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Thread: Trig program - 1 angle 1 side - head against a wall

  1. #1
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    Trig program - 1 angle 1 side - head against a wall

    I can't figure out why the answer sheet says the correct answer is B. There's a diagram of the triangle but I can't copy and paste it. I added one that I made to hopefully help. Please help me figure this out! Thanks

    The vertical support in the center of the suspension bridge in the diagram is 50 feet tall. The angle of elevation of the top of the support from either end of the bridge is 35. Which of the following is the distance across the bridge from point A to point B?

    diagram.jpg
    *Trying to find distance from A to B

    50 feet height
    A 35 35 B

    A. 122.1 ft
    B. 142.8 ft
    C. 174.3 ft
    D. 211.1 ft


    To go through my thinking, with the two 35 degree angles, I know the third angle is 110, but with the 50 height split between them, I use tan55 to try find the opposite side (half distance between A to middle of "vertical support" and then double it to get the entire distance between point A and B) But when I try these options, I keep getting negative answers or D. I've tried sin35, cos55, getting the hypotenuse.. ugh. I know it's a simple mistake. Please help if you can. Thanks.
    Last edited by HCL; 05-18-2013 at 02:34 AM. Reason: add diagram

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    The base of an isosceles triangle (where the top angle is the odd one out) is 2 times the hypotenuse times the cosine of the matching angles (since the hypotenuse times cosine of the angle is the projection of the hypotenuse onto the horizontal axis). If H is the hypotenuse of one of the right triangles, O the opposite side, and B the base then H*sin(35)=O=50, H*cos(35)=B. Use the first equation to get H, and once you have that, you may get B from the second. The answer is 2*B.
    Last edited by daon2; 05-18-2013 at 02:33 AM.

  3. #3
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    Quote Originally Posted by daon2 View Post
    The base of an isosceles triangle (where the top angle is the odd one out) is 2 times the hypotenuse times the cosine of the matching angles (since the hypotenuse times cosine of the angle is the projection of the hypotenuse onto the horizontal axis). If H is the hypotenuse of one of the right triangles, O the opposite side, and B the base then H*sin(35)=O=50, H*cos(35)=B. Use the first equation to get H, and once you have that, you may get B from the second. The answer is 2*B.

    sorry, i'm still not getting it, when i calculate 50/sin(35), I'm getting -116.77 and that's what bothered me because how could it be a negative number? Then when I multiply that H by cos(35), I get 105.52 and that's not one of the given answers... thanks for your help, i don't know what i'm doing wrong or not getting

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    Quote Originally Posted by daon2 View Post
    The base of an isosceles triangle (where the top angle is the odd one out) is 2 times the hypotenuse times the cosine of the matching angles (since the hypotenuse times cosine of the angle is the projection of the hypotenuse onto the horizontal axis). If H is the hypotenuse of one of the right triangles, O the opposite side, and B the base then H*sin(35)=O=50, H*cos(35)=B. Use the first equation to get H, and once you have that, you may get B from the second. The answer is 2*B.
    also, i did exactly what you said and that's when i got 211.1, answer D, but the answer sheet says the answer is B... could it be the answer sheet is wrong?? i'm going to be so mad/relieved if that's the case

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    Quote Originally Posted by HCL View Post
    sorry, i'm still not getting it, when i calculate 50/sin(35), I'm getting -116.77 and that's what bothered me because how could it be a negative number? Then when I multiply that H by cos(35), I get 105.52 and that's not one of the given answers... thanks for your help, i don't know what i'm doing wrong or not getting
    you are in radian mode on your calculator, that is the problem. 35 degrees = 35*pi/180 = 7pi/36 radians.

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    Quote Originally Posted by daon2 View Post
    you are in radian mode on your calculator, that is the problem. 35 degrees = 35*pi/180 = 7pi/36 radians.


    ah, got it. thx. i'm just getting back into this mathematics stuff. thx again

  7. #7
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    No need to use the whole triangle: just use one of the right triangles:

    then other angle = 90 - 35 = 55
    so AB = 2[50SIN(55) / SIN(35)]
    I'm just an imagination of your figment !

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