Solving Right Triangles│Please Help!

Question

Find the missing pieces of the triangle. Round to the nearest tenth.

AC = ?

<A = ?

<B = ?

Image link - https://ibb.co/7vfdnKf

Answer 1

Hi Candi,

Let's break it down by each topic you'd want an answer to. So first, it is side AC of the right triangle.

AC:

We can use the Pythagorean Theorem that is A² + B² = C²

So, let's plug in what we know and determine what we're trying to find:

BC = 13 = A

AB = 19 = C

AC = ? = B

Remember: The Hypotenuse will ALWAYS be represented as C (or easier way to remember what C is...the non straight side of the triangle!)

So...we can substitute A and C in our Theorem and we will get the below:

13² + B² = 19²

169 + B² = 361

-169 + B² = -169

B² = 192

Sqrt(B²) = Sqrt(192)

B ~= 13.9

So, since we said that B = AC... Side AC ~= 13.9

Next, let's find <A

<A:

To find <A, let's use the law of sines which is defined as:

sin(any angle)/side of opposite, examples would be Sin(<A)/A = Sin(<B)/B = Sin(<C)/C

so let's substitute these values into the below equation:

Sin(90)/19 (since these are given to us!) = Sin(<A)/13 (see definition of law of sines above!)

Now...let's substitute and solve!

Sin(90) = 1 (Use the Radius Circle for this!)

19 = Side opposite of Sin(90) in triangle

1/19 = Sin(<A)/13

*13 *13

Sin(<A) = 13/19

In order to find angles, we take the inverse or "arcsin" of what Sin(<A) is equal to...so on a calculator make sure the Sin^-1 is set to DEG and not RAD

Sin^-1(<A) = 13/19 OR Sin^-1(<A) ~= 43.2 degrees OR <A ~= 43.2 Degrees

Finally (the easiest part with what we know now) let's find <B

<B:

Remember: Triangles have a TOTAL of 180 degrees...so we can define a new equation as below

<A + <B + <C = 180 degrees

We know <A AND <C so let's substitute the above equation!

43.2 + <B + 90 = 180

133.2 + <B = 180

-133.2 -133.2

<B = 46.8 Degrees

Answer 2

Candi,

It is always smart to check your answer just in case you made a mistake somewhere along the way.

In this case, it is easy to do so.

Since sin(90º)/19 = sin(∠B)/B,

sin(∠B) = 1* √(19² - 13²) / 19 ≈ 0.7292845506

And then you can take the inverse sine of both sides.

Or to avoid any loss of accuracy, you can directly determine

sin^-1 [ √( 19² - 13² ) / 19 ] ≈º 46.8264488927º

So m∠B = 46.8º to 1 decimal place, thus verifying the answer derived through an alternate method.

By the way, I would always be careful when using any method that loses accuracy before determining the final answer. SAT, for example, typically requires 3 decimal places of accuracy.

Answer 3

Hello,

We can use Pythagorean Thm to find the last missing side AC.

AC^2 + 13^2 = 19^2 => AC = root ( 19^2 - 13^2) = root (192) = 13.85640646 ~~13.9

Now we can use law of sine to find angle A or angle B.

Sin(90)/19 = sin(A)/13

1/19 = sin(A) / 13 => Sin(A) = 13/19 => m<A = arcsin(13/19) = 43.17355 ~ 43.2 deg

[Make sure that you use Deg, not Rad]

Now, we can find m<B => m<A + m<B + 90 = 180 => 43.2 + m<B + 90 = 180

=> m<B = 180 - 133.2 = 46.8 deg

So, AC = 13.9, m<A = 43.2 deg, m<B = 46.8 deg