Trigonometry: The diagram shows a rectangle ABCD inside a semicircle, centre O and radius 5cm, such that |<BOA| = |COD| = theta degrees

[reggiwilliams]

Had problems with this question particularly part two
I attach question and my rough workings to date
Fairly sure about part (I) but not sure about part 2
Please give me your thoughts

 

[Steven G]

Your first statement seems to be wrong. Why is the height 10 sin(theta)?

Angle ABO is NOT angle theta.

You claim that h=5. In a right triangle the hypotenuse, which is 5, is always the LONGEST side. This shows that h<5 and can't be 5.

 

[reggiwilliams]

I said height of rectangle was 5 sintheta
Given ABO is a right angled triangle I assumed angles ABO and AOB were equal
Could I ask you for your workings and solution so I can see where I have gone wrong

 

[Steven G]

I said height of rectangle was 5 sintheta
Given ABO is a right angled triangle I assumed angles ABO and AOB were equal
Could I ask you for your workings and solution so I can see where I have gone wrong

Please re-read your very first line.
In a right triangle, one angle is 90^o while the other two angles add up to 90^o. The other two angles do not have to be equal. For example, in a right triangle the angles can be 30-60-90 or 20-70-90...
The way it works here is you show us your work and then we guide you to the correct answer. Take into account what I said above and see what you can do with that.

 

[reggiwilliams]

Please re-read your very first line.
In a right triangle, one angle is 90^o while the other two angles add up to 90^o. The other two angles do not have to be equal. For example, in a right triangle the angles can be 30-60-90 or 20-70-90...
The way it works here is you show us your work and then we guide you to the correct answer. Take into account what I said above and see what you can do with that.

My first line said height of rectangle was 5 sin theta are you saying that is incorrect?
As far as my assumption that other two angles within right angled triangle were equal I agree that was incorrect
Before I proceed can you confirm from my workings I have got part one of question correct and if not steer me in right direction.
Thanks for your patience

 

[Steven G]

My first line said height of rectangle was 5 sin theta are you saying that is incorrect? Yes, it is incorrect as it does not say what you just wrote.
Why are you debating this with me? In you own handwriting at the very top of the page it says and I quote 'We know the height of the rectangle is 10 sin(theta)'. This is in image IMG_5994.jpeg. Just look at the first thing you wrote on the page. We need to get past this.

 

[reggiwilliams]

Sorry for misunderstanding
I agree height of rectangle AB or CD should be 5sin theta.
I believe length(AD) to be 10 cos theta
So area is 5 sin theta times 10 cos theta
Assuming this is correct not sure how I get to h

 

[The Highlander]

I said height of rectangle was 5 sintheta

No, you didn't; see page two of my amendments to your answers.

Given ABO is a right angled triangle I assumed angles ABO and AOB were equal

No, again, see my amendments (to p.2); the triangle is right-angled but not (necessarily) isosceles so you cannot just assume that it is!)
Has this error arisen because of the mistake I have corrected on p1 of your answers? (qv)

Could I ask you for your workings and solution so I can see where I have gone wrong

It seems to me that the question has been very badly composed!

You will note that "h" does not appear anywhere on the given diagram!

At first, I assumed "h" was probably meant to be the height of the rectangle (ABCD), as you seem to have done too but pursuing an answer to Part (ii) soon disabused me of that notion!

I think the question would have been made a lot clearer (and less error prone for students) if they had marked "h" as the height of the rectangle somewhere on the diagram but then worded Part (ii) as:-

"Find the value of x for which the area of the rectangle is x sin 2θ cm² " (⇒ x = 25)

If you had produced a diagram like the one I have modified the original to (below) you might have found the problem a lot simpler.

 

[reggiwilliams]

Thanks so much for your help
Agreed it would have been far less confusing if they had used x instead of h.

 

[The Highlander]

Thanks so much for your help
Agreed it would have been far less confusing if they had used x instead of h.

You could have saved yourself a lot of paper & ink too.