M203 20260221 Trigonometry in Geometry

#Law_Of_Sines

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$

($R$ is the radius of the circumcircle of triangle $ABC$).

#Law_Of_Cosines

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

$$\text{Area} = \frac{1}{2} ab \sin C$$

17 ) If the area of $\triangle ABC$ is $64$ square inches and the geometric mean (mean proportional) between sides $AB$ and $AC$ is $12$ inches, then $\sin A$ is equal to

(A) $\dfrac{\sqrt{3}}{2}$

(B) $\dfrac{3}{5}$

(C) $\dfrac{4}{5}$

(D) $\dfrac{8}{9}$

(E) $\dfrac{15}{17}$

A sector with acute central angle $\theta$ is cut from a circle of radius $6$. The radius of the circle circumscribed about the sector is

(A) $3 \cos \theta$

(B) $3 \sec \theta$

(C) $3 \cos \frac{1}{2} \theta$

(D) $3 \sec \frac{1}{2} \theta$

(E) $3$

21 ） $ABCD$ is a square and and are the midpoints of and respectively. Then

(A) $\dfrac{\sqrt{5}}{5}$

(B) $\dfrac{3}{5}$

(C) $\dfrac{\sqrt{10}}{5}$

(D) $\dfrac{4}{5}$

(E) none of these

(1987 AHSME Problems/Problem 14)

Method 1: Area Formula

$$1.5 = \frac{1}{2}(\sqrt{5})(\sqrt{5}) \sin \theta$$

$$1.5 = 2.5 \sin \theta \implies \sin \theta = \frac{3}{5}$$

Method 2: Law of Cosines

$$\cos \theta = \frac{(\sqrt{5})^2 + (\sqrt{5})^2 - (\sqrt{2})^2}{2(\sqrt{5})(\sqrt{5})} = \frac{5 + 5 - 2}{10} = \frac{8}{10} = \frac{4}{5}$$

$$\sin \theta = \sqrt{1 - (4/5)^2} = \sqrt{9/25} = \frac{3}{5}$$

Method 3: Double Angle Identity

$$\sin(2\alpha) = 2 \sin \alpha \cos \alpha = 2 \left(\frac{1}{\sqrt{5}}\right) \left(\frac{2}{\sqrt{5}}\right) = \frac{4}{5}$$

$$\sin \theta = \frac{3}{5}$$

20 ) The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is

(A) $\dfrac{3}{4}$ (B) $\dfrac{7}{10}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{9}{14}$ (E) none of these

To find the cosine of the smallest angle in a triangle with consecutive integer sides $(n-1, n, n+1)$ where the largest angle $\alpha$ is twice the smallest angle $\beta$ ($\alpha = 2\beta$):

1. Identify Sides: The smallest side is $a = n-1$ (opposite $\beta$) and the largest side is $c = n+1$ (opposite $\alpha$).

2. Apply Law of Sines:

$$\frac{a}{\sin \beta} = \frac{c}{\sin 2\beta} \implies \frac{n-1}{\sin \beta} = \frac{n+1}{2 \sin \beta \cos \beta}$$

3. Solve for $\cos \beta$:

$$\cos \beta = \frac{n+1}{2(n-1)}$$

4. Apply Law of Cosines:

$$(n+1)^2 = (n-1)^2 + n^2 - 2n(n-1) \cos \beta$$

Substitute $\cos \beta$:

$$(n+1)^2 = (n-1)^2 + n^2 - 2n(n-1) \left[ \frac{n+1}{2(n-1)} \right]$$ $$(n+1)^2 = (n-1)^2 + n^2 - n(n+1)$$

5. Find $n$:$$n^2 + 2n + 1 = n^2 - 2n + 1 + n^2 - n^2 - n$$

6.

$$2n = -3n + n^2 \implies 5n = n^2 \implies n = 5$$

7. Calculate Final Value:

$$\cos \beta = \frac{5+1}{2(5-1)} = \frac{6}{8} = \frac{3}{4}$$

Answer: (A) $\dfrac{3}{4}$

23 ) Two rays with common endpoint $O$ form a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is

(A) $1$      (B) $\dfrac{1 + \sqrt{3}}{\sqrt{2}}$      (C) $\sqrt{3}$      (D) $2$      (E) $\dfrac{4}{\sqrt{3}}$

#Median_Formula #Apollonius_Theorem

Median Formula / Apollonius's Theorem

The length of the median $m_c$ to side $c$ is given by:

$$m_c = \frac{1}{2}\sqrt{2a^2 + 2b^2 - c^2}$$

Proof

Consider $\triangle ABC$ with median $AM$ to side $BC$ (where $BC=a$). Let $AM = m_a$.

1. Apply Law of Cosines to $\triangle ABM$ and $\triangle ACM$:

2. Simplify: Since $\cos(180^\circ - \theta) = -\cos\theta$, the second equation becomes:

3. Sum the equations:

4. Solve for $m_a$:

Or

Proof by Extending the Triangle

$$AB^2 + BD^2 + DC^2 + CA^2 = AD^2 + BC^2$$

$$c^2 + b^2 + c^2 + b^2 = (2m_a)^2 + a^2$$

$$2b^2 + 2c^2 = (2m_a)^2 + a^2$$

By further isolating $m_a$, you reach the standard median length formula:

$$m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$$

25 ) In $\triangle ABC$, we have $AB = 1$ and $AC = 2$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?

( 2002 AMC 12B Problems/Problem 23)

(A) $\dfrac{1 + \sqrt{2}}{2}$      (B) $\dfrac{1 + \sqrt{3}}{2}$      (C) $\sqrt{2}$      (D) $\dfrac{3}{2}$      (E) $\sqrt{3}$

4 ) Equilateral triangle $ABC$ has been creased and folded so that vertex $A$ now rests at $A'$ on $BC$. If $BA' = 1$ and $A'C = 2$, then what is the length of crease $PQ$?

(1991 AHSME Problems/Problem 29)

5 ) In quadrilateral $ABCD$, we have $\angle A = \angle C$, $AB = CD = 180$, and $AD \neq BC$. The perimeter of $ABCD$ is $640$. Find $\cos A$.

Solution for Quadrilateral $ABCD$

1. Set Up Variables:

2. Use Perimeter:

3. Apply Law of Cosines:

$180^2 + x^2 - 360x \cos A = 180^2 + y^2 - 360y \cos A$

4. Solve for $\cos A$:

$x + y = 360 \cos A$

$280 = 360 \cos A$

Answer: $\cos A = \dfrac{7}{9}$

6 ) Quadrilateral $ABCD$ is inscribed in a circle with sides $AB = 2$, $BC = 3$, $CD = 4$, and $DA = 6$. Find $AC$.

Solution

Since $ABCD$ is a cyclic quadrilateral, opposite angles sum to $180^\circ$. Let $\angle ABC = \theta$ and $\angle ADC = 180^\circ - \theta$.

Applying the Law of Cosines to $\triangle ABC$ and $\triangle ADC$:

1. In $\triangle ABC$:

$AC^2 = 2^2 + 3^2 - 2(2)(3)\cos \theta = 13 - 12\cos \theta$

2. In $\triangle ADC$:

$AC^2 = 6^2 + 4^2 - 2(6)(4)\cos(180^\circ - \theta) = 52 + 48\cos \theta$

(Note: $\cos(180^\circ - \theta) = -\cos \theta$)

Equating the two expressions for $AC^2$:

$13 - 12\cos \theta = 52 + 48\cos \theta$

$-39 = 60\cos \theta$

$\cos \theta = -\dfrac{39}{60} = -\dfrac{13}{20}$

Substitute $\cos \theta$ back into the first equation:

$AC^2 = 13 - 12\left(-\dfrac{13}{20}\right)$

$AC^2 = 13 + \dfrac{3 \cdot 13}{5} = 13 + \dfrac{39}{5}$

$AC^2 = \dfrac{65 + 39}{5} = \dfrac{104}{5}$

Final Answer:

$$AC = \sqrt{\frac{104}{5}} = \frac{2\sqrt{130}}{5}$$

8 ) Triangle $ABC$ has a right angle at $B$ and contains a point $P$ for which $PA=10$, $PB=6$ and $\angle APB = \angle BPC = \angle CPA = 120^\circ$. Find $PC$.

Solution

1. Analyze the Angles: Since the angles around point $P$ sum to $360^\circ$ and are equal, $\angle APB = \angle BPC = \angle CPA = 120^\circ$.

2. Law of Cosines in $\triangle APB$:

$$AB^2 = PA^2 + PB^2 - 2(PA)(PB)\cos(120^\circ)$$

$$AB^2 = 10^2 + 6^2 - 2(10)(6)(-0.5) = 100 + 36 + 60 = 196 \implies AB = 14$$

3. Law of Cosines in $\triangle BPC$: Let $PC = x$.

$$BC^2 = PB^2 + PC^2 - 2(PB)(PC)\cos(120^\circ)$$

$$BC^2 = 6^2 + x^2 - 2(6)(x)(-0.5) = 36 + x^2 + 6x$$

4. Law of Cosines in $\triangle CPA$:

$$AC^2 = PA^2 + PC^2 - 2(PA)(PC)\cos(120^\circ)$$

$$AC^2 = 10^2 + x^2 - 2(10)(x)(-0.5) = 100 + x^2 + 10x$$

5. Pythagorean Theorem: Since $\angle ABC = 90^\circ$, $AB^2 + BC^2 = AC^2$.

$$196 + (36 + x^2 + 6x) = 100 + x^2 + 10x$$

$$232 + x^2 + 6x = 100 + x^2 + 10x$$

6. Solve for $x$:

$$232 - 100 = 10x - 6x$$

$$132 = 4x \implies x = 33$$

Answer: $PC = 33$

16 ) In $\triangle ABC$, point $D$ lies on side $BC$. Given $\angle CAD = \angle DAB = 60^\circ$, $AC = 3$, and $AB = 6$, find the length of $AD$.

Solution

Let $AD = x$. The area of $\triangle ABC$ is the sum of the areas of $\triangle ACD$ and $\triangle ABD$.

Using the area formula $\text{Area} = \frac{1}{2}ab \sin C$:

Setting the sum equal to the total area:

$$\frac{3x\sqrt{3}}{4} + \frac{6x\sqrt{3}}{4} = \frac{9\sqrt{3}}{2}$$

$$\frac{9x\sqrt{3}}{4} = \frac{9\sqrt{3}}{2}$$

$$\frac{x}{4} = \frac{1}{2} \implies x = 2$$

Answer: $AD = 2$

3 ) Square $ABCD$ has center $O$ and $AB=900$. Points $E$ and $F$ are on $AB$ with $AE < BF$ and $E$ between $A$ and $F$ such that $\angle EOF = 45^\circ$ and $EF=400$. Find $BF$.

(2005 AIME II Problems/Problem 12)