The measure of an angle of an isosceles triangle is 30°, the shorter side is 20 in. find a distance from point p to the vertex of a triangle, if p is equidistant from all vertices.

the angles in this triangle are 30, 30, and 120. the shorter sides are 20 inches. if P is equidistant from all vertices, it has to be outside of the triangle. We can name the points in the triangle A, B, and C.

You get a quadrilateral if you connect the lines from P to B and P to C. You can get two isosceles triangles from this, APC and APB. In triangle APC, lines AP and CP are equal, and in triangle APB lines AP and BP are equal. The base sides in each of these triangles are 20 inches, and the base angles are 60 degrees, meaning these are equilateral triangles.

HENCE, PA IS EQUAL TO 20 INCHES.

answer: i think it is (1,-2)

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )