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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.

Respuesta :

Answer:

1 unit

Step-by-step 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,  

[tex]\frac{AP}{\sin 36} = \frac{AB}{\sin108}[/tex]

⇒[tex]AP = 1.618 \times\frac{\sin36}{\sin108}=1[/tex] units (Approximate) (Answer)

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