Finding The Measure(s) Of Different Angles

[XxHelpMePlzxX]

I usually am pretty good at these, but it's really late for me, and my ADHD isn't helping. The problem I'm having trouble with is below:

The measures of two angles have the ratio of 5:2. The larger of the angles is 30 more
than half the difference of the angles. Find the measure of each angle.

Thank you so much!
-Zander

 

[Harry_the_cat]

Let the larger angle be X and the smaller angle be Y. Can you come up with 2 equations to solve simultaneously?

 

[The Highlander]

I usually am pretty good at these, but it's really late for me, and my ADHD isn't helping. The problem I'm having trouble with is below:

The measures of two angles have the ratio of 5:2. The larger of the angles is 30 more
than half the difference of the angles. Find the measure of each angle.

Thank you so much!
-Zander

Hi,

Although what "Harry" says (above) is perfectly valid, I would suggest it might be easier to take the larger angle as \(\displaystyle y\) and the smaller as \(\displaystyle x\) (they do everything upside down in Australia you see ?).

That would then give you: \(\displaystyle 5x=2y\) (ie: \(\displaystyle 2y=5x\)) ‘straight away’ (from the ratio).

To help you out with the other piece of information provided, you should get this:-

\(\displaystyle \frac{(y-x)}{2}+30=y\\  \Rightarrow 2y=y-x+60\\\Rightarrow y=60-x  \)​

Now that you have two equations (of the form \(\displaystyle ay=bx+C\)) you should be able to solve these simultaneously and come up with values for the measures of each of the angles.

Be warned: the answers will involve improper fractions (or mixed numbers; which would be preferable) involving sevenths (just a wee hint ?).

 

[XxHelpMePlzxX]

Hi,

Although what "Harry" says (above) is perfectly valid, I would suggest it might be easier to take the larger angle as \(\displaystyle y\) and the smaller as \(\displaystyle x\) (they do everything upside down in Australia you see ?).

That would then give you: \(\displaystyle 5x=2y\) (ie: \(\displaystyle 2y=5x\)) ‘straight away’ (from the ratio).

To help you out with the other piece of information provided, you should get this:-

\(\displaystyle \frac{(y-x)}{2}+30=y\\  \Rightarrow 2y=y-x+60\\\Rightarrow y=60-x  \)​

Now that you have two equations (of the form \(\displaystyle ay=bx+C\)) you should be able to solve these simultaneously and come up with values for the measures of each of the angles.

Be warned: the answers will involve improper fractions (or mixed numbers; which would be preferable) involving sevenths (just a wee hint ?).

Alright, thank you so much!

 

[The Highlander]

Alright, thank you so much!

Do let us know how you get on.
Post your own working(s) & results?