Given: ∆ABC – iso. ∆, m∠BAC = 120° AH ⊥ BC , HD ⊥ AC AD = a cm, HD = b cm Find: P∆ABH P = _ cm

P = 5.4641b cm.

Step-by-step explanation:

If the triangle ABC is isosceles and m∠BAC = 120°, we have that:

Then, in triangle ABH, we have:

If m∠BAC is 120°, we have:

Now we can find the length of AH using the sine relation of angle m∠HAC:

Now, to find the length of HB and AB, we can use the tangent and cosine relation of the angle m∠HAB:

So the perimeter of triangle ABH is:

The relation of a and b can be calculated using the tangent relation of the angle m∠HAC:

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