Triangles

[vampirewitchreine]

This problem is more algebraic in manner but it's in my geometry textbook......


(My math for the angle measures are included on the image in red)

What I need is help solving for variables a and b..... I'm not sure exactly how to begin. Could someone help push me in the right direction?
The one side that already has a measure should say \(\displaystyle 4\sqrt {2}\) (Something I couldn't exactly do on Paint...)

 

[tkhunny]

You may need more help in simply reading your text book or attending class. Are you doing these things?

Did you just cover the Law of Sines?

Did you recently study "Special" Right Triangles, such as 45-45-90 and 30-60-90?

Please provide better information about where you are in your reading or class materials.

 

[vampirewitchreine]

You may need more help in simply reading your text book or attending class. Are you doing these things?

Did you just cover the Law of Sines?

I'm home schooled, so I don't attend classes and my textbook leaves a lot of information out. This is in the section about simple right triangles..... I've not gotten to the part that you just mentioned.

 

[soroban]

Hello, vampirewitchreine!

View attachment 1564

This problem is more algebraic in manner but it's in my geometry textbook......

(My math for the angle measures are included on the image in red)

What I need is help solving for variables a and b.
I'm not sure exactly how to begin.
Could someone help push me in the right direction?

You could use the Law of Sines,
. . but with a little thought, you can virtually "eyeball" the problem.

              B
              *
             *: *
            * :60 *
        _  *45:     *    2h
      4√2 *   :       *
         *    :h        *
        *     :           *
       *45    :         30  *
    A *   *   *   *   *_  *   * C
      :   h   D       √3h     :

Let \(\displaystyle h = BD\), the altitude to side \(\displaystyle AC.\)

Since \(\displaystyle \Delta BDA\) is an isosceles right triangle:. \(\displaystyle AD = BD = h.\)

Pythagorus says: .\(\displaystyle h^2+h^2 \:=\4\sqrt{2})^2 \quad\Rightarrow\quad 2h^2 \:=\:32 \quad\Rightarrow\quad h^2 = 16 \quad\Rightarrow\quad h = 4\)


\(\displaystyle \Delta BDC\) is a 30-60 right triangle.
The side opposite \(\displaystyle 30^o\) is \(\displaystyle h.\)
The side opposite \(\displaystyle 60^o\) is \(\displaystyle \sqrt{3}h.\)
The hypotenuse is \(\displaystyle 2h.\)


Therefore: .\(\displaystyle a \,=\,BC \:=\:2h \:=\:8\)
. . . . and: .\(\displaystyle b \,=\,AC \:=\:h + \sqrt{3}h \:=\:4(1+\sqrt{3})\)

 

[vampirewitchreine]

Hello, vampirewitchreine!


You could use the Law of Sines,
. . but with a little thought, you can virtually "eyeball" the problem.

I have yet to reach the section on Sines and Cosines, but this will be coming up soon. (Most of my problem with working this one out was looking past the fact that it wasn't a pre-labeled 30-60-90 or 45-45-90 triangle.)

              B
              *
             *: *
            * :60 *
        _  *45:     *    2h
      4√2 *   :       *
         *    :h        *
        *     :           *
       *45    :         30  *
    A *   *   *   *   *_  *   * C
      :   h   D       √3h     :

Let \(\displaystyle h = BD\), the altitude to side \(\displaystyle AC.\)

Since \(\displaystyle \Delta BDA\) is an isosceles right triangle:. \(\displaystyle AD = BD = h.\)

Pythagorus says: .\(\displaystyle h^2+h^2 \:=\4\sqrt{2})^2 \quad\Rightarrow\quad 2h^2 \:=\:32 \quad\Rightarrow\quad h^2 = 16 \quad\Rightarrow\quad h = 4\)

Can you explain how you got 32 as the answer to \(\displaystyle 4\sqrt{2}\) (Never mind, I figured out how.... I wasn't putting the whole thing in parenthesis before squaring it.)

\(\displaystyle \Delta BDC\) is a 30-60 right triangle.
The side opposite \(\displaystyle 30^o\) is \(\displaystyle h.\)
The side opposite \(\displaystyle 60^o\) is \(\displaystyle \sqrt{3}h.\)
The hypotenuse is \(\displaystyle 2h.\)


Therefore: .\(\displaystyle a \,=\,BC \:=\:2h \:=\:8\)
. . . . and: .\(\displaystyle b \,=\,AC \:=\:h + \sqrt{3}h \:=\:4(1+\sqrt{3})\)

I understand how you got a, but could you explain b a little more please? (Sorry if it's a dumb thing to ask, but my brain's not entirely working right now)