Thread: [SOLVED] Easy trignoometry question

I am reviewing trigonometry and this question has puzzled me.

isosceles triangle has a base of 10cm long. The height is 4cm. Determine the measures if the three angles the given triangle.

The section I'm on is using TAN. So what I did is draw a perpendicular bisector, bisection the height so I now have right angles so I can use trigonometry "it has to be a right angle triangle right?" anywho, then I used pythagorus to find the length of the perpindicluar bisecter which came out to 9.79cm.

Then I found the angles of the two equal angles "angles P and Q" and I got 78°, so the remaining angle would be 24°.

The answer key has 38.7°, 38.7°, and 102.6°

What am I doing wrong, am I making it harder than it needs to be?


Thanks, any help is greatly appreciated.

I tried it with the two equal sides being 4cm and I still got the wrong answer. "also how do you know which sides are the two equal sides?is it always the height?

tan(v)=opposite/adjacent

Im swedish so my terminology sucks but anywho, you divide the triangle in half from the top to the baseline QR leaving you a new triangle with the sides 4 cm (height), 5 cm (original baseline/2) and the hypotenuse unknown.

Now focusing on TAN, we set the angle QPR/2 as v

tanv=opposite/adjacent=5/4

v=tan^-1(5/4)=51.34 degrees

now you have half of the original triangles QPR angle, double it to get the QPR angle and then subtract that angle from 180 degrees and divide by two to get the remaining angles.


My results:

QPR: 102.6 degrees
RQP: 38.7 degrees
PRQ: 38.7 degrees

Hello, brentwoodbc!

Isosceles triangle has a base of 10cm long.
The height is 4cm.
Determine the measures of the three angles.

Code:

                  P
                  *
               *  |  *
            *     |4    *
         *        |        *
    Q * - - - - - * - - - - - * R
            5     T     5

In right triangle

Similarly: .

Then: .

thanks to both of you, very helpful. For some reason I was thinking on the two lengths as the height "all Ive been doing is right angle triangles, so base height hypotenuse". Thanks again.