How to find the sine of a missing side - ACT Math

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Question

In the triangle below, units and units. What is $\sin C$ ?

Answer

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

Compare your answer with the correct one above

Question

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Answer

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet.

Using the Pythagorean Theorem, we can solve for it:

$8^2+10^2=h^2{}$

$h=\sqrt{164}=2\sqrt{41}{}$

So, the sine of this angle is:

$\frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}{}$

Compare your answer with the correct one above

Question

What is the value of in the triangle above? Round to the nearest hundredth.

Answer

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

Compare your answer with the correct one above

Question

In the triangle below, units and units. What is $\sin C$ ?

Answer

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

Compare your answer with the correct one above

Question

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Answer

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet.

Using the Pythagorean Theorem, we can solve for it:

$8^2+10^2=h^2{}$

$h=\sqrt{164}=2\sqrt{41}{}$

So, the sine of this angle is:

$\frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}{}$

Compare your answer with the correct one above

Question

What is the value of in the triangle above? Round to the nearest hundredth.

Answer

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

Compare your answer with the correct one above

Question

In the triangle below, units and units. What is $\sin C$ ?

Answer

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

Compare your answer with the correct one above

Question

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Answer

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet.

Using the Pythagorean Theorem, we can solve for it:

$8^2+10^2=h^2{}$

$h=\sqrt{164}=2\sqrt{41}{}$

So, the sine of this angle is:

$\frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}{}$

Compare your answer with the correct one above

Question

What is the value of in the triangle above? Round to the nearest hundredth.

Answer

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

Compare your answer with the correct one above

Question

In the triangle below, units and units. What is $\sin C$ ?

Answer

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

Compare your answer with the correct one above

Question

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Answer

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet.

Using the Pythagorean Theorem, we can solve for it:

$8^2+10^2=h^2{}$

$h=\sqrt{164}=2\sqrt{41}{}$

So, the sine of this angle is:

$\frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}{}$

Compare your answer with the correct one above

Question

What is the value of in the triangle above? Round to the nearest hundredth.

Answer

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

Compare your answer with the correct one above

Question

In the triangle below, units and units. What is $\sin C$ ?

Answer

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

Compare your answer with the correct one above

Question

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Answer

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet.

Using the Pythagorean Theorem, we can solve for it:

$8^2+10^2=h^2{}$

$h=\sqrt{164}=2\sqrt{41}{}$

So, the sine of this angle is:

$\frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}{}$

Compare your answer with the correct one above

Question

What is the value of in the triangle above? Round to the nearest hundredth.

Answer

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

Compare your answer with the correct one above

Question

In the triangle below, units and units. What is $\sin C$ ?

Answer

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

Compare your answer with the correct one above

Question

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Answer

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet.

Using the Pythagorean Theorem, we can solve for it:

$8^2+10^2=h^2{}$

$h=\sqrt{164}=2\sqrt{41}{}$

So, the sine of this angle is:

$\frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}{}$

Compare your answer with the correct one above

Question

What is the value of in the triangle above? Round to the nearest hundredth.

Answer

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

Compare your answer with the correct one above

Question

In the triangle below, units and units. What is $\sin C$ ?

Answer

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

Compare your answer with the correct one above

Question

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Answer

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet.

Using the Pythagorean Theorem, we can solve for it:

$8^2+10^2=h^2{}$

$h=\sqrt{164}=2\sqrt{41}{}$

So, the sine of this angle is:

$\frac{10}{2\sqrt{41}} = \frac{5}{\sqrt{41}}{}$

Compare your answer with the correct one above

Tap the card to reveal the answer

How to find the sine of a missing side - ACT Math

In the triangle below, units and units. What is ?

Because , we need to first find the length of BC. Using the Pythagorean theorem, .

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet. Using the Pythagorean Theorem, we can solve for it: So, the sine of this angle is:

What is the value of in the triangle above? Round to the nearest hundredth.

We know that the sine of an angle is: Therefore, we can write for this question: This allows us to solve for easily: Rounding, this gives us .

In the triangle below, units and units. What is ?

Because , we need to first find the length of BC. Using the Pythagorean theorem, .

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet. Using the Pythagorean Theorem, we can solve for it: So, the sine of this angle is:

What is the value of in the triangle above? Round to the nearest hundredth.

We know that the sine of an angle is: Therefore, we can write for this question: This allows us to solve for easily: Rounding, this gives us .

In the triangle below, units and units. What is ?

Because , we need to first find the length of BC. Using the Pythagorean theorem, .

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet. Using the Pythagorean Theorem, we can solve for it: So, the sine of this angle is:

What is the value of in the triangle above? Round to the nearest hundredth.

We know that the sine of an angle is: Therefore, we can write for this question: This allows us to solve for easily: Rounding, this gives us .

In the triangle below, units and units. What is ?

Because , we need to first find the length of BC. Using the Pythagorean theorem, .

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet. Using the Pythagorean Theorem, we can solve for it: So, the sine of this angle is:

What is the value of in the triangle above? Round to the nearest hundredth.

We know that the sine of an angle is: Therefore, we can write for this question: This allows us to solve for easily: Rounding, this gives us .

In the triangle below, units and units. What is ?

Because , we need to first find the length of BC. Using the Pythagorean theorem, .

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet. Using the Pythagorean Theorem, we can solve for it: So, the sine of this angle is:

What is the value of in the triangle above? Round to the nearest hundredth.

We know that the sine of an angle is: Therefore, we can write for this question: This allows us to solve for easily: Rounding, this gives us .

In the triangle below, units and units. What is ?

Because , we need to first find the length of BC. Using the Pythagorean theorem, .

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet. Using the Pythagorean Theorem, we can solve for it: So, the sine of this angle is:

What is the value of in the triangle above? Round to the nearest hundredth.

We know that the sine of an angle is: Therefore, we can write for this question: This allows us to solve for easily: Rounding, this gives us .

In the triangle below, units and units. What is ?

Because , we need to first find the length of BC. Using the Pythagorean theorem, .

A right triangle has leg lengths and . What is the sine of the angle opposite from the side of length ?

Using SOHCAHTOA, the sine of an angle is simply the length of the side opposite to it over the hypotenuse; however, we do not have the length of the hypotenuse yet. Using the Pythagorean Theorem, we can solve for it: So, the sine of this angle is:

In the triangle below, units and units. What is $\sin C$ ?

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

In the triangle below, units and units. What is $\sin C$ ?

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

In the triangle below, units and units. What is $\sin C$ ?

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

In the triangle below, units and units. What is $\sin C$ ?

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

In the triangle below, units and units. What is $\sin C$ ?

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

In the triangle below, units and units. What is $\sin C$ ?

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$

We know that the sine of an angle is:

$\frac{opposite}{hypotenuse}$

Therefore, we can write for this question:

$sin(50)=\frac{x}{14}$

This allows us to solve for easily:

Rounding, this gives us .

In the triangle below, units and units. What is $\sin C$ ?

Because $\sin C =\frac{AB}{BC}$ , we need to first find the length of BC.

Using the Pythagorean theorem,

$BC=\sqrt {AB^2+AC^2}$

$BC=\sqrt {6^2+2^2}$

$BC=\sqrt{36+4} = \sqrt {40} =\sqrt{4\cdot10}$

$BC=2\sqrt10$ .

$\sin C =\frac{AB}{BC}=\frac{2}{2\sqrt10}=\frac{1}{\sqrt10}$