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❌ Wrong Answers — Review These Problems
📐 Algebra 2 — Problems 1–10
Problem 1
Algebra 2
★★★ Hard
📚 Concept: Quadratic Formula & Discriminant
For \(ax^2 + bx + c = 0\), the solutions are \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
The discriminant \(D = b^2 - 4ac\) tells us the nature of roots:
• \(D > 0\): two distinct real roots • \(D = 0\): one repeated real root • \(D < 0\): two complex roots
The discriminant \(D = b^2 - 4ac\) tells us the nature of roots:
• \(D > 0\): two distinct real roots • \(D = 0\): one repeated real root • \(D < 0\): two complex roots
✏️ Example
Solve \(x^2 - 5x + 6 = 0\). Here \(a=1, b=-5, c=6\), so \(D = 25 - 24 = 1 > 0\).
\(x = \dfrac{5 \pm 1}{2}\) → \(x = 3\) or \(x = 2\). ✓
\(x = \dfrac{5 \pm 1}{2}\) → \(x = 3\) or \(x = 2\). ✓
🔷 Your Problem
For the equation \(2x^2 - 3x + k = 0\) to have exactly one real solution, what must be the value of \(k\)?
For exactly one real solution, the discriminant must equal zero: \(D = b^2 - 4ac = 0\).
Here \(a=2, b=-3, c=k\), so \((-3)^2 - 4(2)(k) = 0\) → \(9 - 8k = 0\) → \(k = \dfrac{9}{8}\). ✅ Answer: A
Here \(a=2, b=-3, c=k\), so \((-3)^2 - 4(2)(k) = 0\) → \(9 - 8k = 0\) → \(k = \dfrac{9}{8}\). ✅ Answer: A
Problem 2
Algebra 2
★★★ Hard
📚 Concept: Logarithm Laws
Key log rules: \(\log_b(xy) = \log_b x + \log_b y\), \(\log_b\!\left(\tfrac{x}{y}\right) = \log_b x - \log_b y\), \(\log_b(x^n) = n\log_b x\).
Change of base: \(\log_b x = \dfrac{\ln x}{\ln b}\). Also \(\log_b b = 1\) and \(\log_b 1 = 0\).
Change of base: \(\log_b x = \dfrac{\ln x}{\ln b}\). Also \(\log_b b = 1\) and \(\log_b 1 = 0\).
✏️ Example
Simplify \(\log_2 8 + \log_2 4\) = \(\log_2(8 \times 4)\) = \(\log_2 32\) = 5. ✓
🔷 Your Problem
If \(\log_3 x = 4\) and \(\log_3 y = -2\), what is the value of \(\log_3\!\left(\dfrac{x^2}{y}\right)\)?
\(\log_3\!\left(\dfrac{x^2}{y}\right) = 2\log_3 x - \log_3 y = 2(4) - (-2) = 8 + 2 = 10\). ✅ Answer: B
Problem 3
Algebra 2
★★★ Hard
📚 Concept: Complex Numbers
\(i = \sqrt{-1}\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\) (cycle of 4).
To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator: \((a+bi)(a-bi) = a^2+b^2\).
To divide complex numbers, multiply numerator and denominator by the conjugate of the denominator: \((a+bi)(a-bi) = a^2+b^2\).
✏️ Example
\(\dfrac{1}{1+i} = \dfrac{1-i}{(1+i)(1-i)} = \dfrac{1-i}{2} = \dfrac{1}{2} - \dfrac{1}{2}i\). ✓
🔷 Your Problem
What is the value of \(\dfrac{3+2i}{1-i}\) in the form \(a + bi\)?
Multiply by conjugate: \(\dfrac{(3+2i)(1+i)}{(1-i)(1+i)} = \dfrac{3+3i+2i+2i^2}{1+1} = \dfrac{3+5i-2}{2} = \dfrac{1+5i}{2} = \dfrac{1}{2} + \dfrac{5}{2}i\). ✅ Answer: A
Problem 4
Algebra 2
★★★ Hard
📚 Concept: Polynomial Remainder & Factor Theorem
Remainder Theorem: When \(p(x)\) is divided by \((x-a)\), the remainder is \(p(a)\).
Factor Theorem: \((x-a)\) is a factor of \(p(x)\) if and only if \(p(a)=0\).
Factor Theorem: \((x-a)\) is a factor of \(p(x)\) if and only if \(p(a)=0\).
✏️ Example
Is \((x-2)\) a factor of \(p(x)=x^3-3x+2\)? Check: \(p(2) = 8-6+2 = 4 \ne 0\). No. But \(p(1)=1-3+2=0\), so \((x-1)\) is a factor. ✓
🔷 Your Problem
The polynomial \(p(x) = x^3 + ax^2 - x + 6\) has \((x+3)\) as a factor. What is the value of \(a\)?
Since \((x+3)\) is a factor, \(p(-3)=0\).
\((-3)^3 + a(-3)^2 - (-3) + 6 = 0\)
\(-27 + 9a + 3 + 6 = 0\) → \(9a - 18 = 0\) → \(a = 2\). ✅ Answer: B
\((-3)^3 + a(-3)^2 - (-3) + 6 = 0\)
\(-27 + 9a + 3 + 6 = 0\) → \(9a - 18 = 0\) → \(a = 2\). ✅ Answer: B
Problem 5
Algebra 2
★★★ Hard
📚 Concept: Rational Exponents & Radical Equations
\(a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m\). When solving radical equations, isolate the radical first, then raise both sides to the appropriate power. Always check for extraneous solutions!
✏️ Example
Solve \(\sqrt{x+3} = x-1\). Square both sides: \(x+3 = x^2-2x+1\) → \(x^2-3x-2=0\)... wait, check solutions. Always verify! ✓
🔷 Your Problem
Solve: \(\sqrt{2x+5} - \sqrt{x-1} = 2\). Which of the following is a valid solution?
Let \(u=\sqrt{2x+5}\), \(v=\sqrt{x-1}\). Isolate: \(\sqrt{2x+5}=2+\sqrt{x-1}\). Square: \(2x+5=4+4\sqrt{x-1}+x-1\) → \(x+2=4\sqrt{x-1}\). Square again: \((x+2)^2=16(x-1)\) → \(x^2+4x+4=16x-16\) → \(x^2-12x+20=0\) → \((x-2)(x-10)=0\).
Check \(x=2\): \(\sqrt{9}-\sqrt{1}=3-1=2\) ✓. Check \(x=10\): \(\sqrt{25}-\sqrt{9}=5-3=2\) ✓.
Both work! But the question asks for one valid solution → \(x=2\). ✅ Answer: A
Check \(x=2\): \(\sqrt{9}-\sqrt{1}=3-1=2\) ✓. Check \(x=10\): \(\sqrt{25}-\sqrt{9}=5-3=2\) ✓.
Both work! But the question asks for one valid solution → \(x=2\). ✅ Answer: A
Problem 6
Algebra 2
★★★ Hard
📚 Concept: Conic Sections — Ellipses
Standard form: \(\dfrac{(x-h)^2}{a^2} + \dfrac{(y-k)^2}{b^2} = 1\). Center at \((h,k)\).
If \(a > b\): major axis is horizontal, \(c^2 = a^2 - b^2\). Foci at \((h \pm c, k)\).
If \(a > b\): major axis is horizontal, \(c^2 = a^2 - b^2\). Foci at \((h \pm c, k)\).
✏️ Example
For \(\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1\): \(a=5, b=3, c=\sqrt{25-9}=4\). Foci at \((\pm 4, 0)\). ✓
🔷 Your Problem
The ellipse \(\dfrac{(x-1)^2}{16} + \dfrac{(y+2)^2}{7} = 1\) has two foci. What is the distance between the two foci?
\(a^2=16, b^2=7\), so \(c^2 = a^2 - b^2 = 16-7 = 9\), giving \(c=3\).
Distance between the two foci = \(2c = 6\). ✅ Answer: B
Distance between the two foci = \(2c = 6\). ✅ Answer: B
Problem 7
Algebra 2
★★★ Hard
📚 Concept: Arithmetic & Geometric Sequences
Arithmetic: \(a_n = a_1 + (n-1)d\), \(S_n = \dfrac{n}{2}(a_1 + a_n)\).
Geometric: \(a_n = a_1 \cdot r^{n-1}\), \(S_n = a_1\dfrac{1-r^n}{1-r}\) (for \(r \ne 1\)).
Geometric: \(a_n = a_1 \cdot r^{n-1}\), \(S_n = a_1\dfrac{1-r^n}{1-r}\) (for \(r \ne 1\)).
✏️ Example
Geometric: \(a_1=2, r=3\). Find \(S_4 = 2\cdot\dfrac{1-3^4}{1-3} = 2\cdot\dfrac{-80}{-2} = 80\). ✓
🔷 Your Problem
In a geometric sequence, the 2nd term is 6 and the 5th term is 162. What is the sum of the first 4 terms?
\(a_2 = a_1 r = 6\) and \(a_5 = a_1 r^4 = 162\). Dividing: \(r^3 = 27\), so \(r = 3\).
Then \(a_1 = 6/3 = 2\). Sum of first 4 terms: \(S_4 = 2 \cdot \dfrac{1-3^4}{1-3} = 2 \cdot \dfrac{-80}{-2} = 80\). ✅ Answer: A
Then \(a_1 = 6/3 = 2\). Sum of first 4 terms: \(S_4 = 2 \cdot \dfrac{1-3^4}{1-3} = 2 \cdot \dfrac{-80}{-2} = 80\). ✅ Answer: A
Problem 8
Algebra 2
★★★ Hard
📚 Concept: Systems of Equations — Nonlinear
To solve a system with one linear and one quadratic equation, use substitution: express one variable from the linear equation, substitute into the quadratic, then solve. Intersection points of the graphs are the solutions.
✏️ Example
System: \(y = x^2\) and \(y = x+2\). Substituting: \(x^2 = x+2\) → \(x^2-x-2=0\) → \((x-2)(x+1)=0\). Solutions: \((2,4)\) and \((-1,1)\). ✓
🔷 Your Problem
How many intersection points do the graphs of \(y = x^2 - 4x + 5\) and \(y = 2x - 3\) have?
Set equal: \(x^2-4x+5 = 2x-3\) → \(x^2-6x+8=0\).
Discriminant: \(D = 36-32 = 4 > 0\). Two distinct real solutions → 2 intersection points. ✅ Answer: C
Discriminant: \(D = 36-32 = 4 > 0\). Two distinct real solutions → 2 intersection points. ✅ Answer: C
Problem 9
Algebra 2
★★★ Hard
📚 Concept: Exponential Growth & Decay / Inverse Functions
The inverse of \(f(x) = b^x\) is \(f^{-1}(x) = \log_b x\).
To find the inverse: (1) replace \(f(x)\) with \(y\), (2) swap \(x\) and \(y\), (3) solve for \(y\).
To find the inverse: (1) replace \(f(x)\) with \(y\), (2) swap \(x\) and \(y\), (3) solve for \(y\).
✏️ Example
Inverse of \(f(x)=3^x\): swap to get \(x=3^y\), so \(y=\log_3 x\), i.e. \(f^{-1}(x)=\log_3 x\). ✓
🔷 Your Problem
If \(f(x) = \log_2(x-3) + 1\), what is \(f^{-1}(5)\)?
Find inverse: let \(y = \log_2(x-3)+1\). Swap: \(x = \log_2(y-3)+1\) → \(x-1 = \log_2(y-3)\) → \(2^{x-1}=y-3\) → \(y = 2^{x-1}+3\).
So \(f^{-1}(x) = 2^{x-1}+3\). Plug in \(x=5\): \(f^{-1}(5) = 2^4+3 = 16+3 = 19\). ✅ Answer: A
So \(f^{-1}(x) = 2^{x-1}+3\). Plug in \(x=5\): \(f^{-1}(5) = 2^4+3 = 16+3 = 19\). ✅ Answer: A
Problem 10
Algebra 2
★★★ Hard
📚 Concept: Binomial Theorem
\((a+b)^n = \displaystyle\sum_{k=0}^{n} \binom{n}{k} a^{n-k}b^k\), where \(\binom{n}{k} = \dfrac{n!}{k!(n-k)!}\).
The \((r+1)\)th term is \(\binom{n}{r}a^{n-r}b^r\).
The \((r+1)\)th term is \(\binom{n}{r}a^{n-r}b^r\).
✏️ Example
In \((x+2)^4\), the 3rd term (k=2): \(\binom{4}{2}x^2 \cdot 2^2 = 6 \cdot 4 \cdot x^2 = 24x^2\). ✓
🔷 Your Problem
What is the coefficient of \(x^3\) in the expansion of \((2x - 3)^5\)?
The term with \(x^3\) has \(k=2\) (since \(a^{n-k} = (2x)^3\) means \(n-k=3\), \(k=2\)).
\(\binom{5}{2}(2x)^3(-3)^2 = 10 \cdot 8x^3 \cdot 9 = 720x^3\).
Coefficient is \(\mathbf{720}\). ✅ Answer: B
\(\binom{5}{2}(2x)^3(-3)^2 = 10 \cdot 8x^3 \cdot 9 = 720x^3\).
Coefficient is \(\mathbf{720}\). ✅ Answer: B
📏 Geometry — Problems 11–20
Problem 11
Geometry
★★★ Hard
📚 Concept: Triangle Similarity & Proportions
Similar triangles (AA, SAS, SSS) have equal corresponding angles and proportional corresponding sides.
If \(\triangle ABC \sim \triangle DEF\), then \(\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}\).
If \(\triangle ABC \sim \triangle DEF\), then \(\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}\).
✏️ Example
If \(\triangle ABC \sim \triangle XYZ\) with \(AB=6, XY=9, BC=8\), then \(\dfrac{6}{9}=\dfrac{8}{YZ}\) → \(YZ=12\). ✓
🔷 Your Problem
In the figure, \(\overline{DE} \parallel \overline{BC}\) in triangle \(ABC\), with \(AD = 4\), \(DB = 6\), and \(BC = 15\). What is the length of \(DE\)?
Since \(DE \parallel BC\), triangles \(ADE\) and \(ABC\) are similar (AA similarity).
\(\dfrac{AD}{AB} = \dfrac{DE}{BC}\) → \(\dfrac{4}{10} = \dfrac{DE}{15}\) → \(DE = 6\). ✅ Answer: A
\(\dfrac{AD}{AB} = \dfrac{DE}{BC}\) → \(\dfrac{4}{10} = \dfrac{DE}{15}\) → \(DE = 6\). ✅ Answer: A
Problem 12
Geometry
★★★ Hard
📚 Concept: Circle Theorems — Inscribed Angles & Arcs
Inscribed Angle Theorem: An inscribed angle is half the central angle that subtends the same arc.
Inscribed angle = \(\dfrac{1}{2}\) × intercepted arc.
An inscribed angle in a semicircle = 90°.
Inscribed angle = \(\dfrac{1}{2}\) × intercepted arc.
An inscribed angle in a semicircle = 90°.
✏️ Example
If arc \(AB = 100°\), then the inscribed angle \(\angle ACB = 50°\). ✓
🔷 Your Problem
In a circle, chord \(AB\) and chord \(CD\) intersect inside the circle at point \(P\). If \(\overset{\frown}{AC} = 80°\) and \(\overset{\frown}{BD} = 40°\), what is the measure of \(\angle APD\)?
When two chords intersect inside a circle, the angle formed equals half the sum of the intercepted arcs.
\(\angle APD = \dfrac{1}{2}(\overset{\frown}{AC} + \overset{\frown}{BD}) = \dfrac{1}{2}(80° + 40°) = \dfrac{120°}{2} = 60°\). ✅ Answer: B
\(\angle APD = \dfrac{1}{2}(\overset{\frown}{AC} + \overset{\frown}{BD}) = \dfrac{1}{2}(80° + 40°) = \dfrac{120°}{2} = 60°\). ✅ Answer: B
Problem 13
Geometry
★★★ Hard
📚 Concept: Law of Cosines
\(c^2 = a^2 + b^2 - 2ab\cos C\), where \(C\) is the angle opposite side \(c\).
Used when given: SAS (two sides and included angle) or SSS (all three sides).
Used when given: SAS (two sides and included angle) or SSS (all three sides).
✏️ Example
Triangle with \(a=5, b=7, C=60°\): \(c^2 = 25+49-2(35)(0.5) = 74-35=39\), so \(c=\sqrt{39}\). ✓
🔷 Your Problem
In \(\triangle ABC\), \(AB = 8\), \(AC = 6\), and \(\angle A = 120°\). What is the length of \(BC\)?
\(BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos A\)
\(= 64 + 36 - 2(8)(6)\cos 120° = 100 - 96 \cdot (-\tfrac{1}{2}) = 100 + 48 = 148\)
\(BC = \sqrt{148} = 2\sqrt{37}\). ✅ Answer: B (Note: A and B are equivalent; B is simplified form.)
\(= 64 + 36 - 2(8)(6)\cos 120° = 100 - 96 \cdot (-\tfrac{1}{2}) = 100 + 48 = 148\)
\(BC = \sqrt{148} = 2\sqrt{37}\). ✅ Answer: B (Note: A and B are equivalent; B is simplified form.)
Problem 14
Geometry
★★★ Hard
📚 Concept: Tangent Lines to Circles
A tangent to a circle is perpendicular to the radius at the point of tangency.
Power of a Point: If a tangent from an external point has length \(t\) and a secant has external segment \(e\) and full length \(s\), then \(t^2 = e \cdot s\).
Power of a Point: If a tangent from an external point has length \(t\) and a secant has external segment \(e\) and full length \(s\), then \(t^2 = e \cdot s\).
✏️ Example
External point \(P\), tangent length \(=6\), secant external part \(=3\). Then \(36 = 3 \cdot s\), so \(s = 12\). ✓
🔷 Your Problem
From external point \(P\), a tangent segment touches the circle at \(T\) with \(PT = 12\). A secant from \(P\) passes through the circle, with the nearer intersection at distance 6 from \(P\). What is the distance from \(P\) to the farther intersection?
Power of a point: \(PT^2 = \text{(external)} \times \text{(full secant)}\)
\(12^2 = 6 \times s\) → \(144 = 6s\) → \(s = 24\). ✅ Answer: B
\(12^2 = 6 \times s\) → \(144 = 6s\) → \(s = 24\). ✅ Answer: B
Problem 15
Geometry
★★★ Hard
📚 Concept: Coordinate Geometry — Distance & Midpoint
Distance: \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
Midpoint: \(M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)\).
Perpendicular bisector passes through the midpoint and has slope \(-\dfrac{1}{m}\) where \(m\) is the slope of the segment.
Midpoint: \(M = \left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)\).
Perpendicular bisector passes through the midpoint and has slope \(-\dfrac{1}{m}\) where \(m\) is the slope of the segment.
✏️ Example
Midpoint of \((2,4)\) and \((6,10)\) is \((4,7)\). Slope of segment = \(\tfrac{10-4}{6-2}=\tfrac{3}{2}\). Perpendicular slope = \(-\tfrac{2}{3}\). ✓
🔷 Your Problem
Points \(A(1, 3)\) and \(B(5, 7)\) are endpoints of a diameter of a circle. What is the equation of the circle?
Center = midpoint of diameter = \(\left(\dfrac{1+5}{2}, \dfrac{3+7}{2}\right) = (3,5)\).
Radius = half the diameter: \(d = \sqrt{(5-1)^2+(7-3)^2} = \sqrt{32} = 4\sqrt{2}\), so \(r = 2\sqrt{2}\), \(r^2 = 8\).
Equation: \((x-3)^2+(y-5)^2=8\). ✅ Answer: A
Radius = half the diameter: \(d = \sqrt{(5-1)^2+(7-3)^2} = \sqrt{32} = 4\sqrt{2}\), so \(r = 2\sqrt{2}\), \(r^2 = 8\).
Equation: \((x-3)^2+(y-5)^2=8\). ✅ Answer: A
Problem 16
Geometry
★★★ Hard
📚 Concept: Area & Volume of 3D Solids
Cone: \(V = \dfrac{1}{3}\pi r^2 h\), Lateral Surface Area = \(\pi r l\) (where \(l=\sqrt{r^2+h^2}\) is slant height).
Cylinder: \(V = \pi r^2 h\), Total SA = \(2\pi r^2 + 2\pi r h\).
Cylinder: \(V = \pi r^2 h\), Total SA = \(2\pi r^2 + 2\pi r h\).
✏️ Example
Cone with \(r=3, h=4\): \(l=\sqrt{9+16}=5\). Volume = \(\dfrac{1}{3}\pi(9)(4)=12\pi\). Lateral SA = \(\pi(3)(5)=15\pi\). ✓
🔷 Your Problem
A cone has a radius of 5 and a height of 12. What is its total surface area (including the base)?
Slant height: \(l = \sqrt{5^2+12^2} = \sqrt{25+144} = \sqrt{169} = 13\).
Lateral SA = \(\pi r l = \pi(5)(13) = 65\pi\).
Base area = \(\pi r^2 = 25\pi\).
Total SA = \(65\pi + 25\pi = 90\pi\). ✅ Answer: B
Lateral SA = \(\pi r l = \pi(5)(13) = 65\pi\).
Base area = \(\pi r^2 = 25\pi\).
Total SA = \(65\pi + 25\pi = 90\pi\). ✅ Answer: B
Problem 17
Geometry
★★★ Hard
📚 Concept: Angle Bisector Theorem
If the bisector of angle \(A\) in \(\triangle ABC\) meets \(BC\) at point \(D\), then:
\(\dfrac{BD}{DC} = \dfrac{AB}{AC}\)
This divides the opposite side in the ratio of the adjacent sides.
\(\dfrac{BD}{DC} = \dfrac{AB}{AC}\)
This divides the opposite side in the ratio of the adjacent sides.
✏️ Example
In \(\triangle ABC\), \(AB=4, AC=6, BC=10\). Angle bisector from \(A\) meets \(BC\) at \(D\).
\(\dfrac{BD}{DC}=\dfrac{4}{6}=\dfrac{2}{3}\), so \(BD=4, DC=6\). ✓
\(\dfrac{BD}{DC}=\dfrac{4}{6}=\dfrac{2}{3}\), so \(BD=4, DC=6\). ✓
🔷 Your Problem
In \(\triangle ABC\), \(AB = 9\), \(AC = 12\), and \(BC = 14\). The angle bisector from \(A\) meets \(BC\) at \(D\). What is the length of \(BD\)?
By the Angle Bisector Theorem: \(\dfrac{BD}{DC} = \dfrac{AB}{AC} = \dfrac{9}{12} = \dfrac{3}{4}\).
So \(BD = \dfrac{3}{3+4} \times 14 = \dfrac{3}{7} \times 14 = 6\). ✅ Answer: A
So \(BD = \dfrac{3}{3+4} \times 14 = \dfrac{3}{7} \times 14 = 6\). ✅ Answer: A
Problem 18
Geometry
★★★ Hard
📚 Concept: Geometric Proofs — Parallel Lines & Transversals
When a transversal crosses two parallel lines:
• Alternate interior angles are equal • Corresponding angles are equal
• Co-interior (same-side interior) angles are supplementary (sum = 180°)
• Alternate interior angles are equal • Corresponding angles are equal
• Co-interior (same-side interior) angles are supplementary (sum = 180°)
✏️ Example
Two parallel lines cut by a transversal. One co-interior angle = 70°. The other = 180° − 70° = 110°. ✓
🔷 Your Problem
Lines \(l \parallel m\) are cut by a transversal. One alternate interior angle measures \((3x+15)°\) and the other measures \((5x-9)°\). What is the value of \(x\)?
Alternate interior angles are equal, so: \(3x+15 = 5x-9\)
\(24 = 2x\) → \(x = 12\). ✅ Answer: B
\(24 = 2x\) → \(x = 12\). ✅ Answer: B
Problem 19
Geometry
★★★ Hard
📚 Concept: Trigonometric Ratios in Right Triangles
\(\sin\theta = \dfrac{\text{opp}}{\text{hyp}}\), \(\cos\theta = \dfrac{\text{adj}}{\text{hyp}}\), \(\tan\theta = \dfrac{\text{opp}}{\text{adj}}\).
Special values: \(\sin 30°=\tfrac{1}{2}\), \(\cos 30°=\tfrac{\sqrt{3}}{2}\), \(\tan 45°=1\), \(\sin 60°=\tfrac{\sqrt{3}}{2}\).
Special values: \(\sin 30°=\tfrac{1}{2}\), \(\cos 30°=\tfrac{\sqrt{3}}{2}\), \(\tan 45°=1\), \(\sin 60°=\tfrac{\sqrt{3}}{2}\).
✏️ Example
In a right triangle with hypotenuse 10 and \(\theta=30°\): adjacent = \(10\cos30°=5\sqrt{3}\), opposite = \(10\sin30°=5\). ✓
🔷 Your Problem
A 20-foot ladder leans against a wall. The base of the ladder makes a \(60°\) angle with the ground. How high up the wall does the ladder reach?
The height up the wall is the side opposite the 60° angle. Using sine:
\(h = 20 \sin 60° = 20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\) feet. ✅ Answer: A
\(h = 20 \sin 60° = 20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3}\) feet. ✅ Answer: A
Problem 20
Geometry
★★★ Hard
📚 Concept: Area of a Triangle — Heron's Formula
Given sides \(a, b, c\) and semi-perimeter \(s = \dfrac{a+b+c}{2}\):
\(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\)
Also: \(\text{Area} = \dfrac{1}{2}ab\sin C\) (using two sides and included angle).
\(\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\)
Also: \(\text{Area} = \dfrac{1}{2}ab\sin C\) (using two sides and included angle).
✏️ Example
Triangle with \(a=3, b=4, c=5\): \(s=6\). Area = \(\sqrt{6(3)(2)(1)}=\sqrt{36}=6\). ✓
🔷 Your Problem
In \(\triangle ABC\), \(AB = 7\), \(AC = 10\), and \(\angle A = 30°\). What is the area of the triangle?
Area = \(\dfrac{1}{2} \cdot AB \cdot AC \cdot \sin A = \dfrac{1}{2}(7)(10)\sin 30° = \dfrac{1}{2}(70)\left(\dfrac{1}{2}\right) = \dfrac{70}{4} = 17.5\). ✅ Answer: A