Question

An acute isosceles triangle has an angle with a measure of 36°, the lengths of the two sides are 1 and 1.618. An angle bisector of the base angle intersects the leg at point P. Find the distance between P and the vertex opposite the base.

Answer 1

1 unit

Explanation:

See the attached diagram.

ΔABC is an isosceles triangle having ∠A = 36° and base angles ∠B = ∠C = 72°

Now, the bisector of angle B intersects AC at P.

Now, ∠ABP = 36° and ∠A = 36°, hence, ∠APB = 180° - 36° - 36° = 108°

Now, given that AB = AC = 1.618 units and BC = 1 unit

So, from the property of triangles of ΔABP, we have,

`(AP)/(\sin 36) = (AB)/(\sin108)`

⇒
`AP = 1.618 *(\sin36)/(\sin108)=1` units (Approximate) (Answer)