Math Master — Algebra 2 & Geometry

A1

Quadratic Functions — Vertex Form

🧠 Key Concept

The vertex form of a quadratic is $f(x) = a(x-h)^2 + k$, where $(h, k)$ is the vertex.
• If $a > 0$: parabola opens up, vertex is a minimum.
• If $a < 0$: parabola opens down, vertex is a maximum.
• The axis of symmetry is $x = h$.

✏️ Worked Example

Find the vertex of $f(x) = 2(x-3)^2 - 5$.
Here $h = 3$ and $k = -5$, so the vertex is $(3, -5)$. Since $a = 2 > 0$, the parabola opens up and $(3,-5)$ is a minimum.

Consider the quadratic function $f(x) = -3(x+2)^2 + 7$. Answer the following three questions.

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Step 1 of 3

What is the vertex of $f(x) = -3(x+2)^2 + 7$?

📖 Explanation

In $f(x)=a(x-h)^2+k$, we rewrite $-3(x+2)^2+7$ as $-3(x-(-2))^2+7$. So $h=-2$, $k=7$. Vertex = $\mathbf{(-2,\,7)}$.

Step 2 of 3

Does $f(x)$ have a maximum or minimum value?

📖 Explanation

$a = -3 < 0$ → parabola opens downward. The vertex gives the maximum value. Maximum value = $k = 7$.

Step 3 of 3

What is the axis of symmetry of $f(x)$?

📖 Explanation

The axis of symmetry is always $x = h$. Here $h = -2$, so axis of symmetry: $\boxed{x = -2}$.

A2

Polynomial Long Division

🧠 Key Concept

Polynomial Division: Use long division or synthetic division to divide $p(x)$ by $(x-c)$.
Remainder Theorem: When $p(x)$ is divided by $(x-c)$, the remainder $= p(c)$.
Factor Theorem: $(x-c)$ is a factor of $p(x)$ iff $p(c) = 0$.

✏️ Worked Example

Find the remainder when $p(x) = x^3 - 4x + 5$ is divided by $(x-2)$.
By the Remainder Theorem: $p(2) = 8 - 8 + 5 = \mathbf{5}$.

Let $p(x) = 2x^3 - 3x^2 + x - 6$. Use the Remainder and Factor Theorems to answer the following.

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Step 1 of 3

What is the remainder when $p(x)$ is divided by $(x - 2)$?

📖 Explanation

$p(2) = 2(8) - 3(4) + 2 - 6 = 16 - 12 + 2 - 6 = 0$. Remainder = $\mathbf{0}$.

Step 2 of 3

Since the remainder is $0$, what can we conclude?

📖 Explanation

Factor Theorem: $p(2) = 0$ means $(x - 2)$ is a factor of $p(x)$.

Step 3 of 3

After dividing $p(x)$ by $(x-2)$, the quotient is:

📖 Explanation

Synthetic division with $c=2$, coefficients $[2, -3, 1, -6]$:
$\downarrow \quad 4 \quad 2 \quad 6$ → result: $[2,\,1,\,3,\,0]$ → Quotient $= 2x^2 + x + 3$.

A3

Exponential & Logarithmic Equations

🧠 Key Concept

Log-Exponential Inverse: $\log_b(x) = y \Leftrightarrow b^y = x$
Log Rules: $\log(ab) = \log a + \log b$; $\log(a/b) = \log a - \log b$; $\log(a^n) = n\log a$
Change of Base: $\log_b a = \dfrac{\ln a}{\ln b}$

✏️ Worked Example

Solve $3^{x+1} = 27$.
Write $27 = 3^3$, so $3^{x+1} = 3^3 \Rightarrow x + 1 = 3 \Rightarrow x = 2$.

Solve the following exponential and logarithmic equations.

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Step 1 of 3

Solve: $\log_2(x) = 5$

📖 Explanation

$\log_2(x)=5 \Rightarrow x = 2^5 = \mathbf{32}$.

Step 2 of 3

Solve: $2^{3x} = 64$

📖 Explanation

$64 = 2^6$, so $2^{3x}=2^6 \Rightarrow 3x=6 \Rightarrow x=\mathbf{2}$.

Step 3 of 3

Simplify: $\log_3 9 + \log_3 27$

📖 Explanation

$\log_3 9 = 2$ (since $3^2=9$); $\log_3 27 = 3$ (since $3^3=27$). Sum $= 2+3 = \mathbf{5}$.

A4

Rational Expressions & Equations

🧠 Key Concept

To solve rational equations, multiply both sides by the LCD to clear fractions.
⚠️ Always check for extraneous solutions — values that make any denominator $= 0$ must be excluded.

✏️ Worked Example

Solve $\dfrac{1}{x} + \dfrac{1}{2} = \dfrac{3}{4}$. LCD = $4x$. Multiply: $4 + 2x = 3x \Rightarrow x = 4$.

Solve or simplify the following rational expressions.

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Step 1 of 3

Simplify: $\dfrac{x^2 - 4}{x^2 - x - 2}$

📖 Explanation

$\dfrac{(x-2)(x+2)}{(x-2)(x+1)} = \dfrac{x+2}{x+1}$, $x \neq 2$.

Step 2 of 3

Solve: $\dfrac{3}{x-1} = \dfrac{6}{x+1}$

📖 Explanation

Cross-multiply: $3(x+1)=6(x-1) \Rightarrow 3x+3=6x-6 \Rightarrow 9=3x \Rightarrow x=\mathbf{3}$.

Step 3 of 3

What value of $x$ must be excluded from $\dfrac{2}{x^2-9}$?

📖 Explanation

$x^2-9 = (x-3)(x+3)=0 \Rightarrow x = \pm 3$. Both must be excluded.

A5

Complex Numbers

🧠 Key Concept

$i = \sqrt{-1}$, so $i^2 = -1$, $i^3 = -i$, $i^4 = 1$ (cycle of 4).
Complex number: $a + bi$. To divide, multiply by conjugate: $\dfrac{a+bi}{c+di} \cdot \dfrac{c-di}{c-di}$.

✏️ Worked Example

$(3+2i)(1-i) = 3 - 3i + 2i - 2i^2 = 3 - i - 2(-1) = 5 - i$.

Perform the indicated operations with complex numbers.

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Step 1 of 3

Simplify: $i^{23}$

📖 Explanation

$23 \div 4 = 5$ remainder $3$. So $i^{23} = i^3 = \mathbf{-i}$.

Step 2 of 3

Simplify: $(2+3i) + (5-i)$

📖 Explanation

$(2+5)+(3i-i) = \mathbf{7+2i}$.

Step 3 of 3

Simplify: $(1+i)^2$

📖 Explanation

$(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = \mathbf{2i}$.

A6

Arithmetic & Geometric Sequences

🧠 Key Concept

Arithmetic: $a_n = a_1 + (n-1)d$ | $S_n = \dfrac{n}{2}(a_1 + a_n)$
Geometric: $a_n = a_1 \cdot r^{n-1}$ | $S_n = \dfrac{a_1(1-r^n)}{1-r}$ ($r \neq 1$)

✏️ Worked Example

Arithmetic seq: $a_1=3, d=4$. Find $a_{10}$: $a_{10}=3+(9)(4)=3+36=39$.

Answer the following sequence questions.

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Step 1 of 3

In the arithmetic sequence $5, 11, 17, \ldots$, find $a_{20}$.

📖 Explanation

$d=6$, $a_{20}=5+19(6)=5+114=\mathbf{119}$.

Step 2 of 3

Find the 6th term of the geometric sequence: $2, 6, 18, \ldots$

📖 Explanation

$r=3$, $a_6=2\cdot 3^5=2\cdot 243=\mathbf{486}$.

Step 3 of 3

Sum of the first 5 terms of arithmetic seq: $a_1=10, d=-2$

📖 Explanation

$a_5=10+4(-2)=2$. $S_5=\frac{5}{2}(10+2)=\frac{5}{2}(12)=\mathbf{30}$.

A7

Systems of Equations (3 Variables)

🧠 Key Concept

To solve a system of 3 equations in 3 unknowns, use elimination or substitution:
1. Eliminate one variable from two pairs of equations → get 2 equations in 2 unknowns.
2. Solve the 2×2 system, then back-substitute.

✏️ Worked Example

If $x+y=5$ and $y+z=7$ and $x+z=4$, add all three: $2(x+y+z)=16 \Rightarrow x+y+z=8$.

Solve: $x + y = 5$, $\;y + z = 7$, $\;x + z = 4$.

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Step 1 of 3

What is $x + y + z$?

📖 Explanation

Add all equations: $2(x+y+z)=16 \Rightarrow x+y+z=\mathbf{8}$.

Step 2 of 3

Find $z$. (Use $x+y+z=8$ and $x+y=5$.)

📖 Explanation

$z = 8-5=\mathbf{3}$.

Step 3 of 3

Find $x$.

📖 Explanation

$x+z=4$, $z=3 \Rightarrow x=\mathbf{1}$. Check: $y=5-1=4$. $y+z=7$ ✓.

A8

Radical Equations & Expressions

🧠 Key Concept

To solve radical equations: isolate the radical, then square both sides. Always check for extraneous solutions!
$\sqrt[n]{a^m} = a^{m/n}$ and $a^{m/n} = (\sqrt[n]{a})^m$.

✏️ Worked Example

Solve $\sqrt{x+3} = 4$. Square both sides: $x+3=16 \Rightarrow x=13$. Check: $\sqrt{16}=4$ ✓.

Solve the radical equations below.

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Step 1 of 3

Solve: $\sqrt{2x-1} = 5$

📖 Explanation

$2x-1=25 \Rightarrow 2x=26 \Rightarrow x=\mathbf{13}$.

Step 2 of 3

Simplify: $\sqrt{50}$

📖 Explanation

$\sqrt{50}=\sqrt{25\cdot2}=5\sqrt{2}$.

Step 3 of 3

Solve: $\sqrt{x} = x - 6$. Which value is an extraneous solution?

📖 Explanation

Squaring: $x = (x-6)^2 \Rightarrow x^2-13x+36=0 \Rightarrow x=9$ or $x=4$. Check $x=4$: $\sqrt{4}=2 \neq -2$. So $x=4$ is extraneous. Only $x=9$ is valid.

A9

Inverse & Composition of Functions

🧠 Key Concept

Composition: $(f \circ g)(x) = f(g(x))$ — substitute $g(x)$ into $f$.
Inverse: To find $f^{-1}(x)$: replace $f(x)$ with $y$, swap $x$ and $y$, then solve for $y$.

✏️ Worked Example

If $f(x)=2x+1$ find $f^{-1}(x)$: swap → $x=2y+1 \Rightarrow y=\frac{x-1}{2}$. So $f^{-1}(x)=\frac{x-1}{2}$.

Let $f(x) = 3x - 2$ and $g(x) = x^2 + 1$. Answer the following.

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Step 1 of 3

Find $(f \circ g)(2)$.

📖 Explanation

$g(2)=4+1=5$. $f(5)=3(5)-2=\mathbf{13}$.

Step 2 of 3

Find $f^{-1}(x)$ for $f(x) = 3x - 2$.

📖 Explanation

Swap $x$ and $y$: $x=3y-2 \Rightarrow y=\frac{x+2}{3}$. So $f^{-1}(x)=\mathbf{\dfrac{x+2}{3}}$.

Step 3 of 3

Find $(g \circ f)(1)$.

📖 Explanation

$f(1)=1$. $g(1)=1+1=\mathbf{2}$.

A10

Discriminant & Nature of Roots

🧠 Key Concept

For $ax^2+bx+c=0$, the discriminant $\Delta = b^2-4ac$:
• $\Delta > 0$: two distinct real roots
• $\Delta = 0$: one repeated real root
• $\Delta < 0$: two complex (non-real) conjugate roots

✏️ Worked Example

$x^2-4x+4=0$: $\Delta=16-16=0$ → one repeated root $x=2$.

Determine the nature of the roots using the discriminant.

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Step 1 of 3

For $x^2 - 5x + 6 = 0$, find the discriminant.

📖 Explanation

$\Delta = (-5)^2-4(1)(6)=25-24=\mathbf{1}$.

Step 2 of 3

Since $\Delta = 1 > 0$, what is the nature of the roots?

📖 Explanation

$\Delta = 1 > 0$ → two distinct real roots. (They are $x=2$ and $x=3$.)

Step 3 of 3

For $2x^2 + 3x + 5 = 0$, what type of roots does it have?

📖 Explanation

$\Delta=9-40=-31 < 0$ → two complex (non-real) roots.

G1

Triangle Congruence & Similarity

🧠 Key Concept

Congruence Postulates: SSS, SAS, ASA, AAS, HL (right triangles)
Similarity Theorems: AA, SSS~, SAS~ — corresponding sides are proportional.
If $\triangle ABC \sim \triangle DEF$ with ratio $k$, then $\text{Area ratio} = k^2$.

✏️ Worked Example

$\triangle ABC \sim \triangle DEF$ with $AB=6$, $DE=9$. Scale factor $= \frac{9}{6} = \frac{3}{2}$. Area ratio $= \left(\frac{3}{2}\right)^2 = \frac{9}{4}$.

Answer the following triangle questions.

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Step 1 of 3

Two triangles have sides $3, 4, 5$ and $6, 8, 10$. Are they similar?

📖 Explanation

$\frac{6}{3}=\frac{8}{4}=\frac{10}{5}=2$. All ratios equal → SSS Similarity with scale factor 2.

Step 2 of 3

If the scale factor is 2, what is the ratio of their areas?

📖 Explanation

Area ratio $= k^2 = 2^2 = \mathbf{4:1}$.

Step 3 of 3

Which postulate proves two triangles congruent if two sides and the included angle are equal?

📖 Explanation

SAS — two Sides and the included Angle are congruent.

G2

Pythagorean Theorem & Special Triangles

🧠 Key Concept

$a^2 + b^2 = c^2$ (right triangle, $c$ = hypotenuse)
30-60-90: sides $= x : x\sqrt{3} : 2x$
45-45-90: sides $= x : x : x\sqrt{2}$

✏️ Worked Example

45-45-90 triangle with legs $= 5$: hypotenuse $= 5\sqrt{2}$.

Apply the Pythagorean Theorem and special triangle ratios.

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Step 1 of 3

A right triangle has legs $6$ and $8$. Find the hypotenuse.

📖 Explanation

$c=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=\mathbf{10}$.

Step 2 of 3

In a 30-60-90 triangle, the shorter leg is $5$. What is the hypotenuse?

📖 Explanation

30-60-90 ratio: $x : x\sqrt{3} : 2x$. Hypotenuse $= 2(5) = \mathbf{10}$.

Step 3 of 3

A 45-45-90 triangle has hypotenuse $8\sqrt{2}$. Find each leg.

📖 Explanation

$x\sqrt{2}=8\sqrt{2} \Rightarrow x=\mathbf{8}$. Each leg $= 8$.

G3

Circle Theorems — Arcs & Angles

🧠 Key Concept

• Central angle $=$ intercepted arc.
• Inscribed angle $= \frac{1}{2} \times$ intercepted arc.
• Tangent-Chord angle $= \frac{1}{2} \times$ intercepted arc.
• Two chords inside: angle $= \frac{1}{2}(arc_1 + arc_2)$.

✏️ Worked Example

Inscribed angle intercepts arc of $80°$. Angle $= \frac{80}{2} = 40°$.

Apply circle theorems to find missing angles.

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Step 1 of 3

A central angle intercepts an arc of $130°$. What is the central angle?

📖 Explanation

Central angle $=$ intercepted arc $= \mathbf{130°}$.

Step 2 of 3

An inscribed angle intercepts an arc of $100°$. Find the inscribed angle.

📖 Explanation

Inscribed angle $= \frac{1}{2}(100°) = \mathbf{50°}$.

Step 3 of 3

Two chords intersect inside a circle. They intercept arcs of $80°$ and $60°$. Find the angle formed.

📖 Explanation

Angle $= \frac{1}{2}(80°+60°) = \frac{140°}{2} = \mathbf{70°}$.

G4

Area & Volume of 3D Figures

🧠 Key Concept

Cylinder: $V = \pi r^2 h$, $SA = 2\pi r^2 + 2\pi rh$
Cone: $V = \frac{1}{3}\pi r^2 h$
Sphere: $V = \frac{4}{3}\pi r^3$, $SA = 4\pi r^2$
Pyramid: $V = \frac{1}{3}Bh$ where $B$ = base area

✏️ Worked Example

Cylinder: $r=3, h=5$. $V=\pi(9)(5)=45\pi \approx 141.4$ cubic units.

Calculate volumes of 3D figures. Leave answers in terms of $\pi$ where applicable.

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Step 1 of 3

A sphere has radius $6$. Find its volume.

📖 Explanation

$V=\frac{4}{3}\pi(6^3)=\frac{4}{3}\pi(216)=\mathbf{288\pi}$.

Step 2 of 3

A cone has $r = 4$ and $h = 9$. Find its volume.

📖 Explanation

$V=\frac{1}{3}\pi(4^2)(9)=\frac{1}{3}\pi(144)=\mathbf{48\pi}$.

Step 3 of 3

A square pyramid has base side $6$ and height $4$. Find its volume.

📖 Explanation

$B=6^2=36$. $V=\frac{1}{3}(36)(4)=\mathbf{48}$ cubic units.

G5

Coordinate Geometry — Lines & Distance

🧠 Key Concept

Slope: $m = \dfrac{y_2-y_1}{x_2-x_1}$
Distance: $d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Midpoint: $M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$
Parallel lines: same slope. Perpendicular: slopes are negative reciprocals.

✏️ Worked Example

Distance from $(1,2)$ to $(4,6)$: $d=\sqrt{9+16}=\sqrt{25}=5$.

Use coordinate geometry formulas to answer the following.

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Step 1 of 3

Find the midpoint of the segment joining $(-2, 5)$ and $(4, -1)$.

📖 Explanation

$M=\left(\frac{-2+4}{2},\frac{5+(-1)}{2}\right)=\left(1,\,2\right)$.

Step 2 of 3

A line has slope $\frac{2}{3}$. What is the slope of a perpendicular line?

📖 Explanation

Perpendicular slope $= -\frac{1}{m} = -\frac{3}{2}$.

Step 3 of 3

Find the distance between $(-3, 1)$ and $(1, 4)$.

📖 Explanation

$d=\sqrt{(1-(-3))^2+(4-1)^2}=\sqrt{16+9}=\sqrt{25}=\mathbf{5}$.

G6

Parallel Lines & Transversals

🧠 Key Concept

When a transversal crosses parallel lines:
• Alternate Interior Angles — equal
• Corresponding Angles — equal
• Co-interior (Same-Side Interior) — supplementary (sum $= 180°$)
• Vertical angles — equal

✏️ Worked Example

If $\angle 1 = 65°$ and $\angle 1$ and $\angle 2$ are alternate interior angles, then $\angle 2 = 65°$.

Two parallel lines are cut by a transversal. Use angle relationships to find missing angles.

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Step 1 of 3

One angle measures $110°$. Find its co-interior (same-side interior) angle.

📖 Explanation

Co-interior angles sum to $180°$: $180°-110°=\mathbf{70°}$.

Step 2 of 3

Corresponding angles: one is $(3x+10)°$ and the other is $(5x-20)°$. Find $x$.

📖 Explanation

$3x+10=5x-20 \Rightarrow 30=2x \Rightarrow x=\mathbf{15}$.

Step 3 of 3

Alternate interior angles: $(4x-5)°$ and $(3x+15)°$. Find the angle measure.

📖 Explanation

$4x-5=3x+15 \Rightarrow x=20$. Angle $= 4(20)-5 = \mathbf{75°}$.

G7

Trigonometry in Right Triangles

🧠 Key Concept

SOH-CAH-TOA:
$\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$, $\;\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$, $\;\tan\theta = \dfrac{\text{opp}}{\text{adj}}$
Reciprocals: $\csc = \frac{1}{\sin}$, $\sec = \frac{1}{\cos}$, $\cot = \frac{1}{\tan}$

✏️ Worked Example

Right triangle: opp $= 5$, hyp $= 13$. $\sin\theta = \frac{5}{13}$, $\cos\theta = \frac{12}{13}$, $\tan\theta = \frac{5}{12}$.

Use right triangle trigonometry (SOH-CAH-TOA) to answer the following.

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Step 1 of 3

In a right triangle: opp $= 7$, adj $= 24$, hyp $= 25$. Find $\sin\theta$.

📖 Explanation

$\sin\theta = \frac{\text{opp}}{\text{hyp}} = \frac{7}{25}$.

Step 2 of 3

Using the same triangle, find $\tan\theta$.

📖 Explanation

$\tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{7}{24}$.

Step 3 of 3

A ladder leans against a wall. The angle with the ground is $60°$ and the ladder is $10$ ft long. How high does it reach?

📖 Explanation

$\sin 60° = \frac{\text{height}}{10} \Rightarrow \text{height} = 10\cdot\frac{\sqrt{3}}{2} = \mathbf{5\sqrt{3}}$ ft.

G8

Transformations — Reflections & Rotations

🧠 Key Concept

Reflection:
• Over $x$-axis: $(x,y) \to (x,-y)$
• Over $y$-axis: $(x,y) \to (-x,y)$
• Over $y=x$: $(x,y) \to (y,x)$
Rotation about origin:
• $90°$ CCW: $(x,y)\to(-y,x)$ | $180°$: $(x,y)\to(-x,-y)$

✏️ Worked Example

Reflect $(3,-2)$ over the $y$-axis: $\to(-3,-2)$.

Apply transformation rules to find image coordinates.

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Step 1 of 3

Reflect the point $(4, -3)$ over the $x$-axis.

📖 Explanation

Reflection over $x$-axis: $(x,y)\to(x,-y)$. $(4,-3)\to\mathbf{(4,3)}$.

Step 2 of 3

Rotate $(-2, 5)$ by $90°$ counterclockwise about the origin.

📖 Explanation

$90°$ CCW: $(x,y)\to(-y,x)$. $(-2,5)\to\mathbf{(-5,-2)}$.

Step 3 of 3

Reflect $(3, 7)$ over the line $y = x$.

📖 Explanation

Over $y=x$: $(x,y)\to(y,x)$. $(3,7)\to\mathbf{(7,3)}$.

G9

Polygon Interior & Exterior Angles

🧠 Key Concept

Interior angles sum of $n$-gon: $(n-2) \times 180°$
Each interior angle (regular): $\dfrac{(n-2)\times180°}{n}$
Each exterior angle (regular): $\dfrac{360°}{n}$
Interior + Exterior at each vertex $= 180°$.

✏️ Worked Example

Hexagon ($n=6$): Interior sum $= (6-2)(180°) = 720°$. Each interior $= 120°$. Each exterior $= 60°$.

Answer the following polygon angle questions.

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Step 1 of 3

What is the sum of interior angles of a nonagon (9-gon)?

📖 Explanation

$(9-2)\times180° = 7\times180° = \mathbf{1260°}$.

Step 2 of 3

Each exterior angle of a regular polygon is $24°$. How many sides does it have?

📖 Explanation

$n = \frac{360°}{24°} = \mathbf{15}$ sides.

Step 3 of 3

Each interior angle of a regular polygon is $150°$. How many sides?

📖 Explanation

Exterior $= 180°-150°=30°$. $n=\frac{360°}{30°}=\mathbf{12}$ sides.

G10

Law of Sines & Law of Cosines

🧠 Key Concept

Law of Sines: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$
Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$
Use Law of Sines for AAS/ASA. Use Law of Cosines for SAS/SSS.

✏️ Worked Example

$a=7, b=10, C=60°$. $c^2=49+100-2(7)(10)(0.5)=149-70=79$. $c=\sqrt{79}$.

Apply the Law of Sines or Law of Cosines as appropriate.

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Step 1 of 3

In $\triangle ABC$: $A=30°, B=45°, a=8$. Find $b$ using the Law of Sines.

📖 Explanation

$\frac{b}{\sin 45°}=\frac{8}{\sin 30°} \Rightarrow b=\frac{8\cdot\frac{\sqrt{2}}{2}}{\frac{1}{2}}=\frac{4\sqrt{2}}{\frac{1}{2}}=\mathbf{8\sqrt{2}}$.

Step 2 of 3

$a=5, b=7, C=60°$. Find $c^2$ using Law of Cosines.

📖 Explanation

$c^2=25+49-2(5)(7)\cos60°=74-35=\mathbf{39}$.

Step 3 of 3

When should you use the Law of Cosines instead of the Law of Sines?

📖 Explanation

Use Law of Cosines for SAS (two sides + included angle) and SSS (all three sides known).

Algebra 2 & GeometryPractice Exam

Algebra 2 & Geometry
Practice Exam