# Thread: Bearing and angle between planes

1. ## Bearing and angle between planes

Two questions i have partly done.
1. Prove that in a plane triangle ABC
$\cot A=\frac{a}{b}\csc C-\cot C$
The bearing of an aeroplane from a fixed observation post is $052^o$, the aeroplane being at a horizontal distance of 6 miles from the post. Calculate the flight direction of the aeroplane if, when it passes due east of the post, it is at a horizontal distance of 9 miles from the post. If the height of the aeroplane when its bearing is $052^o$ is h and its angle of elevation from the post is the same in both positions calculate, in terms of h, the height of the aeroplane when due east of the post.
In the figure
I first fin the length AB= $\sqrt{6^2+9^2-2\times 6\times 9\times\cos 38^o}=5.65$
then angle PAX=[math\frac{9\sin 38^o}{5.65}=78.9^o[/tex]
Then using the opposite angles on parallel lines, the direction of the plane would be $180^o-(78.9^o-52^o)=153.1^o$
Answer is supposed to be 131

2. A vertical mast stands on the north bank of a river with straight parallel banks running east-west. The angle of elevation of the top of the mast is $\alpha$ when measured from A and point on the south bank 3a to the east of the mast and $\beta$ when measured from another point B on the south bank distant 5a to the west of the mast. Prove that the height of the mast is $\frac{4a}{\sqrt{\cot^2\beta-\cot^2\alpha}}$ and that the angle of elevation $\theta$ measured from a point midway between A and B is given by the equation $2\cot^2\theta=3\cot^2\alpha-\cot^2\beta$
I have proved $\frac{4a}{\sqrt{\cot^2\beta-\cot^2\alpha}}$.
But i don't know how to begin with the next part
Thanks

2. Hello arze
Originally Posted by arze
Two questions i have partly done.
1. Prove that in a plane triangle ABC
$\cot A=\frac{a}{b}\csc C-\cot C$
The bearing of an aeroplane from a fixed observation post is $052^o$, the aeroplane being at a horizontal distance of 6 miles from the post. Calculate the flight direction of the aeroplane if, when it passes due east of the post, it is at a horizontal distance of 9 miles from the post. If the height of the aeroplane when its bearing is $052^o$ is h and its angle of elevation from the post is the same in both positions calculate, in terms of h, the height of the aeroplane when due east of the post.
In the figure
I first fin the length AB= $\sqrt{6^2+9^2-2\times 6\times 9\times\cos 38^o}=5.65$
then angle PAX=[math\frac{9\sin 38^o}{5.65}=78.9^o[/tex]
Then using the opposite angles on parallel lines, the direction of the plane would be $180^o-(78.9^o-52^o)=153.1^o$
Answer is supposed to be 131
Don't forget that two angles are possible for a given sine: one is $79^o$; the other is $(180-79^o) = 101^o$. You want the second here. This gives the bearing as $131^o$. (You can always tell whether to choose the acute or the obtuse angle, because the biggest angle in a triangle is always opposite the longest side; and so on.)

I'll have a look at the second question now. (Incidentally, it's probably better to post this as a separate thread.)

PS Sorry - haven't time to do question 2 now. Perhaps someone else can take it?

3. Hello again arze
Originally Posted by arze
2. A vertical mast stands on the north bank of a river with straight parallel banks running east-west. The angle of elevation of the top of the mast is $\alpha$ when measured from A and point on the south bank 3a to the east of the mast and $\beta$ when measured from another point B on the south bank distant 5a to the west of the mast. Prove that the height of the mast is $\frac{4a}{\sqrt{\cot^2\beta-\cot^2\alpha}}$ and that the angle of elevation $\theta$ measured from a point midway between A and B is given by the equation $2\cot^2\theta=3\cot^2\alpha-\cot^2\beta$
I have proved $\frac{4a}{\sqrt{\cot^2\beta-\cot^2\alpha}}$.
But i don't know how to begin with the next part
Thanks
I called the foot of the mast $P$, its top $Q$, the point immediately opposite to it $R$, and the mid-point of $AB,\; S$. The height of the mast I denoted by $h$. So, from part (i):
$h = \frac{4a}{\sqrt{\cot^2\beta-\cot^2\alpha}}$

$\Rightarrow \frac{a^2}{h^2}= \frac{\cot^2\beta - \cot^2\alpha}{16}$
...(1)
Then $\angle PSQ = \theta$ and, from $\triangle PQS$:
$PS = h\cot\theta$
and from $\triangle PSR$:
$PS^2 = PR^2+SR^2$

$\Rightarrow h^2\cot^2\theta = h^2\cot^2\beta - 25a^2 + a^2$, using the result from (i) that $PR^2 =h^2\cot^2\beta - 25a^2$
$\Rightarrow \cot^2\theta = \cot^2\beta - \frac{24a^2}{h^2}$
$=\cot^2\beta - \frac{3(\cot^2\beta-\cot^2\alpha)}{2}$, from (1)
$\Rightarrow 2\cot^2\theta = 3\cot^2\alpha - \cot^2\beta$