Right-Triangle Ratios: See What sin, cos and tan Mean

Begin with the right angle

A right triangle contains one angle of $90^\circ$, shown by a small square. The side facing that square is the hypotenuse (कर्ण). It is the longest side. The other two sides meet at the right angle, so they are the perpendicular legs.

Choose either of the remaining corners and call its angle $\theta$. Because the angles of a triangle add to $180^\circ$, the other acute angle is $90^\circ-\theta$. For a genuine right triangle, $0^\circ<\theta<90^\circ$. The right angle does not move when we choose a different acute corner.

Throughout this page, $P$ and $B$ name the two legs and $H$ names the hypotenuse. Pythagoras connects their lengths:

For legs of lengths $3$ and $4$, the square of the hypotenuse is $9+16=25$. Its length is therefore $5$, not $7$. Add the squares of the legs before taking a positive square root.

The units must match: metres with metres, centimetres with centimetres. A diagram helps us organise the calculation, but a picture alone cannot prove that a triangle is right-angled. Use the stated right angle or check the Pythagorean relationship. That small check will prevent many later mistakes in ratios and word problems.

A statement that a triangle is right-angled is information about its geometry. The little square records that information; it is more reliable than how a freehand sketch looks.

P and B depend on the angle you choose

Stand at the marked angle $\theta$ and look across the triangle. The leg you face is the opposite side, which we label $P$ (लंब). The leg beside your angle is the adjacent side, labelled $B$ (आधार). The sloping side is not automatically the adjacent leg: the hypotenuse already has its own role.

This is a relationship to the selected angle, not to the page. A horizontal line is not always $B$, and a vertical line is not always $P$. Turn the whole triangle on the screen and the ratios for the same marked angle remain unchanged.

Now select the other acute corner. The old opposite leg becomes adjacent, and the old adjacent leg becomes opposite. Only the hypotenuse keeps its role. This is why two different angles in the same triangle can have different sine and tangent values.

For example, suppose $\theta$ faces the leg of length $3$ in a $3,4,5$ triangle. Then $P=3$, $B=4$, $H=5$. From the other acute corner, $P=4$, $B=3$, $H=5$. No side has grown; our viewpoint changed.

Before any calculation, make three decisions: locate the right angle, mark the requested acute angle, and label the opposite and adjacent legs. If a question gives $\angle A$, all labels must refer to $A$. If it later asks about the other angle, pause and relabel. The same habit will help when a height-and-distance diagram contains two triangles.

Why a ratio belongs to the angle

Draw two right triangles with the same acute angle $\theta$. Each has a right angle, so their third angles also match. The triangles are similar by the angle–angle criterion. Their corresponding sides can be different lengths, but they all change by the same scale factor.

Start with $P=3$, $B=4$, $H=5$. A triangle twice as large has $P=6$, $B=8$, $H=10$. Compare opposite with hypotenuse:

More generally, scaling by a positive number $k$ changes $P/H$ to $kP/(kH)$. The factor cancels, leaving $P/H$. The same reasoning works for the other side ratios. This proves that the ratios are well defined for an acute angle: choosing a larger similar right triangle does not change them.

Changing the shape is different. If you increase $P$ while keeping $B$ fixed, the marked angle becomes steeper. Its opposite-to-adjacent ratio changes. A ratio tells us about this shape, not the drawing’s size or the unit used to measure it.

Notice also that a ratio has no length unit when the two lengths use the same unit. A value such as $3/5$ is a number, not $3/5$ centimetres. In the triangle lab, compare a triangle with a scaled copy, then change only one leg. Predict which action preserves its angle before looking at the displayed values. The distinction between scale and shape is the key idea to carry into every following topic.

Meet the six ratios

With the sides labelled relative to $\theta$, three ratios come directly from the triangle:

Read $\sin\theta$ as “sine of theta”. It is the name of a function evaluated at an angle, not “sin multiplied by theta”. The three reciprocal ratios reverse these fractions:

For the chosen acute angle in a $3,4,5$ triangle, with $P=3$, the six values are $3/5,4/5,3/4,5/3,5/4,4/3$, in the order shown above. Pair sine with cosecant, cosine with secant, and tangent with cotangent.

There is another useful connection. Dividing $P/H$ by $B/H$ gives $P/B$. Therefore $\tan\theta=\sin\theta/\cos\theta$. The division is valid here because an acute angle has a positive adjacent leg. Every side of a non-degenerate right triangle is positive.

Try the five exact triples in the lab. They are different shapes, so their ratios need not match. When you edit two compatible sides, the third must still satisfy Pythagoras. You cannot independently choose arbitrary values for all three sides and expect a right triangle.

Keep an exact fraction when one is available. A decimal such as $0.75$ is useful, but $3/4$ also shows the side relationship immediately. Check the largest side is $H$, then use the marked angle to decide the numerator and denominator.

Find the missing side first

A ratio question sometimes gives two side lengths rather than all three. Reconstruct the triangle before trying to remember a separate shortcut. Pythagoras provides the missing information.

If the two legs are known, add their squares:

If the hypotenuse and one leg are known, subtract the leg’s square:

Take the positive root because these symbols represent lengths. The hypotenuse must be longer than either leg. If an input claims $H=8$ and a leg is $10$, stop: those values cannot form the stated right triangle.

Suppose $H=15$ and the leg adjacent to $\theta$ is $B=12$. Then $P^2=225-144=81$, so $P=9$. We can now find $\sin\theta=9/15=3/5$, $\cos\theta=12/15=4/5$ and $\tan\theta=9/12=3/4$. This is a scaled $3,4,5$ triangle, which provides a useful independent check.

Not every triangle has integer sides. For $P=1$ and $B=2$, the exact hypotenuse is $\sqrt5$. Its sine is $1/\sqrt5=\sqrt5/5$, not $1/5$. A square root is a complete exact answer when no simpler integer exists.

Keep the drawing and calculation connected: label the known lengths, write the Pythagorean equation, solve for the missing positive length, and only then build the requested ratio. This sequence makes wrong labels much easier to spot.

When a decimal answer seems surprising, return to the squared equation and substitute your result. Both sides should agree in the original units.

One ratio can reveal the whole triangle

Suppose an acute angle satisfies $\sin\theta=12/13$. This tells us a proportion, not that the actual triangle must measure $12$ centimetres by $13$ centimetres. We may choose a convenient similar triangle with $P=12$ and $H=13$, then recover its adjacent leg.

The other ratios now follow: $\cos\theta=5/13$, $\tan\theta=12/5$, $\operatorname{cosec}\theta=13/12$, $\sec\theta=13/5$, and $\cot\theta=5/12$. A larger triangle with sides $12k,5k,13k$, where $k>0$, gives exactly the same answers.

Read the given function before assigning the sides. If $\tan\theta=5/12$, the numbers $5$ and $12$ represent opposite and adjacent legs. The hypotenuse is then $13$. If $\cos\theta=5/13$, the $5$ is adjacent and the $13$ is the hypotenuse. The same numbers can play different roles.

A decimal can also reveal an exact ratio when it is given exactly. For acute $\theta$, $\cos\theta=0.6=3/5$ leads to $B=3,H=5,P=4$. Thus sine is $4/5$, and tangent is $4/3$.

The word acute matters. All six ratios are positive for the right-triangle angles on this page. Later, the unit circle will extend sine and cosine to other angles, where signs require extra information. For now, make a clean reference triangle and use its geometry to justify each result.

Reduce a given fraction before building the reference triangle. A simpler proportional triangle makes the arithmetic shorter while preserving exactly the same angle and ratios.

Choose the ratio that connects what you know

You do not need all three side lengths for every problem. Choose the ratio that contains the quantity you know and the quantity you want. The labels around the selected angle act as a map.

• Opposite and hypotenuse: use $\sin\theta$.
• Adjacent and hypotenuse: use $\cos\theta$.
• Opposite and adjacent: use $\tan\theta$.

Imagine a model ramp of length $10$ units. Its horizontal base is $8$ units, and $\theta$ is the angle between the ramp and that base. The ramp is the hypotenuse, so $\cos\theta=8/10=4/5$. Calling the ramp the “base” just because it is a prominent line would reverse the geometry.

If the height of that ramp is needed, Pythagoras gives $P=\sqrt{100-64}=6$. We may check the same angle using $\sin\theta=6/10=3/5$ or $\tan\theta=6/8=3/4$. The values describe one angle through three different comparisons.

In a new word problem, first mark the angle and decide whether each stated length is horizontal, vertical or along the sloping side. Then write a ratio equation before substituting numbers. Rearrange carefully: if $\sin\theta=P/H$, then $P=H\sin\theta$, while $H=P/\sin\theta$.

Your final check is geometric. A hypotenuse cannot be shorter than its legs. Sine and cosine of an acute angle lie strictly between $0$ and $1$. Tangent can be larger than $1$; that simply means the opposite leg is longer than the adjacent leg. These checks are often more useful than redoing the same arithmetic.