Question

The picture below shows a box sliding down a ramp:A right triangle ABC has measure of angle ABC equal to 90 degrees and measure of angle ACB equal to 65 degrees. The length of AB is 12 feet.What is the distance, in feet, that the box has to travel to move from point A to point C? 12 divided by sec 65 degrees 12 cosec 65° 12 sin 65° 12 divided by cot 65 degrees

Answer 1

The distance that the box has to travel to move from point A to point C is equal to the length of side AC of the right triangle ABC. Since angle ACB is 62 degrees and side AB is opposite this angle, we can use the definition of sine to find the length of side AC. The sine of an angle in a right triangle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. So, we have:

$$\sin(62°)=\frac{AC}{AB}$$

Substituting the known values, we get:

$$\sin(62°)=\frac{AC}{10}$$

Solving for AC, we get:

$$AC=10\sin(62°)$$

So, the distance that the box has to travel to move from point A to point C is **10 sin 62°** feet.

Answer 2

The length of the hypotenuse of the triangle is the solution to the question.

The length of the hypotenuse can be found using the Sine Trigonometric Ratio given to be:

`\sin\theta=(opp)/(hyp)`

From the image, we have the following parameters:

`\begin{gathered} \theta=65\degree \\ opp=12 \\ hyp=AC \end{gathered}`

Therefore, we have:

`\sin65=(12)/(AC)`

To solve for AC, we can cross-multiply:

`AC=(12)/(\sin65)`

Recall the identity:

`\cosec x=(1)/(\sin x)`

Therefore, the equation will be:

`AC=12\cosec65`

The SECOND OPTION is correct.