HI,

I AM AFRAID THE ATTACHED QUESTION, WHICH I HOPE YOU CAN OPEN, HAS ME STUMPED A LITTLE.

I WOULD WELCOME ANY SUGGESTIONS THAT YOU MAY HAVE.

THANKING YOU IN ADVANCE AND HAVE A GOOD WEEKEND,

SHANE.

THE ANSWER TO PART (ii) OF THE QUESTION IS

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- Aug 2nd 2013, 09:36 AMSeaniboyGeometry question
HI,

I AM AFRAID THE ATTACHED QUESTION, WHICH I HOPE YOU CAN OPEN, HAS ME STUMPED A LITTLE.

I WOULD WELCOME ANY SUGGESTIONS THAT YOU MAY HAVE.

THANKING YOU IN ADVANCE AND HAVE A GOOD WEEKEND,

SHANE.

THE ANSWER TO PART (ii) OF THE QUESTION IS - Aug 2nd 2013, 10:18 AMHallsofIvyRe: Geometry question
A few tips for the future- do NOT use all caps- it makes it look like you are "yelling". As much as possible, type what your problem is- many people will not open attached files- there is always a danger of a computer virus. Finally, if this problem has stumped you "a little" then you must have

**tried**something. Show what you have tried so we will know better what hints will help.

Now, as for this problem- you have two right triangles where the hypotenuse of one is a leg of the other. The result you want involves cosines of two of the angles. I would start by assigning labels to each side, write out the definition of "cosine" in terms of the sides and**stare**at them until inspiration hits. - Aug 3rd 2013, 04:32 AMbjhopperRe: Geometry question
Hello Seanboy,

Do you know the properties of an isosceles right triangle and those of a 30-60 right triangle? You should and the answer is a one minute snap. - Aug 3rd 2013, 04:40 AMSeaniboyRe: Geometry question
Hello,

I take your meaning.

Thanks for taking the time to reply.

Shane. - Aug 3rd 2013, 05:02 AMSeaniboyRe: Geometry question
Hello,

If we want to get pedantic, by "a little", I actually meant "a lot".

I feel like a schoolboy who has just been chastised by a teacher.

As I wipe the tears away, I would like to thank you for taking the time to reply. (Giggle)

Your advice is well taken and much appreciated.

THANK YOU,

Shane. - Aug 3rd 2013, 05:39 AMibduttRe: Geometry question
- Aug 3rd 2013, 06:34 AMHallsofIvyRe: Geometry question
- Aug 3rd 2013, 06:57 AMSeaniboyRe: Geometry question
Wipe the teacher away - It may be a while before I am up to that level.

Yes, I worked out the answer, thanks. I only got back to work out the question after I had received the three replies.

All very helpful and insightful. You kindly gave me the template and Ibdutt filled in the blanks.

As Bjhopper pointed out for the second part of the question, it is surprising how I let slip the qualities of 45-45-90 and 30-60-90 triangles.

Thanks once again for your help. - Aug 3rd 2013, 06:59 AMSeaniboyRe: Geometry question
Thanks so much for your reply and for working the problem out for me.

- Aug 6th 2013, 05:43 PMbjhopperRe: Geometry question
Hello Shane,

I am sorry that my reply upset you. All I wanted to tell you and other parties is that solution of special rt triangles is easy if you know the rules.Using Pythagorean theorem or trig is not best.Following are my solutions to part 1 and 2.

Part 1 Label triangles as follows 45 deg tri a,a,c other rt tri c,d,b b is hypot and d is opposite angle y

c = a rad2 property of 45 deg rt tri

cosx =1/2 c /a adj over hyp = 1/2 a rad2/a =rad2/2

cos y =c/b adj over hyp=arad2/b

cosx*cosy = rad2/2*arad2/b = a/b

b= a/cosx*cosy

Part 2

c= arad2

d= arad2/rad3 property of 30 deg rt tri

area pqs c*d*1/2 =(arad2)^2/2rad3=2a^2/2rad3=a^2/rad3 = a^2*rad3 /rad3^2 =a^2rad3/3