How would a triangle for sin 90 degree look

I am studying trigonometry in my school and learned that in a triangle the side opposite to angle theta should be taken as perpendicular side - hypotenuse remains the same and the third remaining side is the base.

The same is required for calculating the sine / cosine etc of the angle theta for below formula for distance / object height in applications of trigonometry -

 sin theta = Perpendicular / hypotenuse
cos theta = Base / hypotenuse


The remaining other can be created using the above two.

So in a Triangle ABC if Angle B is 90 degree it is easy to find sin A or sin C - I mean which side is the Perpendicular, hypotenuse or the base. But then how to find the sides for sin 90 - should we take hypotenuse as the perpendicular side for above triangle as sin B is sin 90 since side opposite to angle will be the perpendicular side - then which side should be taken as hypotenuse and base in such a case

One other way to find angles -

$\sin \theta = \frac{\text{Opposite Side}}{\text{Hyphotenuse}}$

$\cos \theta = \frac{\text{Adjacent Side}}{\text{Hyphotenuse}}$

Edit -

$\sin 90° = \frac{\text{Opposite Side}}{\text{Hyphotenuse}}$

And in this case Opposite Side is Hypotenuse itself.

• I don't understand what you're getting at. Can you think about $\sin 90^\circ$ in terms of a right triangle, as the OP asks? – pjs36 Feb 7 '17 at 5:05
• So in my case considering the triangle ABC having Angle B as 90 degree - sin 90 would be = hypotenuse / hypotenuse ? Is the same true then probably stating that side opposite to angle theta should be taken as perpendicular side is wrong - please re-confirm – Programmer Feb 7 '17 at 5:06
• You don't take that corner with sin/cos. – Kaynex Feb 7 '17 at 5:11
• Actually I was trying to memorize the sine, cosine and other formulas with saying : "some people have curly brown hair" - sin = p / h and cos = b / h (take first alphabet of each word in the saying) ... – Programmer Feb 7 '17 at 5:14
• And we can write opposite side in place of opposite side is perpendicular. – Kanwaljit Singh Feb 7 '17 at 5:23

One could think as follows: Since the side opposite to the angle B is the hypotenuse, then: $$sinB = \frac{hypotenuse}{hypotenuse} = 1$$

However, one must always look carefully and rigorously to definitions. The sine of an angle is defined as the division of the opposite side to the angle by the hypotenuse. The hypotenuse can't be one of the sides. Having a look at the trigonometric circle might help:

The sine is defined here always as the value of $y$ divided by 1 (hypotenuse) When the angle is 90°, then $y=1$. I don't want to sound rude, but there is not a lot of sense in "choosing the sides" for calculating the sine of 90°. Hope i've helped.

The short answer is that you can't construct a right angle triangle to visualise $\sin 90^{\circ}$.

The "opposite" (or "perpendicular" as per your nomenclature) side has to be distinct from the hypotenuse, it cannot be the same side. But in a plane figure with two right angles opposite one another, you will never have an apex as the parallel sides won't ever meet (i.e. you can't construct such a triangle).

You can get arbitrarily close by having one true right angle with the other angle approaching very close to a right angle (with the third angle getting progressively smaller). This will result in a very "tall" and "narrow" right triangle if you orient it a certain way. From this you can tell that the side opposite the right angle will become closer and closer to the length of the hypotenuse, with the sine of that angle approaching $1$ from below. In other words, you can construct plane right triangles allowing you to visualise $\sin x \to 1$ as $x \to 90^{\circ}$, but you can't actually draw the limiting case.

Angle A - First Position

Look at the previous link. The hypotenuse side is the one facing the right angle. It doesn't change as long as the right angle doesn't change. $$SinA = \frac{Opposite\ Side}{Hypotenuse}$$ and $$CosA = \frac{Adjacent\ Side}{Hypotenuse}$$

Opposite Side means opposite to the angle for which you are calculating the sine and cosine. An angle in a right-angled triangle will be formed from two sides of the triangle: Hypotenuse and Adjacent Side. The third side will be the opposite side of this angle.

If we simply change the angle for which the sine and cosine to be calculated, the adjacent and opposite sides will change, but the hypotenuse will remain the same since the right angle did not change.

Look at the following picture:

Angle A Second Position

Did you get it ?

While the idea of measuring the sine of a right angle using a degenerate triangle (hypotenuse coincident with the opposite leg, and a base of zero length) may be intuitive to some, not everyone finds it satisfactory, as you can see from the other answers.

And what about the sine and cosine of angles between $90$ and $180$ degrees? We need them in order to apply the law of sines and law of cosines to the obtuse vertex in an obtuse triangle.

Indeed, as you go on in mathematics you may find it desirable to find sines and cosines of angles greater than $180$ degrees, and even "angles" greater than $360$ degrees.

Mathematicians solve this problem in the definitions of trigonometric functions by extending the definitions of the sine and cosine functions so that they take care of input values of any size. The extensions agree with the hypotenuse-opposite-adjacent definitions for angles that are greater than zero but less than a right angle, and they preserve useful properties such as angle-sum formulas and angle-difference formulas for input values outside that range. Several of these extended definitions are listed in the answers to an earlier question, How many ways are there to define sine and cosine?

My favorite extended definition for use at a level of mathematics where you are just starting to go beyond the elementary right-triangle definition is the unit-circle definition. We observe that if you place a right triangle with unit-length hypotenuse on the Cartesian plane so that one leg is on the positive $x$-axis and one end of the hypotenuse is at the origin of coordinates, $(0,0),$ the $(x,y)$ coordinates of the other end of the hypotenuse are the cosine and sine (respectively) of the angle at the origin. If we then take a sequence of triangles like this with increasing angles at the origin, the other end of the hypotenuse traces out points along the circle of unit radius centered at the origin. The idea of the unit-circle definition is that you use the $(x,y)$ coordinates of these points as the definition of cosine and sine, and to find cosine and sine of $90$ degrees or larger angles, you just keep going around the unit circle. Once you accept this as a definition, there is nothing ambiguous or even special about the sine of $90$ degrees, except perhaps that it is unusually easy to find its value.

• +1 Best answer here imo. This question came up in a class of mine recently, and I did not have a very satisfactory explanation at first. Considering degenerate triangles is, as you say, not satisfactory for many. I think part of the problem is explained away by considering a limiting effect for the ratios, but this doesn't resolve the continuity issue. I found it interesting that in Gelfand's Trigonometry book he makes reference to the limiting idea and then defines $\sin 90^\circ=1$, sort of subtly skirting the perceived continuity problem. [...] – Daniel W. Farlow Feb 18 '18 at 15:12
• [...] I find it depressing that "leading precalculus texts" (i.e., in terms of sales) give no such consideration to problems like these. I think this is a nice situation to highlight the limitations of certain outlooks. The unit circle outlook gives a much clearer picture of everything imo. And it dispenses with many of the little pesky problems that crop up with dealing exclusively with right triangles. – Daniel W. Farlow Feb 18 '18 at 15:14

If we observe $$\sin 30^\circ$$, or $$\sin 60^\circ$$, the length of the opposite side is increasing in relation to the hypotenuse. If we extend the logic to $$\sin 89^\circ$$, the length of the opposite side is almost equivalent to the hypotenuse. At $$\sin 90$$, the opposite and hypotenuse get merged, meaning that both are equal and the ratio $$\sin 90 = \dfrac{\text{opposite}}{\text{hypotenuse}}$$ becomes $$1$$. Hope this clarifies.