Answer:
P = 5.4641b cm.
Step-by-step explanation:
If the triangle ABC is isosceles and m∠BAC = 120°, we have that:
[tex]mACB = mABC = (180-120)/2 = 30\°[/tex]
Then, in triangle ABH, we have:
[tex]mABH + mBHA + mHAB = 180\°[/tex]
[tex]30\° + 90\° + mHAB = 180\°[/tex]
[tex]mHAB = 60\°[/tex]
If m∠BAC is 120°, we have:
[tex]mHAB + mHAC = 120\°[/tex]
[tex]m∠HAC = 60\°[/tex]
Now we can find the length of AH using the sine relation of angle m∠HAC:
[tex]sin(mHAC) = DH / AH[/tex]
[tex]0.866 = b / AH[/tex]
[tex]AH = b / 0.866 = 1.1547b[/tex]
Now, to find the length of HB and AB, we can use the tangent and cosine relation of the angle m∠HAB:
[tex]tan(mHAB) = HB / AH[/tex]
[tex]1.7321 = HB / 1.1547b[/tex]
[tex]HB = 1.7321 * 1.1547b = 2b[/tex]
[tex]cos(mHAB) = AH / AB[/tex]
[tex]0.5 = 1.1547b / AB[/tex]
[tex]AB = 1.1547b / 0.5 = 2.3094b[/tex]
So the perimeter of triangle ABH is:
[tex]P(ABH) = AB + HB + AH[/tex]
[tex]P(ABH) = 2.3094b + 2b + 1.1547b[/tex]
[tex]P(ABH) = 5.4641b[/tex]
The relation of a and b can be calculated using the tangent relation of the angle m∠HAC:
[tex]tan(mHAC) = DH / AD[/tex]
[tex]1.7321 = b / a[/tex]
[tex]b = 1.7321a[/tex]
