Split - Right Triangles and isosceles triangles

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nugget

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Dec 29, 2010
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Hi everyone! thanks for helping me out. Alright, so this is actually an isosceles triangle problem but I couldn't find that on the website so I appreciate your helping me! So here is he problem:
The length of each leg of an isosceles triangle is 10m. The shared angle between the two legs is x. Write the formula for the triangle as a function of x/2.
I believe that you are supposed to use trig. identies but if anyone can find another way to solve it go for it! thank you guys so much and good luck!!! I am really having troule with this problem and i'm not quite sure where to begin. Again, thank you!!!
 
Re: *quick help* on proving identity extra cred problem

You've now posted this problem 3 times!
And hijacked threads each time.

Start your own thread (NEWTOPIC button).
And show your work so we know where you're stuck :idea:
 
nugget said:
Hi everyone! thanks for helping me out. Alright, so this is actually an isosceles triangle problem but I couldn't find that on the website so I appreciate your helping me! So here is he problem:
The length of each leg of an isosceles triangle is 10m. The shared angle between the two legs is x. Write the formula for the triangle as a function of x/2.
I believe that you are supposed to use trig. identies but if anyone can find another way to solve it go for it! thank you guys so much and good luck!!! I am really having troule with this problem and i'm not quite sure where to begin. Again, thank you!!!
Click to expand...

Please start a new thread with a new problem.

Let the triangle be ABC where AB = BC and mABC = x

Drop a perpendicular from the vertex B of the triangle to the base AC. This is your height.

Now continue....
 
"Write the formula for the triangle "

This is not well-stated. The area maybe?

In any case, drop a perpendicular to the base from the apex angle. Contemplate the measure of the two angles you have just created at the apex.
 
yes i meant the area of the triangle sorry. so i put a perpendicular line to make two triangles and I labeled one of the sides 10 leaving a 90 degree angle at its opposite and i know the area for a triangle is found by a=1/2(base x height) but i'm not sure how to find the base or height??? if anyone could help me that would be great! thanks! :)
 


Have you learned the right-triangle definitions for the ratios sine and cosine?

sin(angle) = opposite/hypotenuse

cos(angle) = adjacent/hypotenuse

Look at your picture.

The perpendicular line segment that you drew divides the given triangle into two identical right triangles. Call the length of this segment h.

If we let the symbol b represent the base of the given triangle, then each of these identical right triangles has base b/2.

h is the side adjacent to angle x/2.

b/2 is the side opposite to angle x/2.

Their hypotenuse is 10.

Write the sine and cosine ratios. You can then solve for h and b, and use the resulting expressions in the area formula.

IF you do not yet understand the right-triangle definitions for the ratios sine and cosine, then that's where you need to begin because you're not ready for this exercise.

Let us know what you come up with.

 
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