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Math Help - 2 trig Questions

  1. #1
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    2 trig Questions

    Need help with the following which is due to tomorrow.

    1. A triangular piece of wood has side lenghts 14cm, 19cm and 30cm. The angle between the 14cm and 19cm sides need to be known to see if it will fit properly into a cabinet. What is the angle, accurate to one decimalplace, between the 14cm and 19 cm sides?

    I dont even know how to start here which side should be 14 cm, which 19 and which 30 ?

    2. A hot air balloon is to be attached to the grond using two support ropes. One support is attached on one side of the balloon at an angle of elevation of 42 degrees. Another support is attached on the other side of the balloon at an angle of elevation of 50 degrees. The distance between the locations the supports are attached on the ground is 40m.

    a) How long is the rope that is attached at the 42degrees ?

    b) How high is the balloon.

    Here I dont know how to get the answer for a and b.
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  2. #2
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    Quote Originally Posted by Serialkisser View Post
    Need help with the following which is due to tomorrow.

    1. A triangular piece of wood has side lenghts 14cm, 19cm and 30cm. The angle between the 14cm and 19cm sides need to be known to see if it will fit properly into a cabinet. What is the angle, accurate to one decimalplace, between the 14cm and 19 cm sides?

    I dont even know how to start here which side should be 14 cm, which 19 and which 30 ?

    here you can use the cosine formula

    The Cosine Rule
    c = a + b - 2abcosC
    Last edited by sumit2009; January 19th 2009 at 10:25 PM. Reason: typing error
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  3. #3
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    Quote Originally Posted by Serialkisser View Post
    Need help with the following which is due to tomorrow.


    2. A hot air balloon is to be attached to the grond using two support ropes. One support is attached on one side of the balloon at an angle of elevation of 42 degrees. Another support is attached on the other side of the balloon at an angle of elevation of 50 degrees. The distance between the locations the supports are attached on the ground is 40m.

    a) How long is the rope that is attached at the 42degrees ?

    b) How high is the balloon.

    Here I dont know how to get the answer for a and b.

    See the attached dig.
    solve the eq. to find X and 'h'
    Attached Thumbnails Attached Thumbnails 2 trig Questions-untitled.jpg  
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  4. #4
    Junior Member AlvinCY's Avatar
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    I live in Sydney, Australia.
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    1. A triangular piece of wood has side lenghts 14cm, 19cm and 30cm. The angle between the 14cm and 19cm sides need to be known to see if it will fit properly into a cabinet. What is the angle, accurate to one decimal place, between the 14cm and 19 cm sides?

    Ok, doesn't matter which is 14, 19 or 30, is a triangle, the side with 14 will touch BOTH the sides of length 19 and 30, so you're guaranteed that there will be an angle in between.

    Draw any triangle and label the sides 14, 19 and 30.

    Let the angle between 14 and 19 be A

    Using the cosine formula:

    cosA=\frac{c^2+b^2-a^2}{2bc}, we get:

    cosA=\frac{19^2+14^2-30^2}{2\times14\times19}

    cosA=-\frac{343}{532}

    cosA=-\frac{49}{76}

    A=cos^{-1}(-\frac{49}{76})

    A=130.1^\circ


    2. A hot air balloon is to be attached to the ground using two support ropes. One support is attached on one side of the balloon at an angle of elevation of 42 degrees. Another support is attached on the other side of the balloon at an angle of elevation of 50 degrees. The distance between the locations the supports are attached on the ground is 40m.

    First draw a triangle:
    xxxxxxxxxxxxxxx*
    xxxxxxxxxxxx* 78 *
    xxxxxxxb *xxxxxxxxx* c
    xxxxxx*xxxxxxxxxxxxxx*
    xxx*42xxxxxxxxxxxxxx50*
    ********************
    xxxxxxxxxxx40

    *ignore the x's if you see them*

    First of all we know the TOP angle is 180 - 42 - 50 = 78 degrees

    a) How long is the rope that is attached at the 42degrees ?

    Using SINE RULE:

    \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}

    \frac{40}{sin78^\circ}=\frac{b}{sin50^\circ}

    b=\frac{40\times sin50^\circ}{sin78^\circ}

    b=31.3m

    b) How high is the balloon.

    Using the SINE AREA RULE:

    A=\frac{1}{2}absinC

    A=\frac{1}{2}\times 14\times 19 sin42^\circ

    A=88.99m^2 (store this in your calculator)

    Using the NORMAL triangle AREA FORMULA:

    A=\frac{1}{2}bh

    Here, let b = the ground distance = 40

    88.99=\frac{1}{2}\times 40\times h

    88.99=20\times h

    h=88.99\div 20

    h=4.45m

    So, to 1 decimal place, the hot air balloon is 4.5m off the ground.
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