6. The ratio of the base of an isosceles triangle to its altitude is \( 3: 4 \). Find the measures of the angles of the triangle.

Let the base of the isosceles triangle be \( b \) and the altitude be \( h \). According to the problem, the ratio of the base to the altitude is given as \( 3:4 \). This can be expressed mathematically as: \[ \frac{b}{h} = \frac{3}{4} \] From this ratio, we can express \( b \) in terms of \( h \): \[ b = \frac{3}{4}h \] In an isosceles triangle, the altitude from the vertex opposite the base bisects the base. Therefore, the length of each half of the base is: \[ \frac{b}{2} = \frac{3}{8}h \] Now, we can use the right triangle formed by the altitude, half of the base, and the side of the triangle. The right triangle has: - One leg as the altitude \( h \) - The other leg as half the base \( \frac{b}{2} = \frac{3}{8}h \) - The hypotenuse as the side of the triangle, which we will denote as \( s \) Using the Pythagorean theorem, we have: \[ s^2 = h^2 + \left(\frac{b}{2}\right)^2 \] Substituting \( b = \frac{3}{4}h \): \[ s^2 = h^2 + \left(\frac{3}{8}h\right)^2 \] Calculating \( \left(\frac{3}{8}h\right)^2 \): \[ \left(\frac{3}{8}h\right)^2 = \frac{9}{64}h^2 \] Now substituting this back into the equation for \( s^2 \): \[ s^2 = h^2 + \frac{9}{64}h^2 = \left(1 + \frac{9}{64}\right)h^2 = \frac{64}{64}h^2 + \frac{9}{64}h^2 = \frac{73}{64}h^2 \] Thus, we have: \[ s = \sqrt{\frac{73}{64}}h = \frac{\sqrt{73}}{8}h \] Next, we will find the angles of the triangle. Let \( A \) be the angle at the vertex opposite the base, and \( B \) be the angles at the base. Since the triangle is isosceles, we have: \[ \angle B = \angle C \] Using the tangent function to find \( \angle B \): \[ \tan B = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{\frac{b}{2}} = \frac{h}{\frac{3}{8}h} = \frac{8}{3} \] Now, we can find \( \angle B \): \[ B = \tan^{-1}\left(\frac{8}{3}\right) \] To find \( A \): \[ A = 180^\circ - 2B \] Thus, the measures of the angles of the triangle are: \[ B = \tan^{-1}\left(\frac{8}{3}\right) \quad \text{and} \quad A = 180^\circ - 2\tan^{-1}\left(\frac{8}{3}\right) \] Calculating \( B \) and \( A \): Using a calculator, we find: \[ B \approx 69.44^\circ \] Then, \[ A \approx 180^\circ - 2 \times 69.44^\circ \approx 41.12^\circ \] Thus, the measures of the angles of the triangle are approximately: \[ A \approx 41.12^\circ, \quad B \approx 69.44^\circ, \quad C \approx 69.44^\circ \] In conclusion, the angles of the isosceles triangle are approximately: \[ \boxed{41.12^\circ, 69.44^\circ, 69.44^\circ} \]