Question

Consider the right triangle ABC given below:

A. What is the length of the hypotenuse of triangle ABC?

B. What is the length of the shorter leg of triangle ABC?

C. What is the length of the longer leg of triangle ABC?

Answer 1

I do not know what you mean by the wording of some of these questions, so I will just give you the formula and tell you how to apply it. The formula of the Pythagorean Theorem is a^2 + b^2 = c^2. a and b are the two sides adjecent to the right angle (in the corner). c is the side opposite the right angle. So, to get c you would have to add sides a and b but with each of them to the second power. Then, you would find the square root of the sum (you can do this with a calculator). To find sides a or b, subtract a or b (whichever one available) to the second power from c to the second power. Then find the square root of the sum. That is all I know how to do, so combine this knowledge with yours to see if you can solve it. If you have any questions about what I already know, feel free to ask. You can also ask for an example. If you have a question about something that I have not explained, look it up on the internet, or hopefully you can find someone who can help you. I hope this helps.

Answer 2

Part A) The length of the hypotenuse of triangle ABC is

`AB=(40√(3))/(3)\ units`

Part B) The length of the shorter leg of triangle ABC is

`AC=(20√(3))/(3)\ units`

Part C) The length of the longer leg of triangle ABC is

`BC=20\ units`

Explanation:

see the attached figure to better understand the problem

Step 1

Find the length of the longer leg of triangle ABC (side BC)

we know that

In the right triangle DBC

`sin(30\°)=(DC)/(BC)`

substitute the values and solve for BC

`sin(30\°)=(10)/(BC)`

`BC=10/sin(30\°)=20\ units`

Step 2

Find the length of the shorter leg of triangle ABC (side AC)

we know that

In the right triangle ACD

`sin(60\°)=(DC)/(AC)`

substitute the values and solve for AC

`sin(60\°)=(10)/(AC)`

`AC=10/sin(60\°)=(20√(3))/(3)\ units`

Step 3

Find the length of the hypotenuse of triangle ABC (side AB)

Applying the Pythagoras theorem

`AB^(2)=BC^(2)+AC^(2)`

we have

`AC=(20√(3))/(3)\ units`

`BC=20\ units`

substitutes

`AB^(2)=20^(2)+((20√(3))/(3))^(2)`

`AB^(2)=400+((400)/(3))`

`AB^(2)=((1600)/(3))`

`AB=(40√(3))/(3)\ units`