Introduction to sine
This is an introductory lesson about sine, one of the trigonometric functions, for grades 8-9. We start out by studying several similar right triangles and the ratio of certain two sides in them.
The sine is simply a RATIO of certain two sides in a right triangle. The triangles below all have the same shape, which means they are SIMILAR figures. That means they have the SAME ANGLE MEASURES but the lengths of the sides are different.
Ask students to measure the sides s1, h1, s2, h2, s3, h3 as accurately as possible (or ask them to draw several similar right triangles on their own).
|Then let them calculate the following ratios:||
What do you note?
Those ratios should all be the same (or close to same due to measuring errors). That is so because the triangles are similar (have the same shape), which means their respective sides are PROPORTIONAL. That is why the ratio of those side lengths remains the same.
Now ask the students what would happen if we had a fourth triangle with the same shape. The answer of course is that even in that fourth triangle this ratio would be the same.
The ratio you calculated remains the same for all these triangles, because they are similar. This means that in ALL right triangles that are similar to each other this ratio is the same, too. And if triangles are similar, their angles are congruent (the same). We associate this ratio with the angle α (see the image to find angle α). THAT RATIO IS CALLED THE SINE OF THAT ANGLE α.
What follows is that if you know the ratio, you can find what the angle α is. Or the other way: if you know what the angle α is, you can find this ratio (called sine of that angle) – and then when you know this ratio and one side of a right triangle, you can find the other lengths of sides.
|= sin α = 0.57358|
In our pictures the angle α is 35 degrees. So sin(35) = 0.57358 (rounded to five decimals).
We can use this fact when dealing with OTHER right triangles that have a 35° angle. See, other such triangles are, again, similar to these ones we see here, so the ratio of the opposite side to the hypotenuse, WHICH IS THE SINE OF THE 35 ANGLE, is the same!
Suppose we have a triangle that has the same shape as the triangles above. The side opposite to the 35° angle is 5 cm. How long is the hypotenuse?
Solution: Let h be the hypotenuse. Then
|= sin 35 ≈ 0.57358|
|From this equation one can easily solve that h =||
|≈ 8.72 cm.|
The two triangles in the image are pictured both overlapping and separate. We can find H3 simply by the fact that these two triangles are similar. Since the triangles are similar,
|, from which H3 =||
6 × 3.9
We didn't even need the sine to solve that, but note how closely it ties in with similar triangles: The triangles have the same angle α.
|Sin α would of course be the ratio||
Now we can find the actual angle α from a calculator: Since sin α = 0.4333, then α = sin-10.4333 ≈ 25.7 degrees.
Test your understanding
1. Draw a right triangle that has a 40° angle. Then measure the opposite side and the hypotenuse and use those measurements to calculate sin 40°. Check your answer with a calculator (remember the calculator has to be in the "degrees" mode instead of "radians" mode).
2. Draw two right triangles that have a 70° angle but that are of different sizes. Use the first triangle to find sin 70° (like you did in problem 1). Then measure the hypotenuse of your second triangle. Use sin 70° and the measurement of the hypotenuse to find the opposite side in your second triangle. Check by measuring the opposite side from your triangle.
3. Draw a right triangle that has a 48° angle. Measure the hypotenuse. Then use sin 48° (from a calculator) and your measurement to calculate the length of the opposite side. Check by measuring the opposite side from your triangle.
When I type in sine in my graphic calculator, why does it give me a wave?"
Read my answer: How do we get the familiar sine wave?.
My question is that if someone gave us only the length of sides of a triangle, how can we draw a triangle?
sajjad ahmed shah
This is a construction problem and can be solved easily. Please see Constructing a triangle when the lengths of 3 sides are known for a solution.
If I am in a plane flying at 30000 ft, how many linear miles of ground can I see? And please explain how that answer is generated. Does it have anything to do with right triangles and the Pythagorean theorem?
The image below is NOT to scale – it is just to help in the problem. The angle α is much smaller in reality. In the image, x is the amount of linear miles you can see from the plane, and r is the radius of the Earth.
Yes, you have a right triangle. Actually, the radius of the Earth is not constant but varies because Earth is not a perfect sphere. For this problem, I was using the mean radius 3,959.871 miles. This also means our answer will be only an approximation. I also converted 30,000 feet to 5.6818182 miles.
First we calculate α using cosine. You should get cos α = (adjacent/hypotenuse) = r/(r + 5.6818182 mi). From that, α = 3.067476356 degrees. Then, we use a proportion with the ratios α/360° and x/r. You will get x ≈ 212 miles. Even that result might be too 'exact'.