Triangle sidelengths

[gduren]

Ok, so I saw this problem today and I am completely confused. Triangle with Vertices A. B. C. The triangle is split or disected to form 90 degree angles at the base. This of course now forms two triangles. The one on the left is a 30-60-90 triangle. With vertices mentioned above A=60, B= 30. On the right side of the 30 degree andle where it has been cut in two is an angle of 45 degrees. so I have a 90-45-45 triangle. The Height of the triangle is sqrt of 6. Placeing the letter D where the disecting line is perpendicular to CA.

The question is what is the side lengths of AB, BC, CD, DA

Just curious.[sup:1967dwhr][/sup:1967dwhr]

 

[wjm11]

Ok, so I saw this problem today and I am completely confused. Triangle with Vertices A. B. C. The triangle is split or disected to form 90 degree angles at the base. This of course now forms two triangles. The one on the left is a 30-60-90 triangle. With vertices mentioned above A=60, B= 30. On the right side of the 30 degree andle where it has been cut in two is an angle of 45 degrees. so I have a 90-45-45 triangle. The Height of the triangle is sqrt of 6. Placeing the letter D where the disecting line is perpendicular to CA.

The question is what is the side lengths of AB, BC, CD, DA

Since you have two right triangles and you know all the angles, you can use SOH-CAH-TOA (basic trig). If you have one side and all the angles, you can always use this to solve all the other sides.

Ex: side AB: cos(30) = adj./hyp. = (6^.5)/AB

AB = (6^.5)/(cos(30)) = 2(2^.5) or about 2.828

Note: if you aren't allowed to use a calculator, you would have to know the ratios of the sides of the triangles you are working with. For the 45-45-90, it's 1:1:2^.5 ; for the 30-60-90, it's 1:2:3^.5

Hope that helps.

 

[soroban]

Hello, gduren!

Triangle ABC with altitude BD.
Triangle ABD is a 30-60-90 right triangle, angle A = 60 degrees.
Triangle CBD is a 45-45-90 right triange, BD = sqrt{6}.

Find the lengths: AB, BC, CD, AD.

              B
              o
             *| *
            * |45 *
           *30|     *
          *   |  _    *
         *    | /6      *
        *     |           *
       * 60   |          45 *
    A o - - - o - - - - - - - o C
              D

You're expected to be familiar with those two special right triangles.

\(\displaystyle \text{In a 30-60 right triangle, the ratio of the sides opposite }(30^o,\:60^o,\:90^o)\,\text{ is }\1:\sqrt{3}:2)\)

\(\displaystyle \text{In }\Delta ABD\!:\;\;(AD:BD:AB) \:=\\sqrt{2}:\sqrt{6}:2\sqrt{2})\)


\(\displaystyle \text{In a 45-45 right triangle, the ratio of the sides opposite }(45^o,\:45^o,\:90^o)\,\text{ is }\,(1:1:\sqrt{2})\)

\(\displaystyle \text{In }\Delta CBD\!:\;\;(BD:CD: BC) \:=\\sqrt{6},\:\sqrt{6},\:2\sqrt{3})\)


Got it?