From Euclidean elegance to the power of the unit circle — explore shapes, angles, distances, and the identities that bind them.
Drag the vertices to see how the squares on each side relate. Watch the areas change in real time.
Explore sine, cosine, and tangent on the unit circle. Move the point around to see how the values change.
Experiment with translations, rotations, reflections, and dilations interactively.
Visualize ellipses, parabolas, and hyperbolas. Adjust parameters to see how conic sections are formed.
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.]
Step 1: Apply the Law of Cosines to find side \(c\): $$c^2 = a^2 + b^2 - 2ab\cos C = 7^2 + 8^2 - 2(7)(8)\cos 60°$$
Step 2: Since \(\cos 60° = \frac{1}{2}\): \(c^2 = 49 + 64 - 112 \times \frac{1}{2} = 113 - 56 = 57\).
Step 3: So \(c = \sqrt{57}\).
Step 4: For the area, use \(A = \frac{1}{2}ab\sin C = \frac{1}{2}(7)(8)\sin 60° = 28 \times \frac{\sqrt{3}}{2} = 14\sqrt{3}\).
Answer: \(c = \sqrt{57} \approx 7.55\) and Area \(= 14\sqrt{3} \approx 24.25\) square units
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.]
Step 1: The perpendicular from the center \(O\) to chord \(AB\) bisects the chord. Let \(M\) be the midpoint of \(AB\). Then \(AM = \frac{10}{2} = 5\).
Step 2: Triangle \(OMA\) is a right triangle with \(\angle OMA = 90°\), \(OA = 13\) (radius), and \(AM = 5\).
Step 3: By the Pythagorean theorem: $$OM^2 + AM^2 = OA^2 \Rightarrow OM^2 + 25 = 169 \Rightarrow OM^2 = 144$$
Step 4: \(OM = 12\).
Answer: The distance from the center to the chord is \(12\)
Prove that \(\dfrac{1 + \tan^2\theta}{1 + \cot^2\theta} = \tan^2\theta\) for all \(\theta\) where defined.
Step 1: Use the Pythagorean identities: \(1 + \tan^2\theta = \sec^2\theta\) and \(1 + \cot^2\theta = \csc^2\theta\).
Step 2: Substitute: $$\frac{1 + \tan^2\theta}{1 + \cot^2\theta} = \frac{\sec^2\theta}{\csc^2\theta}$$
Step 3: Rewrite in terms of sine and cosine: $$\frac{\sec^2\theta}{\csc^2\theta} = \frac{1/\cos^2\theta}{1/\sin^2\theta} = \frac{\sin^2\theta}{\cos^2\theta} = \tan^2\theta$$
Answer: Proven: \(\dfrac{1 + \tan^2\theta}{1 + \cot^2\theta} = \tan^2\theta\)
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.]
Step 1: By symmetry, \(A(1,2)\) and \(C(5,2)\) have the same \(y\)-coordinate. The center lies on the perpendicular bisector of \(AC\), which is \(x = 3\). So center \(= (3, k)\) for some \(k\).
Step 2: The distance from center to \(A\) equals the distance to \(B\): $$\sqrt{(3-1)^2 + (k-2)^2} = \sqrt{(3-3)^2 + (k-4)^2}$$
Step 3: Square both sides: \(4 + (k-2)^2 = (k-4)^2\). Expand: \(4 + k^2 - 4k + 4 = k^2 - 8k + 16\).
Step 4: Simplify: \(8 - 4k = -8k + 16 \Rightarrow 4k = 8 \Rightarrow k = 2\). Center = \((3, 2)\).
Step 5: Radius \(= \sqrt{(3-1)^2 + (2-2)^2} = 2\).
Step 6: Equation: \((x - 3)^2 + (y - 2)^2 = 4\).
Answer: \((x - 3)^2 + (y - 2)^2 = 4\)
Find the area of a triangle with sides 13, 14, and 15.
Step 1: Compute the semi-perimeter: \(s = \frac{13 + 14 + 15}{2} = 21\).
Step 2: Compute the factors: \(s - a = 21 - 13 = 8\), \(s - b = 21 - 14 = 7\), \(s - c = 21 - 15 = 6\).
Step 3: Apply Heron's formula: $$A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21 \times 8 \times 7 \times 6}$$
Step 4: Compute: \(21 \times 8 = 168\), \(7 \times 6 = 42\), \(168 \times 42 = 7056\).
Step 5: \(\sqrt{7056} = 84\).
Answer: Area \(= 84\) square units
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, \(\angle ADB = ?\)]
Step 1: The inscribed angle at \(C\) on the major arc subtends arc \(AB\) (the minor arc). By the inscribed angle theorem, \(\angle ACB = \frac{1}{2} \times \text{arc } AB\). So arc \(AB = 2 \times 35° = 70°\).
Step 2: The major arc \(AB = 360° - 70° = 290°\).
Step 3: Point \(D\) is on the minor arc, so the inscribed angle at \(D\) subtends the major arc: \(\angle ADB = \frac{1}{2} \times 290° = 145°\).
Step 4: Alternatively, inscribed angles on opposite arcs subtending the same chord are supplementary: \(\angle ADB = 180° - 35° = 145°\).
Answer: \(\angle ADB = 145°\)
Find all solutions in \([0, 2\pi)\) for \(2\sin^2 x - 3\sin x + 1 = 0\).
Step 1: Let \(u = \sin x\). The equation becomes \(2u^2 - 3u + 1 = 0\).
Step 2: Factor: \((2u - 1)(u - 1) = 0\), so \(u = \frac{1}{2}\) or \(u = 1\).
Step 3: For \(\sin x = \frac{1}{2}\): \(x = \frac{\pi}{6}\) or \(x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\).
Step 4: For \(\sin x = 1\): \(x = \frac{\pi}{2}\).
Answer: \(x = \dfrac{\pi}{6},\; \dfrac{\pi}{2},\; \dfrac{5\pi}{6}\)
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.]
Step 1: When two chords intersect inside a circle at point \(P\), the intersecting chords theorem (power of a point) states: $$PA \times PB = PC \times PD$$
Step 2: Substitute: \(3 \times 8 = 4 \times PD\).
Step 3: \(24 = 4 \times PD \Rightarrow PD = 6\).
Answer: \(PD = 6\)
A regular tetrahedron has edge length 6. Find its height, surface area, and volume.
Step 1 (Surface Area): A regular tetrahedron has 4 equilateral triangular faces. Area of each face \(= \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}(36) = 9\sqrt{3}\). Total surface area \(= 4 \times 9\sqrt{3} = 36\sqrt{3}\).
Step 2 (Height): The height of a regular tetrahedron is \(h = a\sqrt{\frac{2}{3}}\). With \(a = 6\): $$h = 6\sqrt{\frac{2}{3}} = 6 \cdot \frac{\sqrt{2}}{\sqrt{3}} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt{6}$$
Step 3 (Volume): The base is an equilateral triangle with area \(9\sqrt{3}\). $$V = \frac{1}{3} \times \text{base area} \times h = \frac{1}{3} \times 9\sqrt{3} \times 2\sqrt{6} = \frac{18\sqrt{18}}{3} = 6 \times 3\sqrt{2} = 18\sqrt{2}$$
Answer: Height \(= 2\sqrt{6} \approx 4.90\), Surface Area \(= 36\sqrt{3} \approx 62.35\), Volume \(= 18\sqrt{2} \approx 25.46\)
Simplify: \(\sin 75° + \sin 15°\) without a calculator. Express as a single simplified radical.
Step 1: Apply the sum-to-product identity: $$\sin A + \sin B = 2\sin\!\left(\frac{A+B}{2}\right)\cos\!\left(\frac{A-B}{2}\right)$$
Step 2: With \(A = 75°\) and \(B = 15°\): $$\frac{A+B}{2} = 45°, \quad \frac{A-B}{2} = 30°$$
Step 3: Substitute: $$\sin 75° + \sin 15° = 2\sin 45° \cos 30° = 2 \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{6}}{2}$$
Answer: \(\dfrac{\sqrt{6}}{2}\)
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.]
Step 1: Stewart's Theorem: \(b^2 m + c^2 n - a(mn + d^2) = 0\), where \(a = BC = 10\), \(b = AC = 9\), \(c = AB = 7\), \(m = BD = 4\), \(n = DC = 6\), \(d = AD\).
Step 2: Substitute: $$9^2(4) + 7^2(6) - 10(4 \times 6 + d^2) = 0$$
Step 3: Compute: \(81(4) + 49(6) - 10(24 + d^2) = 0\).
\(324 + 294 - 240 - 10d^2 = 0\).
\(378 - 10d^2 = 0\).
Step 4: \(d^2 = \frac{378}{10} = 37.8\), so \(d = \sqrt{37.8} = \sqrt{\frac{189}{5}} = \frac{3\sqrt{21}}{\sqrt{5}} = \frac{3\sqrt{105}}{5}\).
Answer: \(AD = \dfrac{3\sqrt{105}}{5} \approx 6.15\)
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.]
Step 1 (Centroid): \(G = \left(\frac{0+0+8}{3},\, \frac{6+0+0}{3}\right) = \left(\frac{8}{3},\, 2\right)\).
Step 2 (Circumcenter): Since \(\angle B = 90°\) (right angle at origin), the circumcenter is the midpoint of the hypotenuse \(AC\): $$O = \left(\frac{0+8}{2},\, \frac{6+0}{2}\right) = (4, 3)$$
Step 3 (Orthocenter): In a right triangle, the orthocenter is at the vertex of the right angle: \(H = B = (0, 0)\).
Step 4 (Verify collinearity): Check that \(G, H, O\) are collinear. The slope from \(H(0,0)\) to \(G(\frac{8}{3}, 2)\) is \(\frac{2}{8/3} = \frac{3}{4}\). The slope from \(H(0,0)\) to \(O(4,3)\) is \(\frac{3}{4}\). Same slope, so they are collinear.
Step 5: The Euler line has equation \(y = \frac{3}{4}x\). Also note \(OG:GH = 1:2\), confirming the Euler line property.
Answer: Circumcenter \(= (4, 3)\), Centroid \(= \left(\frac{8}{3}, 2\right)\), Orthocenter \(= (0, 0)\). All lie on the line \(y = \frac{3}{4}x\).