = 6\sqrt{2}\text{ sm}$, $\alpha = 60^\circ$ va $\beta = 45^\circ$. Qolgan burchak va tomonlarni toping.

• $\gamma = 180^\circ - (60^\circ + 45^\circ) = 75^\circ$.
• Sinuslar teoremasi: $\dfrac{b}{\sin 45^\circ} = \dfrac{6\sqrt{2}}{\sin 60^\circ} \implies b = \dfrac{6\sqrt{2} \cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{3}}{2}} = \dfrac{6}{\frac{\sqrt{3}}{2}} = 4\sqrt{3}\text{ sm}$

Javob: $\gamma = 75^\circ$, $b = 4\sqrt{3}\text{ sm}$.

2-misol. Ikki tomon va ular orasidagi burchak

Uchburchakda $a = 5\text{ sm}$, $b = 8\text{ sm}$ va $\gamma = 60^\circ$. $c$ tomonni va $\cos \alpha$ ni toping.

• Kosinuslar teoremasi: $c^2 = 5^2 + 8^2 - 2(5)(8)\cos 60^\circ = 25 + 64 - 40 = 49 \implies c = 7\text{ sm}$
• $\alpha$ burchak kosinusi: $\cos \alpha = \dfrac{8^2 + 7^2 - 5^2}{2(8)(7)} = \dfrac{64 + 49 - 25}{112} = \dfrac{88}{112} = \dfrac{11}{14}$

Javob: $c = 7\text{ sm}$, $\cos \alpha = \dfrac{11}{14}$.

3-misol. Uchta tomon berilganda

Tomonlari $a = 13\text{ sm}, b = 14\text{ sm}, c = 15\text{ sm}$ bo'lgan uchburchakning kichik burchagi kosinusini va yuzini toping.

• Kichik tomon $a = 13\text{ sm}$ qarshisida $\alpha$ yotadi: $\cos \alpha = \dfrac{14^2 + 15^2 - 13^2}{2(14)(15)} = \dfrac{196 + 225 - 169}{420} = \dfrac{252}{420} = 0{,}6$
• $\sin \alpha = \sqrt{1 - 0{,}6^2} = 0{,}8$.
• Yuza: $S = \dfrac{1}{2}bc \sin \alpha = \dfrac{1}{2} \cdot 14 \cdot 15 \cdot 0{,}8 = 84\text{ sm}^2$.

Javob: $\cos \alpha = 0{,}6$, $S = 84\text{ sm}^2$.

4-misol. Amaliy masofa masalasi

Daryo bo'yida $AB = 30\text{ m}$ kesma belgilandi va narigi sohilidagi $C$ daraxtgacha burchaklar o'lchandi: $\angle CAB = 60^\circ$, $\angle CBA = 90^\circ$. $AC$ va $BC$ masofalarni toping.

• $\triangle ABC$ to'g'ri burchakli ($\angle B = 90^\circ$).
• $\angle ACB = 30^\circ$.
• $30^\circ$ qarshisidagi katet $AB = 30\text{ m}$ bo'lgani uchun gipotenuza $AC = 2 \cdot 30 = 60\text{ m}$.
• $BC = 30 \cdot \operatorname{tg} 60^\circ = 30\sqrt{3}\text{ m}$.

Javob: $AC = 60\text{ m}$, $BC = 30\sqrt{3}\text{ m}$.

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Mashq

1. $ABC$ uchburchakda $a = 8\text{ sm}$, $\angle A = 30^\circ$, $\angle B = 45^\circ$. $b$ tomonni va $\angle C$ burchakni toping.
2. Uchburchakda $a = 6\text{ sm}, b = 10\text{ sm}$ va $\angle C = 120^\circ$. $c$ tomonni toping.
3. Tomonlari $a = 4\text{ sm}, b = 5\text{ sm}, c = 6\text{ sm}$ bo'lgan uchburchakda $\angle B$ kosinusini hisoblang.
4. $ABC$ uchburchakda $a = 7\text{ sm}, b = 8\text{ sm}, c = 9\text{ sm}$. Katta burchak kosinusini toping.
5. Uchburchakda $a = 12\text{ sm}$, $\angle A = 60^\circ$, $\angle B = 60^\circ$. Qolgan tomonlari va burchagini aniqlang.
6. To'g'ri burchakli uchburchakda $a = 5\text{ sm}$, $b = 5\sqrt{3}\text{ sm}$. Gipotenuza va o'tkir burchaklarini toping.
7. Uchburchakda $a = 4\text{ sm}, b = 6\text{ sm}$ va $\angle C = 60^\circ$. Uchburchak yuzini toping.
8. Minora tagidan $30\text{ m}$ masofada uning uchi $30^\circ$ burchak ostida ko'rinadi. Minora balandligini toping.
9. Tomonlari $5\text{ sm}, 6\text{ sm}, 7\text{ sm}$ bo'lgan uchburchak o'tkir burchaklimi yoki o'tmas burchaklimi?
10. $ABC$ uchburchakda $b = 8\text{ sm}, c = 5\text{ sm}$ va $\angle A = 60^\circ$. $a$ tomonni toping.

Javoblar va yechim yo'llari

1. $\angle C = 180^\circ - 75^\circ = 105^\circ$. $b = \dfrac{8 \sin 45^\circ}{\sin 30^\circ} = 8\sqrt{2}\text{ sm}$. Javob: $b = 8\sqrt{2}\text{ sm}, \angle C = 105^\circ$.
2. $c^2 = 36 + 100 - 120(-0{,}5) = 196 \implies c = 14\text{ sm}$. Javob: $14\text{ sm}$.
3. $\cos B = \dfrac{16 + 36 - 25}{48} = \dfrac{27}{48} = \dfrac{9}{16} = 0{,}5625$. Javob: $\dfrac{9}{16}$.
4. Katta tomon $9$: $\cos \gamma = \dfrac{49 + 64 - 81}{112} = \dfrac{32}{112} = \dfrac{2}{7}$. Javob: $\dfrac{2}{7}$.
5. $\angle C = 60^\circ$. Muntazam uchburchak: $b = c = 12\text{ sm}$. Javob: $\angle C = 60^\circ, b = 12\text{ sm}, c = 12\text{ sm}$.
6. $c = \sqrt{25 + 75} = 10\text{ sm}$. $\sin A = 0{,}5 \implies \angle A = 30^\circ, \angle B = 60^\circ$. Javob: $c = 10\text{ sm}, \angle A = 30^\circ, \angle B = 60^\circ$.
7. $S = \dfrac{1}{2} \cdot 4 \cdot 6 \cdot \sin 60^\circ = 6\sqrt{3}\text{ sm}^2$. Javob: $6\sqrt{3}\text{ sm}^2$.
8. $H = 30 \cdot \operatorname{tg} 30^\circ = 10\sqrt{3}\text{ m}$. Javob: $10\sqrt{3}\text{ m}$.
9. $5^2 + 6^2 = 61 > 7^2 = 49$. O'tkir burchakli. Javob: O'tkir burchakli.
10. $a^2 = 64 + 25 - 2(8)(5)(0{,}5) = 49 \implies a = 7\text{ sm}$. Javob: $7\text{ sm}$.

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Keng tarqalgan xatolar

1. Uchta burchak orqali tomonlarni topishga urinish. ✗ $\angle A, \angle B, \angle C$ berilganda tomonlarni bir qiymatli hisoblash. ✓ Uchta burchak faqat burchaklar nisbatini (o'xshashlikni) beradi. Kamida bitta tomon ma'lum bo'lishi shart.

2. Uchburchak tengsizligini tekshirmaslik. ✗ Tomonlari $2, 3, 6$ bo'lgan uchburchakni yechish. ✓ $2 + 3 = 5 < 6$ bo'lgani sababli bunday uchburchak mavjud emas.

3. Sinuslar teoremasida burchakning o'tmaslik ehtimolini unutish. ✗ $\sin \beta = 0{,}5$ bo'lganda faqat $\beta = 30^\circ$ deb olish. ✓ $\beta = 150^\circ$ ham bo'lishi mumkin. Shu sababli burchaklarni kosinuslar teoremasi orqali topish xavfsizroq, chunki kosinus burchak turini aniq ko'rsatadi.

4. Burchaklar yig'indisini tekshirmaslik. ✗ Burchaklarni hisoblab, ularning yig'indisi $180^\circ$ bo'lishini nazorat qilmaslik. ✓ Doimo yakunda $\alpha + \beta + \gamma = 180^\circ$ bajarilishini tekshiring.

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DTM uslubidagi test

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

Keyingi qadam

Uchburchakda trigonometriya bo'limini muvaffaqiyatli yakunladingiz! Endi siz to'g'ri burchakli va ixtiyoriy uchburchaklarni to'liq yecha olasiz. Keyingi bo'limda geometriyadagi eng muhim shakllardan biri — Aylana va doira geometriyasi bilan tanishamiz.