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(D) Let the angles of the triangle be $\alpha, \beta, \gamma$. We are given two angles as $A = 2x + 7$ and $B = 7x + 10$.For an isosceles triangle,at

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$3$

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(D) Let the angles of the triangle be $\alpha, \beta, \gamma$. We are given two angles as $A = 2x + 7$ and $B = 7x + 10$.For an isosceles triangle,at least two angles must be equal. We consider three cases:Case $1$: $A = B$$2x + 7 = 7x + 10$ $\Rightarrow 5x = -3$ $\Rightarrow x = -0.6$.The angles are $A = 5.8^\circ, B = 5.8^\circ, C = 180 - 11.6 = 168.4^\circ$. This is a valid triangle.Case $2$: $A = C$Since $A + B + C = 180^\circ$,we have $2A + B = 180^\circ$.$2(2x + 7) + (7x + 10) = 180$ $\Rightarrow 4x + 14 + 7x + 10 = 180$ $\Rightarrow 11x = 156$ $\Rightarrow x = \frac{156}{11} \approx 14.18$.The angles are $A = 35.36^\circ, B = 109.26^\circ, C = 35.36^\circ$. This is a valid triangle.Case $3$: $B = C$Since $A + B + C = 180^\circ$,we have $A + 2B = 180^\circ$.$(2x + 7) + 2(7x + 10) = 180$ $\Rightarrow 2x + 7 + 14x + 20 = 180$ $\Rightarrow 16x = 153$ $\Rightarrow x = \frac{153}{16} = 9.5625$.The angles are $A = 26.125^\circ, B = 76.9375^\circ, C = 76.9375^\circ$. This is a valid triangle.Since all three cases yield valid real values for $x$,the total number of such real numbers is $3$.

The number of real numbers $x$ such that there exists an isosceles triangle having two of its angles measured in degrees equal to $2x + 7$ and $7x + 10$ is:

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