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What rules do I use to solve this triangle question? watch

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    Howdy,

    Unsure of how to get started on the below question. Name:  Triangle.jpg
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    I can find the missing angle as all angles add up to 180. Do I use the 6m and use the triangle to the right to find the details of the main triangle?

    Thanks
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    (Original post by jojo55)
    Howdy,

    Unsure of how to get started on the below question. Name:  Triangle.jpg
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    I can find the missing angle as all angles add up to 180. Do I use the 6m and use the triangle to the right to find the details of the main triangle?

    Thanks
    So what have you got so far? What is the question asking you to do?
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    (Original post by Muttley79)
    So what have you got so far? What is the question asking you to do?
    It asks to calculate the lengths of the sides A B and C.
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    (Original post by jojo55)
    Howdy,

    Unsure of how to get started on the below question. Name:  Triangle.jpg
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    I can find the missing angle as all angles add up to 180. Do I use the 6m and use the triangle to the right to find the details of the main triangle?

    Thanks
    What's the question? If it's asking you to find the length of the sides, you have to use trigonometry.
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    Is the angle next to the 50 going to be 40?
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    (Original post by jojo55)
    Is the angle next to the 50 going to be 40?
    Use trig on the triangle on the right-angled triangle to work out length B then just apply the sine rule on the main triangle.
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    (Original post by jojo55)
    It asks to calculate the lengths of the sides A B and C.
    OK - so how can you find the lengths in terms of 6m and the angles?
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    (Original post by jojo55)
    Howdy,

    Unsure of how to get started on the below question. Name:  Triangle.jpg
Views: 76
Size:  19.0 KB.

    I can find the missing angle as all angles add up to 180. Do I use the 6m and use the triangle to the right to find the details of the main triangle?

    Thanks
    So you find the missing angle. With that, you can find all the angles of the right-angled triangle.

    Then you find the hypotenuse by using the formula

    sin H/length of H = sin angle you found/6

    Once you find the hypotenuse, you use the same formula to find the other sides of that triangle. I hope you can make sense of that
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    (Original post by Muttley79)
    OK - so how can you find the lengths in terms of 6m and the angles?
    The angle next to the 50 seems to be 40, due to it making a right angle in total.

    So I use cos 40 = 6/h to find the length of B?

    This comes to 7.832443736...
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    (Original post by jojo55)
    The angle next to the 50 seems to be 40, due to it making a right angle in total.

    So I use cos 40 = 6/h to find the length of B?

    This comes to 7.832443736...
    You could have used sin 50 = 6/h .... how many dps do you require?

    OK now repeat for the other sides; one will need two 'bits' which can be added.
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    (Original post by jojo55)
    Howdy,

    Unsure of how to get started on the below question. Name:  Triangle.jpg
Views: 76
Size:  19.0 KB.

    I can find the missing angle as all angles add up to 180. Do I use the 6m and use the triangle to the right to find the details of the main triangle?

    Thanks
    You can.
    There's many way but either way i think you gonna have to use your calculator so some simple trig will do

    See a z shape? The angle at the top of the "extra" drawn in triangle is also 50° and from there if you drawn a line from the main triangle to make 2 smaller right angled triangles the small right angled triangle on the right is 40°

    If course you'll have to use the drawn in triangle to fine some sides using sin and cos but other than that you should have enough angles and info to work stuff out ^-^
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    (Original post by will'o'wisp)
    You can.
    There's many way but either way i think you gonna have to use your calculator so some simple trig will do

    See a z shape? The angle at the top of the "extra" drawn in triangle is also 50° and from there if you drawn a line from the main triangle to make 2 smaller right angled triangles the small right angled triangle on the right is 40°

    If course you'll have to use the drawn in triangle to fine some sides using sin and cos but other than that you should have enough angles and info to work stuff out ^-^
    (Original post by Muttley79)
    You could have used sin 50 = 6/h .... how many dps do you require?

    OK now repeat for the other sides; one will need two 'bits' which can be added.
    (Original post by pinkisthefloyd)
    So you find the missing angle. With that, you can find all the angles of the right-angled triangle.

    Then you find the hypotenuse by using the formula

    sin H/length of H = sin angle you found/6

    Once you find the hypotenuse, you use the same formula to find the other sides of that triangle. I hope you can make sense of that
    (Original post by RDKGames)
    Use trig on the triangle on the right-angled triangle to work out length B then just apply the sine rule on the main triangle.
    Thanks to everyone that helped .

    I can now see the z shape making an alternate angle and that I can use trig. Also, I can see the right angle that the 50 makes with the other triangle, thus getting a 40 degree angle. I can use the SINE rule to get the other two sides, which I am comfortable using.

    Thanks so much and have a lovely Sunday.
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    (Original post by RDKGames)
    Use trig on the triangle on the right-angled triangle to work out length B then just apply the sine rule on the main triangle.
    You don't need the sine rule at all - just use trig on the two right angles triangles.
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    (Original post by jojo55)
    Thanks to everyone that helped .

    I can now see the z shape making an alternate angle and that I can use trig. Also, I can see the right angle that the 50 makes with the other triangle, thus getting a 40 degree angle. I can use the SINE rule to get the other two sides, which I am comfortable using.

    Thanks so much and have a lovely Sunday.
    You don't need the sine rule at all - just use trig on the two right angles triangles.
    The third side is just the sum of two 'bits'.
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    (Original post by Muttley79)
    You don't need the sine rule at all - just use trig on the two right angles triangles.
    The third side is just the sum of two 'bits'.
    Cheers. I've just done the sine rule. I am checking my answers with the trig rule now.
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    (Original post by Muttley79)
    You don't need the sine rule at all - just use trig on the two right angles triangles.
    I know, that's just what I initially thought when I first saw it. Either approach works.
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    (Original post by RDKGames)
    I know, that's just what I initially thought when I first saw it. Either approach works.
    I know but the trig approach is quicker.
 
 
 
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