1. **Calculate the height of a tree whose shadow is 20 m shorter when the sun's elevation angle changes from 30° to 60°**. Let the height of the tree be $h$ meters, the shadow length at 30° be $s$ meters, and at 60° be $s-20$ meters. We use the tangent function: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{\text{shadow length}}$. At 30°, $\tan 30^\circ = \frac{h}{s}$; since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we get $$h = \frac{s}{\sqrt{3}}.$$ At 60°, $\tan 60^\circ = \frac{h}{s-20}$; since $\tan 60^\circ = \sqrt{3}$, we get $$h = \sqrt{3} (s - 20).$$ Set equal since the height is the same: $$\frac{s}{\sqrt{3}} = \sqrt{3} (s - 20).$$ Multiply both sides by $\sqrt{3}$: $$s = 3(s - 20).$$ Expand right side: $$s = 3s - 60.$$ Bring terms together: $$3s - s = 60 \implies 2s = 60 \implies s = 30.$$ Now, find height: $$h = \frac{s}{\sqrt{3}} = \frac{30}{\sqrt{3}} = 10\sqrt{3}.$$ Answer: The height of the tree is $10\sqrt{3}$ meters. 2. **Calculate lengths $x$ and $y$ in Fig. 1.5 (composite trapezoid split into triangles)**. - For the equilateral triangle with sides 6 and $x$ (all sides equal in equilateral). Given one side is 6, all sides including $x$ equal 6. - The right triangle with 45° angle has sides in ratio $1:1:\sqrt{2}$. Let the side adjacent be $y$; then the opposite side is also $y$. Answer: $x=6$, $y =$ length of side in the 45° triangle (not numerically given). 3. **Find the height of point $X$ above $Y$ given angle of elevation 30° from $Z$, and $|XY|=40$ m (Fig. 1.7).** Given $\sin 30^\circ = \frac{|XN|}{|XZ|} = \frac{z}{v}$ with $\sin 30^\circ = \frac{1}{2}$. We have: $$\frac{1}{2} = \frac{4\sqrt{3}}{v} \implies v = 2 \times 4\sqrt{3} = 8\sqrt{3}.$$ Explanation: - $v$ is the hypotenuse length. - $z=4\sqrt{3}$ is the side opposite the 30° angle. Therefore, the height $z = 4\sqrt{3}$ meters. --- Final answers: - Tree height: $10\sqrt{3}$ m - $x = 6$ (equilateral triangle side) - Height $z = 4\sqrt{3}$ m