In △ABC, ∠ABC = 60 degrees .

P is a point inside △ABC such that ∠APB = ∠BPC = ∠CPA = 120 degrees, PA = 8, and PC = 6. Find PB.

Respuesta :

A given shape that is bounded by three sides and has got three internal angles is referred to as a triangle. Thus the value of PB is 8.0 units.

A given shape that is bounded by three sides and has got three internal angles is referred to as a triangle. Types of triangles include right angle triangle, isosceles triangle, equilateral triangle, acute angle triangle, etc. The sum of the internal angles of any triangle is [tex]180^{o}[/tex].

In the given question, point P is center of the given triangle such that <APB = <APC = <BPC = [tex]120^{o}[/tex]. Such that line PB bisects <ABC into two equal measures of [tex]30^{o}[/tex].

Thus;

<ABP = [tex]30^{o}[/tex]

Thus,

<ABP + <APB + <BAP = [tex]180^{o}[/tex]

30 + 120 + <BAP = [tex]180^{o}[/tex]

<BAP = [tex]180^{o}[/tex] - 150

<BAP = [tex]30^{o}[/tex]

Apply the Sine rule to determine the value of PB, such that;

[tex]\frac{a}{SinA}[/tex] = [tex]\frac{b}{SinB}[/tex]

[tex]\frac{8}{Sin 30}[/tex] = [tex]\frac{PB}{Sin 30}[/tex]

BP = [tex]\frac{8*Sin 30}{Sin 30}[/tex]

    = 8

BP = 8.0

Therefore, the value of PB = 8 units.

For more clarifications on Sine rule, visit: https://brainly.com/question/27174058

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