Question

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

Answer 1

Answer:

P = 5.4641b cm.

Step-by-step 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:

$tan(mHAC) = DH / AD$

$1.7321 = b / a$

$b = 1.7321a$