Question

URGENT HELP NEEDED WITH RIGHT TRIANGLES! (MULTIPLE PROBLEMS)!!! Please give an in depth explanation :D

Answer 1

Remember "SOH CAH TOA"

Sine = (Opposite/Hypotenuse)

Cosine = (Adjacent/Hypotenuse)

Tangent = (Opposite/Adjacent)

1st Problem:

Remember that cosine is (Adjacent/Hypotenuse). And also know that we are using inverse trig (regular trig solves for sides, inverse solves for angles.)

$cos^{-1}(\frac{17}{24}) =44.9005$

2nd Problem:

For this we are using Cosine (Adjacent/Hypotenuse) to solve for the adjacent side. Your equation will look like this: $\frac{cos(49)}{1} =\frac{x}{10}$

For this, all you need to do is cross multiply, and your answer will be $6.6=x$

3rd Problem:

So this is actually a special triangle. Because its an isosceles right triangle, it follows the 45-45-90 rule, which is if one of the legs is x, then the other leg is also x and the hypotenuse is x√2.

In this case, EF is going to be 16√2, or 22.63 cm, because 1. for the hypotenuse to not have √2 means that the leg does. Remember that √2(√2) = 2, which means that we would need to know what times 2 makes 32, which is 16.

4th Problem:

So for this we will be using the pythagorean theorem, which is $leg^2+leg^2=hypotenuse^2$ . In this case, the equation is going to be $15^2+19^2=GH^2$

Firstly, solve the exponents: $225+361=GH^2$

Next, combine 225 and 361 together: $586=GH^2$

Lastly, square root each side, and your answer should be $24.2=GH$

5th Problem:

So you'll be using the pythagorean theorem like before, except with new numbers and your solving for one of the legs.

$19^2+FG^2=25^2$

Like before, solve the exponents: $361+FG^2=625$

Next, subtract 361 on both sides: $FG^2=264$

Lastly, square root both sides, and your answer should be $FG=16.25$