Key Formulas & Theorems
- Pythagorean Theorem: $$a^2 + b^2 = c^2$$
- Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$
- Law of Cosines: $$c^2 = a^2 + b^2 - 2ab\cos C$$
- Triangle Area (Heron's): $$A = \sqrt{s(s-a)(s-b)(s-c)}, \quad s = \frac{a+b+c}{2}$$
- Pythagorean Identity: $$\sin^2\theta + \cos^2\theta = 1$$
- Double-Angle: $$\sin 2\theta = 2\sin\theta\cos\theta, \quad \cos 2\theta = \cos^2\theta - \sin^2\theta$$
- Inscribed Angle Theorem: An inscribed angle is half the central angle subtending the same arc.
- Circle Area: $$A = \pi r^2, \quad \text{Circumference} = 2\pi r$$
- Coordinate Distance: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$
In triangle \(ABC\), sides \(a = 7\), \(b = 8\), and angle \(C = 60°\). Find side \(c\) and the area of the triangle.
[Diagram: Triangle with vertices A (top), B (bottom-left), C (bottom-right). Side \(a = 7\) opposite A, side \(b = 8\) opposite B, angle \(C = 60°\) at vertex C.]
Show Hint
Use the Law of Cosines: \(c^2 = 49 + 64 - 2(7)(8)\cos 60° = 113 - 56 = 57\). For area use \(\frac{1}{2}ab\sin C\).
A chord of length 10 is drawn in a circle of radius 13. Find the distance from the center of the circle to the chord.
[Diagram: Circle with center O. Horizontal chord AB of length 10. Perpendicular from O meets chord at midpoint M. OM is the unknown distance.]
Show Hint
The perpendicular from center to chord bisects the chord. You get a right triangle with hypotenuse 13 and one leg 5. Use the Pythagorean theorem.
Prove that \(\dfrac{1 + \tan^2\theta}{1 + \cot^2\theta} = \tan^2\theta\) for all \(\theta\) where defined.
Show Hint
Use the identities \(1 + \tan^2\theta = \sec^2\theta\) and \(1 + \cot^2\theta = \csc^2\theta\). Then \(\frac{\sec^2\theta}{\csc^2\theta} = \frac{1/\cos^2\theta}{1/\sin^2\theta} = \frac{\sin^2\theta}{\cos^2\theta}\).
Find the equation of the circle passing through points \(A(1, 2)\), \(B(3, 4)\), and \(C(5, 2)\).
[Diagram: Three points forming an isosceles triangle. A at (1,2), B at (3,4) top, C at (5,2). A circle passes through all three.]
Show Hint
Use the general equation \(x^2 + y^2 + Dx + Ey + F = 0\). Substitute each point to get 3 equations in \(D, E, F\). Alternatively, note the symmetry: center lies on \(x = 3\).
Find the area of a triangle with sides 13, 14, and 15.
Show Hint
\(s = \frac{13+14+15}{2} = 21\). Area \(= \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056}\). Simplify.
In a circle, chord \(AB\) subtends an inscribed angle of \(35°\) at point \(C\) on the major arc. What is the inscribed angle subtended at point \(D\) on the minor arc?
[Diagram: Circle with chord AB. Point C on major arc, \(\angle ACB = 35°\). Point D on minor arc, \(\angle ADB = ?\)]
Show Hint
An inscribed angle on the major arc and one on the minor arc subtending the same chord are supplementary. So the angle at \(D = 180° - 35°\).
Find all solutions in \([0, 2\pi)\) for \(2\sin^2 x - 3\sin x + 1 = 0\).
Show Hint
Let \(u = \sin x\). Factor: \((2u - 1)(u - 1) = 0\), giving \(u = \frac{1}{2}\) or \(u = 1\). Find all angles in \([0, 2\pi)\) with these sine values.
Two chords \(AB\) and \(CD\) of a circle intersect at point \(P\) inside the circle. If \(PA = 3\), \(PB = 8\), and \(PC = 4\), find \(PD\).
[Diagram: Circle with two chords crossing at interior point P. \(PA = 3\), \(PB = 8\) along one chord; \(PC = 4\), \(PD = ?\) along the other.]
Show Hint
By the intersecting chords theorem: \(PA \times PB = PC \times PD\). So \(3 \times 8 = 4 \times PD\).
A regular tetrahedron has edge length 6. Find its height, surface area, and volume.
Show Hint
Height \(h = a\sqrt{\frac{2}{3}}\) where \(a\) is the edge length. Surface area \(= 4 \times \frac{\sqrt{3}}{4}a^2\). Volume \(= \frac{a^3}{6\sqrt{2}}\).
Simplify: \(\sin 75° + \sin 15°\) without a calculator. Express as a single simplified radical.
Show Hint
Use sum-to-product: \(\sin A + \sin B = 2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)\). Here: \(2\sin 45°\cos 30° = 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2}\).
In triangle \(ABC\) with \(BC = 10\), a cevian \(AD\) is drawn to side \(BC\) with \(BD = 4\) and \(DC = 6\). If \(AB = 7\) and \(AC = 9\), find the length of \(AD\).
[Diagram: Triangle ABC. Point D on BC with \(BD = 4\), \(DC = 6\). Cevian AD drawn from A to D.]
Show Hint
Stewart's Theorem: \(b^2 m + c^2 n - a(mn + d^2) = 0\) where \(m = BD\), \(n = DC\), \(a = BC\), \(d = AD\), \(b = AC\), \(c = AB\). Alternatively: \(man + dad = bmb + cnc\) (the mnemonic).
In triangle \(ABC\) with vertices \(A(0, 6)\), \(B(0, 0)\), \(C(8, 0)\), find the circumcenter, centroid, and orthocenter. Verify they are collinear (lie on the Euler line).
[Diagram: Right triangle with B at origin, C at (8,0), A at (0,6). Right angle at B.]
Show Hint
Centroid = average of vertices. Circumcenter of a right triangle is the midpoint of the hypotenuse. For the orthocenter, note that in a right triangle it lies at the vertex of the right angle.