You're on a roll — 8.1 Step 1: Identify the relevant right-angled triangle. Since D is directly above A, $\triangle DAB$ is a right-angled triangle with $\angle DAB = 90^\circ$. Step 2: Use trigonometry in $\triangle DAB$. We are given $AD = p$ and $\angle DBA = y$. We want to find $AB$. In $\triangle DAB$: $$\tan(\angle DBA) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AD}{AB}$$ $$\tan y = \frac{p}{AB}$$ Step 3: Solve for $AB$. $$AB = \frac{p}{\tan y}$$ The length of AB in terms of $p$ and $y$ is $\boxed{AB = \frac{p}{\tan y}}$. 8.2 Step 1: Use the given information about $\triangle ABC$. We are given that A, B, C lie in the same horizontal plane, and $AB = AC$. This means $\triangle ABC$ is an isosceles triangle. We are also given $\angle ABC = x$. Since $AB=AC$, then $\angle ACB = x$. We are given $2AD = BC$. Since $AD = p$, then $BC = 2p$. Step 2: Construct an altitude in $\triangle ABC$. Draw a perpendicular line from A to BC, let the intersection point be M. Since $\triangle ABC$ is isosceles with $AB=AC$, the altitude AM bisects BC. Therefore, $BM = MC = \frac{1}{2} BC = \frac{1}{2}(2p) = p$. Step 3: Use trigonometry in $\triangle AMB$. $\triangle AMB$ is a right-angled triangle with $\angle AMB = 90^\circ$. In $\triangle AMB$: $$\cos(\angle ABM) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{BM}{AB}$$ $$\cos x = \frac{p}{AB}$$ Step 4: Equate expressions for $AB$ and simplify. From 8.1, we found $AB = \frac{p}{\tan y}$. From Step 3, we found $AB = \frac{p}{\cos x}$. Equating these two expressions for $AB$: $$\frac{p}{\tan y} = \frac{p}{\cos x}$$ Divide both sides by $p$ (since $p \ne 0$): $$\frac{1}{\tan y} = \frac{1}{\cos x}$$ Therefore, $\boxed{\cos x = \tan y}$. 8.3 Step 1: Use the relationship derived in 8.2. We have $\cos x = \tan y$. Step 2: Substitute the given value of $x$. Given $x = 60^\circ$. $$\cos 60^\circ = \tan y$$ Step 3: Calculate the value of $\cos 60^\circ$. We know that $\cos 60^\circ = \frac{1}{2}$. $$\frac{1}{2} = \tan y$$ Step 4: Solve for $y$. $$y = \arctan\left(\frac{1}{2}\right)$$ $$y \approx 26.565^\circ$$ Rounding to one decimal place: $$y \approx 26.6^\circ$$ The size of $y$ is $\boxed{26.6^\circ}$. Send me the next one 📸