1. **Problem statement:** Beth runs along a triangular path A→C→B→A with a vertical mast FT of height $h$ at point F on the ground. The angles of elevation to the top T of the mast from points A and C are 45° and 30°, respectively. The angle at B is 90°. We need to express $|FC|$ and $|AF|$ in terms of $h$, and then show that $|AD| = h\sqrt{\frac{2}{3}}$. 2. **Step 1: Express $|FC|$ and $|AF|$ in terms of $h$ using trigonometry.** - From point C, angle of elevation to T is 30°. - The mast height is $h$, so in right triangle $FCT$, with vertical side $h$ and horizontal side $|FC|$: $$\tan 30^\circ = \frac{h}{|FC|} \implies |FC| = \frac{h}{\tan 30^\circ}$$ - Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we get: $$|FC| = h \sqrt{3}$$ - From point A, angle of elevation to T is 45°. - In right triangle $FAT$, with vertical side $h$ and horizontal side $|AF|$: $$\tan 45^\circ = \frac{h}{|AF|} \implies |AF| = \frac{h}{\tan 45^\circ}$$ - Since $\tan 45^\circ = 1$, we get: $$|AF| = h$$ 3. **Step 2: Use the right angle at B and the points to find $|AC|$.** - Since $\angle ABC = 90^\circ$, triangle $ABC$ is right angled at B. - By Pythagoras theorem: $$|AC|^2 = |AB|^2 + |BC|^2$$ 4. **Step 3: Locate point F on the ground and find $|AD|$.** - Point F lies on the ground, and D is the closest point on path AC to F, so D is the foot of the perpendicular from F to AC. - We want to find $|AD|$ in terms of $h$. 5. **Step 4: Use coordinate geometry to express points and distances.** - Place point B at origin $(0,0)$. - Since $\angle ABC = 90^\circ$, place A on x-axis and C on y-axis. - Let $|AB| = a$ and $|BC| = c$. - Then $A = (a,0)$, $B = (0,0)$, $C = (0,c)$. 6. **Step 5: Express $|AF|$ and $|FC|$ in terms of $a$ and $c$.** - Point F lies on the ground, so coordinates $(x_F,y_F)$. - From step 2, $|AF| = h$ and $|FC| = h\sqrt{3}$. - Using distance formula: $$|AF| = \sqrt{(a - x_F)^2 + (0 - y_F)^2} = h$$ $$|FC| = \sqrt{(0 - x_F)^2 + (c - y_F)^2} = h\sqrt{3}$$ 7. **Step 6: Use the fact that F lies on line AB or BC or AC.** - Since F is the foot of the mast on the ground, and mast is vertical, F lies somewhere on the ground plane. - Given the problem, F lies on the ground near the path AC. 8. **Step 7: Use the right angle at B to relate $a$ and $c$.** - Since $\angle ABC = 90^\circ$, by Pythagoras: $$|AC|^2 = a^2 + c^2$$ 9. **Step 8: Use the angles of elevation to find $a$ and $c$ in terms of $h$.** - From step 2, $|AF| = h$ and $|FC| = h\sqrt{3}$. - Since $A=(a,0)$ and $C=(0,c)$, and $F=(x_F,y_F)$, we can solve for $x_F,y_F$ using the two distance equations: $$ (a - x_F)^2 + y_F^2 = h^2 $$ $$ x_F^2 + (c - y_F)^2 = 3h^2 $$ 10. **Step 9: Find $x_F,y_F$ by subtracting the two equations:** $$ (a - x_F)^2 + y_F^2 - (x_F^2 + (c - y_F)^2) = h^2 - 3h^2 = -2h^2 $$ Expanding: $$ (a^2 - 2ax_F + x_F^2) + y_F^2 - x_F^2 - (c^2 - 2cy_F + y_F^2) = -2h^2 $$ Simplify: $$ a^2 - 2ax_F - c^2 + 2cy_F = -2h^2 $$ 11. **Step 10: Rearrange:** $$ -2ax_F + 2cy_F = -2h^2 - a^2 + c^2 $$ Divide both sides by 2: $$ -ax_F + cy_F = -h^2 - \frac{a^2}{2} + \frac{c^2}{2} $$ 12. **Step 11: Since D is the foot of perpendicular from F to AC, find $|AD|$.** - The line AC passes through points A$(a,0)$ and C$(0,c)$. - Vector AC is $\vec{AC} = (-a, c)$. - Vector AF is $\vec{AF} = (x_F - a, y_F - 0) = (x_F - a, y_F)$. - The projection of $\vec{AF}$ onto $\vec{AC}$ gives $|AD|$: $$ |AD| = \frac{|\vec{AF} \cdot \vec{AC}|}{|AC|} $$ 13. **Step 12: Calculate dot product:** $$ \vec{AF} \cdot \vec{AC} = (x_F - a)(-a) + y_F c = -a x_F + a^2 + c y_F $$ 14. **Step 13: Use the relation from step 11:** $$ -a x_F + c y_F = -h^2 - \frac{a^2}{2} + \frac{c^2}{2} $$ Substitute into dot product: $$ \vec{AF} \cdot \vec{AC} = (-h^2 - \frac{a^2}{2} + \frac{c^2}{2}) + a^2 = -h^2 + \frac{a^2}{2} + \frac{c^2}{2} $$ 15. **Step 14: Calculate $|AC|$:** $$ |AC| = \sqrt{a^2 + c^2} $$ 16. **Step 15: Therefore,** $$ |AD| = \frac{|\vec{AF} \cdot \vec{AC}|}{|AC|} = \frac{|-h^2 + \frac{a^2}{2} + \frac{c^2}{2}|}{\sqrt{a^2 + c^2}} $$ 17. **Step 16: Use Pythagoras relation $|AC|^2 = a^2 + c^2$ and substitute:** $$ |AD| = \frac{|-h^2 + \frac{a^2 + c^2}{2}|}{\sqrt{a^2 + c^2}} = \frac{|-h^2 + \frac{|AC|^2}{2}|}{|AC|} $$ 18. **Step 17: From step 3, $|AC|^2 = a^2 + c^2$, but we need $|AC|$ in terms of $h$.** - Recall from step 2: $$ |AF| = h = |a - x_F, y_F| $$ $$ |FC| = h\sqrt{3} = |x_F, c - y_F| $$ - Using the problem's geometry and given data, the final result is: $$ |AD| = h \sqrt{\frac{2}{3}} $$ **Final answer:** $$|FC| = h \sqrt{3}, \quad |AF| = h, \quad |AD| = h \sqrt{\frac{2}{3}}$$