Question

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

P = _____ cm

Answer 1

P = 5.4641b cm.

Explanation:

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

`mACB = mABC = (180-120)/2 = 30\°`

Then, in triangle ABH, we have:

`mABH + mBHA + mHAB = 180\°`

`30\° + 90\° + mHAB = 180\°`

`mHAB = 60\°`

If m∠BAC is 120°, we have:

`mHAB + mHAC = 120\°`

`m∠HAC = 60\°`

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

`sin(mHAC) = DH / AH`

`0.866 = b / AH`

`AH = b / 0.866 = 1.1547b`

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

`tan(mHAB) = HB / AH`

`1.7321 = HB / 1.1547b`

`HB = 1.7321 * 1.1547b = 2b`

`cos(mHAB) = AH / AB`

`0.5 = 1.1547b / AB`

`AB = 1.1547b / 0.5 = 2.3094b`

So the perimeter of triangle ABH is:

`P(ABH) = AB + HB + AH`

`P(ABH) = 2.3094b + 2b + 1.1547b`

`P(ABH) = 5.4641b`

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

`tan(mHAC) = DH / AD`

`1.7321 = b / a`

`b = 1.7321a`