Sine - ACT Math

We can solve for the length of the chord by drawing a line the bisects the angle and the chord, shown below as .

In this circle, we can see the triangle has a hypotenuse equal to the radius of the circle (), an angle equal to half the angle made by the chord, and a side that is half the length of the chord. By using the sine function, we can solve for .

The length of the entire chord is twice the length of , so the entire chord length is .

sin(30º) = ½

sine = opposite / hypotenuse

½ = opposite / 8

Opposite = 8 * ½ = 4

We can solve for the length of the chord by drawing a line the bisects the angle and the chord, shown below as .

In this circle, we can see the triangle has a hypotenuse equal to the radius of the circle (), an angle equal to half the angle made by the chord, and a side that is half the length of the chord. By using the sine function, we can solve for .

The length of the entire chord is twice the length of , so the entire chord length is .

Begin by drawing out this scenario using a little right triangle:

Note importantly: We are looking for as the the distance to the top of the building. We know that the sine of an angle is equal to the ratio of the side opposite to that angle to the hypotenuse of the triangle. Thus, for our triangle, we know:

Using your calculator, solve for :

This is . Now, take the decimal portion in order to find the number of inches involved.

Thus, rounded, your answer is feet and inches.

Recall that the sine of an angle is the ratio of the opposite side to the hypotenuse of that triangle. Thus, for this triangle, we can say:

Solving for , we get:

or

To find the sine of an angle, remember the mnemonic SOH-CAH-TOA.
This means that

.

We are asked to find the . So at point we see that side is opposite, and the hypotenuse never changes, so it is always . Thus we see that

In right triangles, SOHCAHTOA tells us that , and we know that and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.

Use a calculator or reference to approximate cosine.

Isolate the variable term.

Thus, .

In right triangles, SOHCAHTOA tells us that , and we know that and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.

Isolate the variable term.

Thus, .

In right triangles, SOHCAHTOA tells us that , and we know that and hypotenuse . Therefore, a simple substitution and some algebra gives us our answer.

Use a calculator or reference to approximate cosine.

Isolate the variable term.

Thus, .

Using the foil method, multiply. Simplify using the Pythagorean identity sin2Θ + cos2Θ = 1 and the double angle identity sin2Θ = 2sinΘcosΘ.

To solve this equation, it is best to remember the mnemonic SOHCAHTOA which translates to Sin = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent. Looking at the problem statement, we are given the side opposite of the angle we are trying to find as well as the hypotenuse. Therefore, we will be using the SOH part of our mnemonic. Inserting our values, this becomes . Then, we can write . Solving this, we get

You can draw your scenario using the following right triangle:

Recall that the sine of an angle is equal to the ratio of the oppositeside to the hypotenuse of the triangle. You can solve for the angle by using an inverse sine function:

or .

Recall that the sine of an angle is equal to the ratio of the opposite side to the hypotenuse of the triangle. You can solve for the angle by using an inverse sine function:

or .

Recall that .

Therefore, we are looking for or .

Now, this has a reference angle of , but it is in the third quadrant. This means that the value will be negative. The value of is . However, given the quadrant of our angle, it will be .

Recall that the triangle appears as follows in radians:

Now, the sine of is . However, if you rationalize the denominator, you get:

Since , we know that must be represent an angle in the third quadrant (where the sine function is negative). Adding its reference angle to , we get:

Therefore, we know that:

, meaning that

You can begin by imagining a little triangle in the fourth quadrant to find your reference angle. It would look like this:

Now, to find the sine of that angle, you will need to find the hypotenuse of the triangle. Using the Pythagorean Theorem, , where and are leg lengths and is the length of the hypotenuse, the hypotenuse is , or, for our data:

The sine of an angle is:

For our data, this is:

Since this is in the fourth quadrant, it is negative, because sine is negative in that quadrant.

You can begin by imagining a little triangle in the third quadrant to find your reference angle. It would look like this:

Now, to find the sine of that angle, you will need to find the hypotenuse of the triangle. Using the Pythagorean Theorem, , where and are leg lengths and is the length of the hypotenuse, the hypotenuse is , or, for our data:

The sine of an angle is:

For our data, this is:

Since this is in the third quadrant, it is negative, because sine is negative in that quadrant.

Recall that the sine wave is symmetrical with respect to the origin. Therefore, for any value , the value for is . Therefore, if is , then for , it will be .

In a right triangle, for sides a and b, with c being the hypotenuse, . Thus if cos(A) is , then c = 14, and the side adjacent to A is 11. Therefore, the side opposite of angle A is the square root of , which is Since sin is , sin(A) is .