Each face of a pyramid is an isosceles triangle with a 66 degrees vertex angle. What are the measures of the base​ angles?

Answer 1

Let $ABC$ be the base of the pyramid, and let $P$ be the vertex of the pyramid. Then $\angle APB = \angle BPC = \angle CPA = 66^\circ$. Let $D$ be the foot of the altitude from $P$ to $\overline{ABC}$.

[asy]
unitsize (0.8 cm);

pair A, B, C, D, P;

D = (0,0);
C = (2,0);
B = rotate(95)*(0.5,0);
A = extension(B,(D + B)/2,C,(D + C)/2);
P = (5,3);

draw(A--B--C--D--P--A);
draw(A--P,dashed);
draw(B--P,dashed);
draw(C--P,dashed);

label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, SE);
label("$D$", D, S);
label("$P$", P, N);
[/asy]

Since $\triangle DPC$ and $\triangle DPA$ are isosceles, $\angle APC = (180^\circ - 66^\circ)/2 = 57^\circ$ and $\angle CPB = (180^\circ - 66^\circ)/2 = 57^\circ$. Therefore, $\angle CPB = \angle APC = 57^\circ$.

Therefore, the base angles of $\triangle ABC$ are $180^\circ - 2(57^\circ) = \boxed{66^\circ}$.