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Math Help - I dont understand this triangle math question

  1. #1
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    Exclamation I dont understand this triangle math question

    If its not too much trouble

    The Directions said: two triangles can be formed using the given measurements. Solve both triangles.

    A= 60 degrees, side a = 12, side b = 13

    I drew the triangle, found all sides and angles, but i dont get how "two triangles can be formed" (i cant visualize it, because i dont know how to draw it with 2 triangles?)
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  2. #2
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    Triangles: the Ambiguous Case

    Hello yeunju
    Quote Originally Posted by yeunju View Post
    If its not too much trouble

    The Directions said: two triangles can be formed using the given measurements. Solve both triangles.

    A= 60 degrees, side a = 12, side b = 13

    I drew the triangle, found all sides and angles, but i dont get how "two triangles can be formed" (i cant visualize it, because i dont know how to draw it with 2 triangles?)
    A question like this, where you are given two sides of a triangle and the non-included angle, and two triangles are possible, is called the Ambiguous Case. The attached diagram shows how you can draw the two triangles with the measurements you were given.

    You've probably found the one where \angle B is acute (that's B_2 in my diagram). But there's also a possibility that \angle B is obtuse (like B_1). Since \triangle B_1CB_2 is isosceles, you can probably see why \angle AB_1C = 180^o - \angle AB_2C.

    Can you complete the solution of the second triangle now?

    Grandad
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  3. #3
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    Triangles

    So the direction is :
    Two triangles can be formed using the given measurements. Solve both triangles.

    angle A= 60 degrees, side a = 12, side b = 13

    I used the law of sines to find the rest of the angles and sides

    Angle B= 70 degrees, angle c= 50 degrees, side c = 10.6 (i drew it with c as the base)


    My issue was knowing how to form two triangles with the measurements. I drew the triangle, found all the sides and angles..but i dont understand forming two triangles?

    My work to get the rest of the measurements:

    sin(60)/12 = sin(x)/side 13 (cross multiplied)
    x= angle b = 69.75 about 70 degrees

    180-(angle B+angle A)= 180-(70+60) = 50 degrees (angle C)

    Sin 60/side 12 = sin50/ x (cross multiplied)
    x = side c = 10.6
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  4. #4
    MHF Contributor red_dog's Avatar
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    \sin x=\sin(180-x), \ x\in(0,90)

    So, B=x or B=180-x
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