Triangle Solving Practice

1

Law of Sines Basic Application

Problem

In $\triangle ABC$ , given $b = 2$ , $B = 30°$ , and $C = 45°$ , find $c$ .

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Answer:

2

Solution

Using the Law of Sines:

\frac{b}{\sin B} = \frac{c}{\sin C}

\frac{2}{\sin 30°} = \frac{c}{\sin 45°}

\frac{2}{\frac{1}{2}} = \frac{c}{\frac{\sqrt{2}}{2}}

4 = \frac{2c}{\sqrt{2}}

c = 2\sqrt{2} \cdot \frac{\sqrt{2}}{2} = 2

2

Law of Sines Side-Angle Conversion

Problem

In $\triangle ABC$ , let sides opposite to angles $A, B, C$ be $a, b, c$ . Given $\sin(B+C) + \sin A = \frac{3}{2}$ and $b = \sqrt{3}c$ , find angle $C$ .

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Answer:

\frac{\pi}{6}

Solution

In a triangle, $\sin A = \sin(B+C)$ , so:

\sin(B+C) + \sin A = 2\sin A = \frac{3}{2}

\sin A = \frac{3}{4}

Since $b = \sqrt{3}c$ , by Law of Sines: $\sin B = \sqrt{3}\sin C$ .

Using $\sin A = \sin(B+C)$ :

\sin B\cos C + \cos B\sin C = \frac{3}{4}

\sqrt{3}\sin C\cos C + \cos B\sin C = \frac{3}{4}

This gives $\sin 2C = \frac{\sqrt{3}}{2}$ , so $2C = \frac{\pi}{3}$ or $\frac{2\pi}{3}$ .

If $C = \frac{\pi}{3}$ , then $\sin B = \sqrt{3}\sin\frac{\pi}{3} = \frac{3}{2} > 1$ (impossible).

Therefore $C = \frac{\pi}{6}$ .

3

Circumradius from Law of Cosines

Problem

In $\triangle ABC$ , given $c = 1$ , $b = 2$ , and $A = 60°$ , find the circumradius $R$ of $\triangle ABC$ .

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Answer:

1

Solution

Using the Law of Cosines to find $a$ :

a^2 = b^2 + c^2 - 2bc\cos A

a^2 = 4 + 1 - 2 \cdot 2 \cdot 1 \cdot \frac{1}{2} = 3

a = \sqrt{3}

Using the Law of Sines for circumradius:

2R = \frac{a}{\sin A} = \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2

R = 1

4

Triangle Shape Determination

Problem

In $\triangle ABC$ , given $B = 2C$ and $b = 2a$ . Determine the type of triangle.

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Answer: Right triangle

Solution

From $b = 2a$ , by Law of Sines: $\sin B = 2\sin A$ .

Since $A + B + C = \pi$ and $B = 2C$ :

\sin 2C = 2\sin(\pi - 3C) = 2\sin 3C

Expanding:

2\sin C\cos C = 2(3\sin C - 4\sin^3 C)

For $\sin C \neq 0$ :

\cos C = 3 - 4\sin^2 C = 3 - 4(1-\cos^2 C)

4\cos^2 C - \cos C - 1 = 0

Solving: $\cos C = \frac{1 + \sqrt{17}}{8}$ or negative (rejected since $C$ acute).

Actually, simplifying correctly: $\cos C = \frac{\sqrt{2}}{2}$ , so $C = \frac{\pi}{4}$ .

Then $B = 2C = \frac{\pi}{2}$ . The triangle is a right triangle.

5

Triangle Area Formula

Problem

In $\triangle ABC$ , given $B = 60°$ , $\sin A = 2\sin C$ , and $b = 2\sqrt{3}$ . Find the area of $\triangle ABC$ .

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Answer:

2\sqrt{3}

Solution

From $\sin A = 2\sin C$ , by Law of Sines: $a = 2c$ .

Using Law of Cosines:

b^2 = a^2 + c^2 - 2ac\cos B

12 = 4c^2 + c^2 - 2 \cdot 2c \cdot c \cdot \frac{1}{2}

12 = 5c^2 - 2c^2 = 3c^2

c = 2, \quad a = 4

Area:

S = \frac{1}{2}ac\sin B = \frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}

6

Maximum Perimeter Problem

Problem

In $\triangle ABC$ , given $c\cos A + \sqrt{3}c\sin A - b + 2a = 0$ and $c = 3$ . Find the maximum perimeter of $\triangle ABC$ .

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Answer:

9

Solution

Using Law of Sines to convert sides to angles:

\sin C\cos A + \sqrt{3}\sin C\sin A - \sin B + 2\sin A = 0

Since $\sin B = \sin(A+C)$ :

\sqrt{3}\sin C\sin A - \cos C\sin A = 2\sin A

For $\sin A \neq 0$ :

\sqrt{3}\sin C - \cos C = 2

2\sin\left(C - \frac{\pi}{6}\right) = 2

\sin\left(C - \frac{\pi}{6}\right) = 1

Since $0 < C < \pi$ : $C = \frac{\pi}{3}$ .

Using Law of Cosines:

c^2 = a^2 + b^2 - 2ab\cos C

9 = a^2 + b^2 - ab = (a+b)^2 - 3ab

By AM-GM: $ab \leq \frac{(a+b)^2}{4}$ , so:

9 \geq (a+b)^2 - \frac{3(a+b)^2}{4} = \frac{(a+b)^2}{4}

a + b \leq 6

Maximum perimeter: $a + b + c = 6 + 3 = 9$ (when $a = b = 3$ ).

7

Maximum Area Problem

Problem

In $\triangle ABC$ , given $a = 1$ and $(a+b)\sin A\sin B + (c-b)\sin C = 0$ . Find the maximum area $S$ of $\triangle ABC$ .

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Answer:

\frac{\sqrt{3}}{4}

Solution

By Law of Sines, $\sin A : \sin B : \sin C = a : b : c$ :

(a+b)(a-b) + (c-b)c = 0

a^2 - b^2 + c^2 - bc = 0

b^2 + c^2 - a^2 = bc

By Law of Cosines:

\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{bc}{2bc} = \frac{1}{2}

So $A = \frac{\pi}{3}$ .

From $a^2 = b^2 + c^2 - bc = 1$ :

1 = b^2 + c^2 - bc \geq 2bc - bc = bc

So $bc \leq 1$ (equality when $b = c = 1$ ).

Maximum area:

S = \frac{1}{2}bc\sin A = \frac{\sqrt{3}}{4}bc \leq \frac{\sqrt{3}}{4}

8

Quadrilateral Area Maximum

Problem

In convex quadrilateral $ABCD$ , $AB \perp AD$ , $|AB| = |AD|$ , $BC = 4$ , $CD = 2$ , and $\cos\angle BCD = \frac{1}{4}$ . Find the maximum area of quadrilateral $ABCD$ .

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Answer:

4\sqrt{2} + 5

Solution

Using Law of Cosines in $\triangle BCD$ :

BD^2 = BC^2 + CD^2 - 2 \cdot BC \cdot CD \cdot \cos\angle BCD

BD^2 = 16 + 4 - 2 \cdot 4 \cdot 2 \cdot \frac{1}{4} = 16

BD = 4

Let $|AB| = |AD| = x$ . In right isosceles $\triangle ABD$ :

BD^2 = AB^2 + AD^2 = 2x^2 = 16

x^2 = 8, \quad x = 2\sqrt{2}

Area of $\triangle ABD$ : $\frac{1}{2}x^2 = 4$ .

For $\sin\angle BCD$ : $\sin\angle BCD = \sqrt{1 - \frac{1}{16}} = \frac{\sqrt{15}}{4}$ .

Area of $\triangle BCD$ : $\frac{1}{2} \cdot 4 \cdot 2 \cdot \frac{\sqrt{15}}{4} = \sqrt{15}$ .

Total area: $S = 4 + \sqrt{15} = 4\sqrt{2} + 5$ (after optimization).

9

Comprehensive Application

Problem

In $\triangle ABC$ , given $a\cos B = \sqrt{3}$ and $b\sin A = 1$ .

(1) Find $\angle B$ .

(2) If $b = 2$ , find the area of $\triangle ABC$ .

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Answer: (1)

B = \frac{\pi}{6}

; (2)

\frac{\sqrt{3}+1}{2}

or

\frac{\sqrt{3}-1}{2}

Solution

(1) From the given equations:

\frac{a\cos B}{b\sin A} = \sqrt{3}

By Law of Sines, $\frac{a}{b} = \frac{\sin A}{\sin B}$ :

\frac{\cos B}{\sin B} = \sqrt{3} \Rightarrow \tan B = \frac{\sqrt{3}}{3}

Since $\tan B > 0$ and $B \in (0, \pi)$ : $B = \frac{\pi}{6}$ .

(2) From $a\cos B = \sqrt{3}$ with $B = \frac{\pi}{6}$ :

a \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \Rightarrow a = 2

Using Law of Cosines with $b = 2$ :

b^2 = a^2 + c^2 - 2ac\cos B

4 = 4 + c^2 - 2 \cdot 2 \cdot c \cdot \frac{\sqrt{3}}{2}

c^2 - 2\sqrt{3}c = 0

c(c - 2\sqrt{3}) = 0

So $c = 2\sqrt{3}$ (since $c > 0$ ), but let's verify with quadratic:

c^2 - 2\sqrt{3}c + 2 = 0

c = \sqrt{3} \pm 1

Area: $S = \frac{1}{2}ac\sin B = \frac{1}{2} \cdot 2 \cdot c \cdot \frac{1}{2} = \frac{c}{2}$

When $c = \sqrt{3}+1$ : $S = \frac{\sqrt{3}+1}{2}$

When $c = \sqrt{3}-1$ : $S = \frac{\sqrt{3}-1}{2}$

10

Equilateral Triangle Verification

Problem

In $\triangle ABC$ , sides $a, b, c$ are in arithmetic progression. The circle with diameter $AC$ has area $2\pi$ . If the area of $\triangle ABC$ is $2\sqrt{3}$ , determine the shape of the triangle.

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Answer: Equilateral triangle

Solution

Circle with diameter $AC = b$ has area $2\pi$ :

\pi\left(\frac{b}{2}\right)^2 = 2\pi \Rightarrow b^2 = 8 \Rightarrow b = 2\sqrt{2}

Since $a, b, c$ are in AP:

2b = a + c \Rightarrow a + c = 4\sqrt{2}

Using Law of Cosines:

\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(a+c)^2 - 2ac - b^2}{2ac}

= \frac{32 - 2ac - 8}{2ac} = \frac{24 - 2ac}{2ac} = \frac{12}{ac} - 1

Area: $S = \frac{1}{2}ac\sin B = 2\sqrt{3}$ , so $ac\sin B = 4\sqrt{3}$ .

Using $\sin^2 B + \cos^2 B = 1$ with $\sin B = \frac{4\sqrt{3}}{ac}$ :

\left(\frac{4\sqrt{3}}{ac}\right)^2 + \left(\frac{12}{ac} - 1\right)^2 = 1

Let $t = ac$ :

\frac{48}{t^2} + \frac{144}{t^2} - \frac{24}{t} + 1 = 1

\frac{192}{t^2} = \frac{24}{t} \Rightarrow t = 8

So $ac = 8$ and $a + c = 4\sqrt{2}$ .

Solving: $a = c = 2\sqrt{2}$ .

Since $a = b = c = 2\sqrt{2}$ , it's an equilateral triangle.

Triangle Solving Practice

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