MHB How do I find missing angles in trigonometry and geometry?

[clhrhrklsr]

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

 

[MarkFL]

For the convenience of our helpers, I have combined the images and resized a bit so that no one has to open an attachment:

View attachment 2880

 

[MarkFL]

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

 

[clhrhrklsr]

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.

 

[MarkFL]

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?

 

[clhrhrklsr]

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+?

 

[MarkFL]

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?

 

[clhrhrklsr]

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?

 

[MarkFL]

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.