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Trigonometric ratios and solving right triangles

Angle chooses the role of the side Choose an acute angle $\alpha$ in a right triangle. In relation to it, one side is opposite and the other is adjacent. If we change angles, these roles are reversed; the hypotenuse remains the same. If two triangles have the same acute angle, they are similar. Therefore, the ratios between their corresponding sides are equal:

October 22, 2026 Read
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Pythagoras, height and projections in the right triangle

Name each segment In a right triangle, the side opposite the right angle is the hypotenuse. Call her $a$ and the boys $b,c$. The other angles are acute and add up to $90°$. Draw the height $h$ to the hypotenuse. It divides $a$ into two projections: $m$, associated with the side $b$, and $n$, associated with the side $c$. Thus, $m+n=a$. Demonstration by similarity The two smaller triangles have a right angle and an acute angle the same as the original triangle. They are similar to him. Comparing corresponding sides:

October 22, 2026 Read
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Demonstrations of the values ​​of 30, 45 and 60 degrees

Two drawings replace memorization The trigonometric ratios of an acute angle are: $$\sin\alpha=\frac{\text{cateto oposto}}{\text{hipotenusa}},\quad \cos\alpha=\frac{\text{cateto adjacente}}{\text{hipotenusa}},$$ $$\tan\alpha=\frac{\text{cateto oposto}}{\text{cateto adjacente}}.$$They can be calculated exactly for some angles using Pythagoras. See the metric relations class in this track for the proof of the theorem. The 45 degree triangle The diagonal of a square with side 1 divides it into two isosceles right triangles. The acute angles are equal and add up to $90°$; each measures $45°$.

October 21, 2026 Read
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Inscribed polygons and the sine of a central half-angle

A rope between two spokes Inscribe a regular polygon with $n\ge3$ sides in a circle of radius $R$. Connect two neighboring vertices to the center O. The triangle formed has two sides equal to $R$ and a base equal to the side $\ell$ of the polygon. The $n$ equal central angles complete one turn. Each measures $360°/n$. Draw the height of this triangle from O. For symmetry, it cuts the base and the central angle in half.

October 21, 2026 Read
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Measurement of angles in degrees, minutes and seconds

A unit to compare spins Dividing a turn into 360 equal parts defines the degree. A full turn measures $360°$; half-turn, $180°$; a room, $90°$. Angles between $0°$ and $90°$ are acute; between $90°$ and $180°$, obtuse. On a circle, the measure of an arc in degrees is the measure of its corresponding central angle. To distinguish between two arcs with the same endpoints, indicate which one is being measured.

October 21, 2026 Read
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Arcs, chords, rays and angles

Start with the drawing Mark distinct points A and B on a circle. They divide it into two arcs, both including A and B. The chord AB is the straight segment between the points. An arc is not a segment. If the extremes coincide, we distinguish the null arc, which is reduced to the point, and the arc of a turn, which covers the entire circumference. This distinction will be important when comparing the final position with the turn performed.

October 21, 2026 Read
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Triangles that grow without changing shape

A comic-drawn ramp On a mesh, draw a path 3 comics to the right and 4 comics to the top. Connect the beginning to the end by a segment. You have formed a triangle with a straight corner: a right triangle. Now make a copy with all lengths folded. The horizontal path will have 6 comics; the vertical, 8. The lengths changed, but the corner openings remained the same. $$3\cdot2=6,\qquad4\cdot2=8.$$Why does bending just one side change the shape? Compare two copies: in the first, fold the horizontal and vertical path; in the second, fold only the horizontal. In the second copy the ramp is more laid down. Growing without changing shape requires expanding in all directions in the same proportion.

October 21, 2026 Read
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Corners, spokes and paths on the wheel

Two paths between the same points Draw a wheel and mark two points on it, A and B. Connect them with a straight line: it is a rope. Now follow the edge of the wheel from A to B: you have covered an arc. There are two possible paths around the edge; indicate which one you are using. The center is the point in the middle of the wheel. A segment from the center to the edge is a radius. All spokes on the same wheel have the same length.

October 21, 2026 Read
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Whole turns, half turns and quarters

Take a turn with your body Look at a wall. Spin around until you’re looking at the back wall: you’ve made a u-turn. Another half turn takes you back to the start. Now try to do the same spin in four equal parts. Each part is a quarter turn. A lap can be split in several ways. Two halves and four quarters complete the same lap: $$\frac12+\frac12=1,\qquad \frac14+\frac14+\frac14+\frac14=1.$$Here the number 1 represents a full turn. We are not measuring the length of the path your feet travel, but the turn of your direction.

October 21, 2026 Read
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Solving triangles and discussing the ambiguous case

What data determines a triangle? Solving a triangle is getting its sides and angles from enough data. Always use the convention: $a,b,c$ opposite $A,B,C$. A non-degenerate triangle has positive sides and angles between 0 and $\pi$. One side and two angles Knowing $a,B,C$, calculate $A=\pi-B-C$. $B+C<\pi$ is required. The law of sines gives: $$b=\frac{a\sin B}{\sin A},\qquad c=\frac{a\sin C}{\sin A}.$$The data determines a triangle, minus position and reflection. Two sides and the angle between them Knowing $b,c,A$, with $0

October 21, 2026 Read
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Heights, Heron, Medians, angle bisectors and Radii

Get special measurements from the sides Consider sides $a,b,c$, area $K$ and semiperimeter $s=(a+b+c)/2$. We require a non-degenerate triangle: positive sides and each side smaller than the sum of the other two. Demonstrate Heron’s formula From the law of cosines, $\cos A=(b^2+c^2-a^2)/(2bc)$. In $K=bc\sin A/2$, use $\sin^2A=1-\cos^2A$: $$\begin{aligned} 16K^2&=4b^2c^2-(b^2+c^2-a^2)^2\\ &=((b+c)^2-a^2)(a^2-(b-c)^2)\\ &=(a+b+c)(b+c-a)\\ &\quad\cdot(a+b-c)(a-b+c)\\ &=16s(s-a)(s-b)(s-c). \end{aligned}$$The area is positive, so: $$K=\sqrt{s(s-a)(s-b)(s-c)}.$$Heights As $K=ah_a/2=bh_b/2=ch_c/2$: $$h_a=\frac{2K}{a},\qquad h_b=\frac{2K}{b},\qquad h_c=\frac{2K}{c}.$$Substituting Heron’s expression gives each height as a function of the sides only. $h_a=b\sin C=c\sin B$ and its permutations are also valid.

October 21, 2026 Read
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Laws of sines, cosines, projections and areas

Fix the notation In a non-degenerate triangle, the angles $A,B,C$ are opposite the sides $a,b,c$, respectively. All sides are positive and $A+B+C=\pi$. Demonstration of the law of cosines Place vertex A at the origin, B at $(c,0)$ and C at $(b\cos A,b\sin A)$. The distance between B and C is $a$. By Pythagoras’ theorem applied to coordinate differences: $$\begin{aligned} a^2&=(b\cos A-c)^2+(b\sin A)^2\\ &=b^2(\cos^2A+\sin^2A)+c^2-2bc\cos A\\ &=b^2+c^2-2bc\cos A. \end{aligned}$$The argument also applies to obtuse angles. If $A=\pi/2$, the cosine term is zero and we find Pythagoras again. Permuting the letters gives the other two cosine laws.

October 21, 2026 Read