Exploring Trigonometric Ratios with SOH CAH TOA

The video tutorial introduces the SOHCAHTOA acronym, which helps remember trigonometric ratios for sine, cosine, and tangent in right triangles. It explains how to calculate these ratios using examples of right triangles, focusing on the relationships between the sides and angles. The tutorial also demonstrates solving problems by applying these ratios to find missing sides in triangles, providing practical examples and tips for efficient calculations.

What does the 'S' in SOHCAHTOA stand for?

• Sine

• Segment

• Solution

• Scalar

In the context of a right triangle, what does cosine represent?

• Opposite over hypotenuse

• Opposite over adjacent

• Adjacent over opposite

• Adjacent over hypotenuse

How is the tangent of an angle in a right triangle calculated?

• Hypotenuse over opposite

• Hypotenuse over adjacent

• Adjacent over opposite

• Opposite over adjacent

In trigonometry, what does the acronym SOHCAHTOA help you remember?

• The names of trigonometric functions

• The formulas for calculating angles

• The method for solving linear equations

• The relationships between the sides of a right triangle

If the hypotenuse of a right triangle is 13 and the side opposite to the angle is 5, what is the sine of the angle?

• 13/5

• 5/13

• 12/13

• 5/12

For angle B in a right triangle, if the adjacent side is 5 and the hypotenuse is 13, what is the cosine of angle B?

• 12/13

• 5/12

• 13/5

• 5/13

Using SOHCAHTOA, how would you find the tangent of an angle in a right triangle?

• Adjacent over hypotenuse

• Opposite over hypotenuse

• Opposite over adjacent

• Hypotenuse over opposite

To solve for a missing side in a right triangle using the tangent ratio, which sides do you need?

• Opposite and adjacent

• Adjacent and hypotenuse

• Opposite and hypotenuse

• None of the above

When solving for a side using the cosine ratio, what sides of the triangle are involved?

• Adjacent and hypotenuse

• Opposite and adjacent

• Opposite and hypotenuse

• None of the above

If the sine of a 50-degree angle is given as 4 divided by Y, how do you solve for Y?

• sin(50) = Y / 4

• Y = 4 / sin(50)

• sin(50) = 4 * Y

• Y = 4 * sin(50)