Let the height of the viewpoint be \( H = 80 \text{ m} \). Let the height of the flag be \( h \text{ m} \). Let the horizontal distance between the base of the viewpoint and the base of the flag be \( d \text{ m} \). The angle of depression to the bottom of the flag is \( 45^\circ \). By alternating interior angles, the angle of elevation from the bottom of the flag to the top of the viewpoint is also \( 45^\circ \). In the right-angled triangle formed with the base: \[ \tan(45^\circ) = \frac{H}{d} \] \[ 1 = \frac{80}{d} \implies d = 80 \text{ m} \] The angle of depression to the top of the flag is \( 30^\circ \). Draw a horizontal line from the top of the flag to the viewpoint structure. This creates a smaller right-angled triangle at the top. The height of this smaller triangle is \( H - h = 80 - h \). The base of this smaller triangle is equal to \( d = 80 \text{ m} \). Using tangent for this triangle: \[ \tan(30^\circ) = \frac{80 - h}{d} \] \[ \frac{1}{\sqrt{3}} = \frac{80 - h}{80} \] Now, solve for \( h \): \[ 80 - h = \frac{80}{\sqrt{3}} \] \[ h = 80 - \frac{80}{\sqrt{3}} \] \[ h = 80 \left( 1 - \frac{1}{\sqrt{3}} \right) \] Therefore, the height of the flag is \( 80 \left( 1 - \frac{1}{\sqrt{3}} \right) \text{ metres} \).