I want to solve parts (ii) and (iii) of this problem without using vectors
In the region of 3 fixed buoys A, B and C at sea there is a plane stratum of oil-bearing rock. The depths of the rock below A, B and C are 900m, 800m and 1,000m respectively. B is 600m due east of A and the bearings of C from A and B are 190° and 235° respectively.
Calculate
(i) the distance BC
(ii) the direction of the horizontal projection of the line of greatest slope of the plane
(iii) the angle this plane makes with the horizontal
(It May be helpful to consider a horizontal plane at depth 900 m)
This is my diagram for the problem:
where D, E and F are the oil-bearing rocks under A, B & C respectively.
Using the sine and cosine rules I have worked out the lengths of the sides of $\triangle$ ABC and $\triangle$ DEF but it is parts (ii) and (iii) I cannot do.
Best Answer
We can write $$A(0,0,0),\quad B(600,0,0),\quad C(p,q,0)$$ $$D(p,q,-1000),\quad E(0,0,-900),\quad F(600,0,-800)$$
Let $G$ be a point on the line $AB$ such that $CG\perp AB$.
Solving the system$$\begin{cases}\tan\angle{CBG}=\tan(35^\circ)=\frac{CG}{BG}=\frac{-q}{600-p}\\\\\tan\angle{ACG}=\tan(10^\circ)=\frac{AG}{CG}=\frac{-p}{-q}\end{cases}$$ gives $$p=\frac{600\tan(10^\circ)\tan(35^\circ)}{\tan(10^\circ)\tan(35^\circ)-1},\qquad q=\frac{600\tan(35^\circ)}{\tan(10^\circ)\tan(35^\circ)-1}$$ where both $p$ and $q$ are negative.
(ii)
Let $H(600,0,-900),I(p,q,-900)$. Also, let $J$ be the intersection point of the line $HI$ with the line $FD$. (Note that $J$ is the midpoint of the line segment $FD$.) Then, the line $EJ$ is the intersection of the plane $DEF$ with the horizontal plane at depth $900\ \text{m}$.
So, the direction of the horizontal projection of the line of greatest slope of the plane is the direction which is perpendicular to the line $EJ$.
(This page explains what "the line of greatest slope on a plane" is.)
(iii)
Since $J(\frac{p+600}{2},\frac q2,-900)$, the equation of the line $EJ$ is given by $$qx-(p+600)y=0,\quad z=-900\tag1$$
Let $K$ be a point on the the horizontal plane at depth $900\ \text{m}$ such that $KH\perp EJ$. Then, the equation of the line $KH$ is given by $$(p+600)x+qy-600(p+600)=0,\quad z=-900\tag2$$ Solving $(1)(2)$ gives $$K\bigg(\frac{600(p+600)^2}{q^2+(p+600)^2},\frac{600q(p+600)}{q^2+(p+600)^2},-900\bigg)$$
Let $\theta$ be the angle the plane $DEF$ makes with the horizontal. Then, we get $$\tan\theta=\frac{FH}{KH}=-\frac{\sqrt{q^2+(p+600)^2}}{6q}$$
Added :
$C(p,q,0)$ means that the $x$ coordinate of $C$ is $p$, the $y$ coordinate of $C$ is $q$, and the $z$ coordinate of $C$ is $0$.
Now, note that $C$ exists in the third quadrant. So, both $p$ and $q$ are negative. So, for example, the distance between $C$ and $G$ (note that the coordinates of $G$ is $(p,0,0)$) is given by $$(\text{$y$ coordinate of $G$})-(\text{$y$ coordinate of $C$})=0-q=-q$$ which is positive since $q$ is negative. Similarly, the distance between $A$ and $G$ is given by $$(\text{$x$ coordinate of $A$})-(\text{$A$ coordinate of $G$})=0-p=-p$$ which is positive since $p$ is negative.
We want to solve the system $$\tan(35^\circ)=\frac{-q}{600-p}\tag1$$ $$\tan(10^\circ)=\frac{-p}{-q}\tag2$$ From $(1)$, we get $$q=(p-600)\tan(35^\circ)\tag3$$ From $(2)$, we get $$q=\frac{p}{\tan(10^\circ)}\tag4$$ From $(3)(4)$, we get $$(p-600)\tan(35^\circ)=\frac{p}{\tan(10^\circ)}$$ from which you can get $p$, and get $q$ from $(4)$.
By "Since $J(\frac{p+600}{2},\frac q2,-900)$", I mean that "Since the coordinates of $J$ is given by $(\frac{p+600}{2},\frac q2,-900)$".