Please help with Trig problems

[mshepstone]

Please determine the area of the Quadrilateral attached...(WITHOUT CALCULATOR) we know it can be done with Sin/Cos/Tan but this question appeared in a school maths olympiad and was wondering what the simple solution would look like?

 

[Steven G]

What have you tried? Where are you stuck? Can you improve labeling the figure?

You can use a calculator to get some/all the values and then try to figure out how to get those values without a calculator.

 

[Deleted member 4993]

Please determine the area of the Quadrilateral attached...(WITHOUT CALCULATOR) we know it can be done with Sin/Cos/Tan but this question appeared in a school maths olympiad and was wondering what the simple solution would look like?

View attachment 21569

I do not see a"clever" solution to the problem without a scientific calculator.

 

[jonah2.0]

Beer soaked request follows

Please determine the area of the Quadrilateral attached...(WITHOUT CALCULATOR) we know it can be done with Sin/Cos/Tan but this question appeared in a school maths olympiad and was wondering what the simple solution would look like?

View attachment 21569

We also know that but we would also first like to see what you've done with Sin/Cos/Tan.

 

[Dr.Peterson]

I don't see an elegant solution either; the best I can think of would be to express everything in terms of trig functions and hope the area simplified to something that doesn't need a calculator.

I did construct the figure in GeoGebra, and the area appears to be a nice exact integer. I can't see how that could be, but it does encourage the assumption that an exact calculation is possible, which is the first step in finding one.

 

[Cubist]

I just found the simple way of doing this (it took a long time for me to see it).

I drew a picture to post here, but then I decided that it gives too much away. So I'll give a subtle hint...

35°+10°=? can you redraw the shape, keeping the same area, to take advantage of this?

 

[Dr.Peterson]

That's the perfect hint! Suddenly it becomes obvious.

 

[jonah2.0]

Beer soaked declaration follows.

I just found the simple way of doing this (it took a long time for me to see it).

I drew a picture to post here, but then I decided that it gives too much away. So I'll give a subtle hint...

35°+10°=? can you redraw the shape, keeping the same area, to take advantage of this?

I love you Cubist!
I'd buy you a beer (or a six pack) if you lived near me.

 

[joshuaa]

area [MATH]= 200 \ cm^2[/MATH]