Ratios in a Right Triangle

For an acute angle $\theta$ of a right triangle:

• $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$;
• $\operatorname{cosec}\theta$, $\sec\theta$, $\cot\theta$ are their reciprocals.

Pythagoras supplies the third side. Know the common triples by sight: $3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$, $7$-$24$-$25$, $20$-$21$-$29$ and their multiples.

Given one ratio $\dfrac{a}{b}$ of an acute angle:

• draw a right triangle with those two sides and find the third by Pythagoras;
• read off any other ratio;
• if the angle is not acute, keep the sizes and fix each sign from the quadrant.

For a ratio given as $\dfrac{m^2 - n^2}{m^2 + n^2}$, the third side is $2mn$, because $(m^2+n^2)^2 - (m^2-n^2)^2 = 4m^2n^2$. When a question gives $\dfrac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta}$ or similar, divide top and bottom by $\cos\theta$ to get an equation in $\tan\theta$ alone.

Places a right angle hides:

• the corner of a rectangle or square (a diagonal makes two right triangles);
• the angle in a semicircle;
• the foot of a perpendicular — from the centre of a circle to a chord it bisects the chord, so a chord subtending $2\theta$ at the centre of a circle of radius $r$ has length $2r\sin\theta$;
• the altitude to the hypotenuse, which splits the triangle into two triangles similar to it.

Area $= \dfrac12 ab\sin C$ for any two sides and the angle between them. For a sum and a hypotenuse, square: $(AB + BC)^2 - (AB^2 + BC^2) = 2\,AB \cdot BC$.