Turn on thread page Beta
    • Thread Starter
    Offline

    18
    ReputationRep:
    Hey everyone I'm back with another question (hopefully it's as good as my others). I hope you like it and can give me feedback below. Feel free to attempt in and please spoiler your answers. This is aimed at Grade 9 students but anyone is welcome. I'd probably give it 7 marks because it's slightly lengthy. This question is calculator. Anyways I hope you enjoy and have a great day! :hello:

    Offline

    15
    ReputationRep:
    (Original post by Y11_Maths)
    Hey everyone I'm back with another question (hopefully it's as good as my others). I hope you like it and can give me feedback below. Feel free to attempt in and please spoiler your answers. This is aimed at Grade 9 students but anyone is welcome. I'd probably give it 7 marks because it's slightly lengthy. This question is calculator. Anyways I hope you enjoy and have a great day! :hello:

    Not sure but is it these?
    Spoiler:
    Show

    i) 414 square miles
    ii) 40.3
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by 3pointonefour)
    Not sure but is it these?
    Spoiler:
    Show

    i) 414 square miles
    ii) 40.3
    Spoiler:
    Show
    Part I) is correct however ii) is incorrect
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    15
    ReputationRep:
    (Original post by Y11_Maths)
    Spoiler:
    Show

    Part I) is correct however ii) is incorrect
    Spoiler:
    Show

    Is part ii) 50.8?
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by 3pointonefour)
    Spoiler:
    Show

    Is part ii) 50.8?
    Spoiler:
    Show
    No it isn’t. You’re quite far away from it. Would you like a hint?
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    15
    ReputationRep:
    (Original post by Y11_Maths)
    Spoiler:
    Show

    No it isn’t. You’re quite far away from it. Would you like a hint?
    I'll get there in the end
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by 3pointonefour)
    I'll get there in the end
    Perseverance is key. Side note: It also took me longer to find the angle because I couldn’t spot it haha. I was bashing myself in the head because I couldn’t find it for ages lol
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    15
    ReputationRep:
    (Original post by Y11_Maths)
    Perseverance is key. Side note: It also took me longer to find the angle because I couldn’t spot it haha. I was bashing myself in the head because I couldn’t find it for ages lol
    Spoiler:
    Show

    Not sure where I went wrong:

    CG = 22.8034
    GD = 4
    Angle CGD = 43
    So used cosine rule to get CD = 20.0643
    Then got a right-angled triangle with the perpendicular length being 15.5519, and CD.
    Used sin to get theta = 50.8?
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by 3pointonefour)
    Spoiler:
    Show

    Not sure where I went wrong:

    CG = 22.8034
    GD = 4
    Angle CGD = 43
    So used cosine rule to get CD = 20.0643
    Then got a right-angled triangle with the perpendicular length being 15.5519, and CD.
    Used sin to get theta = 50.8?
    Spoiler:
    Show
    hmm let me have a think. I’ve got 129 degrees as I used a different method. I worked out angle GCD and subtracted it from 137. Bear with me
    Edit: If you think about it, if both methods are legitimate, can ask the student to find both possible value of angle BCD? Which would make it better lol
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    15
    ReputationRep:
    (Original post by Y11_Maths)
    Spoiler:
    Show

    hmm let me have a think. I’ve got 129 degrees as I used a different method. I worked out angle GCD and subtracted it from 137. Bear with me
    Spoiler:
    Show

    Yeah I know where you went wrong. I got that as part of my working. It can't be 129 since the angle is clearly acute. Here's how it works:

    If you look at the graph of sin(x), you see it is positive from 0 to 180
    This means there's 2 possible solutions for angles in a triangle.
    For example, both sin(30) and sin(150) = 0.5
    The rule is sin(x) = sin(180-x).
    So you notice that my answer (51) = 180 - your answer (129)?
    It's because we both used sine, but we took different roots.

    https://brilliant.org/wiki/sine-rule...ambiguous-case
    Offline

    15
    ReputationRep:
    (Original post by Y11_Maths)
    Spoiler:
    Show

    hmm let me have a think. I’ve got 129 degrees as I used a different method. I worked out angle GCD and subtracted it from 137. Bear with me

    Edit: If you think about it, if both methods are legitimate, can ask the student to find both possible value of angle BCD? Which would make it better lol
    Spoiler:
    Show

    Try finding out the remaining angles of triangle CGD. I don't think it adds to 180 using your angle.
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by 3pointonefour)
    Spoiler:
    Show

    Yeah I know where you went wrong. I got that as part of my working. It can't be 129 since the angle is clearly acute. Here's how it works:

    If you look at the graph of sin(x), you see it is positive from 0 to 180
    This means there's 2 possible solutions for angles in a triangle.
    For example, both sin(30) and sin(150) = 0.5
    The rule is sin(x) = sin(180-x).
    So you notice that my answer (51) = 180 - your answer (129)?
    It's because we both used sine, but we took different roots.

    https://brilliant.org/wiki/sine-rule...ambiguous-case
    Spoiler:
    Show
    Aah thank you so much that’s a fantastic explanation! I should’ve know it couldnt be obtuse haha. How did you find it? Any feedback?
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    15
    ReputationRep:
    (Original post by Y11_Maths)
    Spoiler:
    Show

    Aah thank you so much that’s a fantastic explanation! I should’ve know it couldnt be obtuse haha. How did you find it? Any feedback?
    Spoiler:
    Show

    Well I think it's one of the best questions you've made! Like your other questions, it's too hard and doesnt give enough hints for GCSE (Its also maybe too long to be put with 20 other GCSE questions lol), but it's great to give as a challenge. Very nice .
    I'd just say the 27.2799344 side just seems a bit dodgy and out of place. Can you maybe round it to still get the side of the rhombus as 40?
    But otherwise, well done! Great test of problem solving and trigonometry!
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by 3pointonefour)
    Spoiler:
    Show

    Well I think it's one of the best questions you've made! Like your other questions, it's too hard and doesnt give enough hints for GCSE (Its also maybe too long to be put with 20 other GCSE questions lol), but it's great to give as a challenge. Very nice .
    I'd just say the 27.2799344 side just seems a bit dodgy and out of place. Can you maybe round it to still get the side of the rhombus as 40?
    But otherwise, well done! Great test of problem solving and trigonometry!
    Spoiler:
    Show
    Thank you so much. Yes I dislike the height I used but I originally gave the side of 40 but turns out it’s better if I make them work it out themselves. Thank you for completing it and giving me feedback. Have an awesome day!
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    6
    ReputationRep:
    Can someone please explain this to me? I've been getting 8s and 9s but I've never seen a question as bad as this - thank you!
    • Thread Starter
    Offline

    18
    ReputationRep:
    (Original post by IReckonItsDanyal)
    as bad as this
    Haha.
    Spoiler:
    Show
    Start by using the perpendicular height and a bit of trig to work out the length of the rhombus since they’re all the same side. Then use the cosine rule to work out the length of the diagonal and subtract 32 to get one length of the trapezium. It’s all about spotting triangles and using repeated sine, consine rules and trig.
    Posted on the TSR App. Download from Apple or Google Play
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: June 2, 2018

University open days

  1. Loughborough University
    General Open Day Undergraduate
    Fri, 21 Sep '18
  2. University of Cambridge
    Churchill College Undergraduate
    Fri, 21 Sep '18
  3. Richmond, The American International University in London
    Undergraduate Open Day Undergraduate
    Fri, 21 Sep '18
Poll
Which accompaniment is best?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.