How do I find missing angles in trigonometry and geometry?

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Discussion Overview

The discussion revolves around finding missing angles in trigonometry and geometry, specifically focusing on right triangles. Participants seek assistance with understanding the application of trigonometric functions to determine angles based on given side lengths.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how to start solving the problems related to missing angles.
  • Another participant suggests identifying the appropriate trigonometric function based on known side lengths (opposite and adjacent) for angles K and L.
  • There is a clarification regarding the identification of sides in a right triangle, with one participant stating the opposite and adjacent sides for angle L.
  • A participant provides the definitions of sine, cosine, and tangent functions, explaining their relevance to finding angles.
  • Another participant corrects the identification of sides, asserting that the opposite side for angle L is 10' 3" and the adjacent side is 5' 6".
  • Using the corrected side lengths, a participant calculates the tangent of angle L and expresses it in terms of fractions.
  • There is a discussion about how to input the calculated fraction into a calculator to find the angle for L.
  • A participant mentions the use of an online tool (Wolfram|Alpha) to find the angle based on the tangent value.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial steps to solve the problems, with some expressing uncertainty and others providing differing interpretations of the triangle's sides. The discussion remains unresolved regarding the best approach to finding the angles.

Contextual Notes

Some participants express confusion about the order of the sides and how to apply trigonometric functions, indicating a potential misunderstanding of the relationships in right triangles. There are also unresolved questions about calculator usage and input methods.

clhrhrklsr
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View attachment 2879
Someone please take a look at the attachment above. These two questions have me stumped and it ask on each question to find three missing angles. I don't even know where to start. Could someone be so kind and explain to me how to figure each of these. I need help with this asap to help my studying of trigonometry and geometry. Thanks
 

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View attachment 2880
 

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For angles $K$ and $L$, you know the sides opposite and adjacent...so which trig. function do you think will come into play here?
 
MarkFL said:
For angles $K$ and $L$, you know the sides opposite and adjacent...so which trig. function do you think will come into play here?

Honestly, I'm not sure. I don't even know where to begin or how to start this. I apologize.
 
clhrhrklsr said:
Honestly, I'm not sure. I don't even know where to begin or how to start this. I apologize.

Consider the following right triangle:

View attachment 2881

With regards to the angle $\theta$, we say that side $x$ is adjacent, side $y$ is opposite, and side $z$ is the hypotenuse (the longest side, which is opposite the $90^{\circ}$ angle).

Now, in trigonometry, we define the three primary trigonometric functions as follows:

$$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{y}{z}$$

$$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{z}$$

$$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{y}{x}$$

So, we can see that when we know the opposite and adjacent sides, it is the tangent function which we can use to find the angle in question. Actually, it is the inverse of the tangent function that we will use, but in order to know where to begin, we decide which of the sides relative to the angle we know.

So, let's look at angle $L$. What are the opposite and adjacent sides?
 

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  • triangletrig.png
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MarkFL said:
Consider the following right triangle:

View attachment 2881

With regards to the angle $\theta$, we say that side $x$ is adjacent, side $y$ is opposite, and side $z$ is the hypotenuse (the longest side, which is opposite the $90^{\circ}$ angle).

Now, in trigonometry, we define the three primary trigonometric functions as follows:

$$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{y}{z}$$

$$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{x}{z}$$

$$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{y}{x}$$

So, we can see that when we know the opposite and adjacent sides, it is the tangent function which we can use to find the angle in question. Actually, it is the inverse of the tangent function that we will use, but in order to know where to begin, we decide which of the sides relative to the angle we know.

So, let's look at angle $L$. What are the opposite and adjacent sides?

The opposite site is the 5' - 6". The adjacent side is the 10' 3". how would I input this into my hp 48g+?
 
clhrhrklsr said:
The opposite site is the 5' - 6". The adjacent side is the 10' 3". how would I input this into my hp 48g+?

You have those reversed...the side opposite angle $L$ is 10' 3" while the side adjacent is 5' 6". So, if we convert from mixed measures, and using feet as the measure, we find:

$$\text{opposite}=10.25=\frac{41}{4}$$

$$\text{adjacent}=5.5=\frac{11}{2}$$

And so we may write:

$$\tan(L)=\frac{\frac{41}{4}}{\frac{11}{2}}=\frac{41}{4}\cdot\frac{2}{11}=\frac{41}{22}$$

And this implies:

$$L=\tan^{-1}\left(\frac{41}{22}\right)$$

Make sure your calculator is is degree mode if you wish to get a result in degrees.

Now, we could do the same for angle $K$, but if we observe that angles $K$ and $L$ are complementary, that is their sum is $90^{\circ}$, we can just use:

$$K=90^{\circ}-L$$

So, what do you get?
 
MarkFL said:
You have those reversed...the side opposite angle $L$ is 10' 3" while the side adjacent is 5' 6". So, if we convert from mixed measures, and using feet as the measure, we find:

$$\text{opposite}=10.25=\frac{41}{4}$$

$$\text{adjacent}=5.5=\frac{11}{2}$$

And so we may write:

$$\tan(L)=\frac{\frac{41}{4}}{\frac{11}{2}}=\frac{41}{4}\cdot\frac{2}{11}=\frac{41}{22}$$

And this implies:

$$L=\tan^{-1}\left(\frac{41}{22}\right)$$

Make sure your calculator is is degree mode if you wish to get a result in degrees.

Now, we could do the same for angle $K$, but if we observe that angles $K$ and $L$ are complementary, that is their sum is $90^{\circ}$, we can just use:

$$K=90^{\circ}-L$$

So, what do you get?

how do you input that 41/22 in the calculator? How does that turn into the angle for L?
 
clhrhrklsr said:
how do you input that 41/22 in the calculator? How does that turn into the angle for L?

I can't tell you how your particular calculator operates, but the inverse tangent of 41/22 represents the angle in question. I can point you to a site that will give you the angle though

arctan'('41'/'22')' - Wolfram|Alpha

If you scroll down a little, you will see it gives the angle in degrees, rounded to two decimal places.
 

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