Proof: Height of Vertical Mast on River Banks

[Appleton]

Homework Statement

A vertical mast stands on the north bank of a river with straight parallel banks running from east to west. The angle of elevation of the top of the mast is α when measured from a point A on the south bank distant 3a to the east of the mast and β when measured from another point B on the south bank distant 5a to the west mast. prove that the height of the mast is
4a/(cot^2\beta-cot^2\alpha)^\frac{1}{2}

Homework Equations

The Attempt at a Solution

Let the height of the mast be h.
Let C be the base of the mast.
Let D be the point at which the perpendicular from BA to C divides BA.

<br /> BC = \frac{h}{tan β}\\\\<br /> CA = \frac{h}{tan α}\\\\<br />
By Pythagoras' theorem

<br /> CD = \sqrt{(\frac{h}{tan β})^2 - (5a)^2)}\\\\<br /> CD = \sqrt{(\frac{h}{tan α})^2 - (3a)^2)}\\\\<br />
So

<br /> \sqrt{(\frac{h}{tan β})^2 - (5a)^2)} = \sqrt{(\frac{h}{tan α})^2 - (3a)^2)}\\\\\<br /> (\frac{h}{tan β})^2 - (\frac{h}{tan α})^2 = (5a)^2 - (3a)^2\\\\<br /> \frac{h^2 tan^2 α - h^2 tan^2 β }{(tan^2 β) (tan^2 α)} = 16a^2\\<br /> h = 4a\sqrt{\frac{(tan^2 β) (tan^2 α)}{tan^2 α - tan^2 β }}<br />

At this point I figure that either the question is floored or I've made a mistake. Usually it's the latter.

 

[SammyS]

Homework Statement

A vertical mast stands on the north bank of a river with straight parallel banks running from east to west. The angle of elevation of the top of the mast is α when measured from a point A on the south bank distant 3a to the east of the mast and β when measured from another point B on the south bank distant 5a to the west mast. prove that the height of the mast is
4a/(cot^2\beta-cot^2\alpha)^\frac{1}{2}

Homework Equations

The Attempt at a Solution

Let the height of the mast be h.
Let C be the base of the mast.
Let D be the point at which the perpendicular from BA to C divides BA.

<br /> BC = \frac{h}{tan β}\\\\<br /> CA = \frac{h}{tan α}\\\\<br />
By Pythagoras' theorem

<br /> CD = \sqrt{(\frac{h}{tan β})^2 - (5a)^2)}\\\\<br /> CD = \sqrt{(\frac{h}{tan α})^2 - (3a)^2)}\\\\<br />
So

<br /> \sqrt{(\frac{h}{tan β})^2 - (5a)^2)} = \sqrt{(\frac{h}{tan α})^2 - (3a)^2)}\\\\\<br /> (\frac{h}{tan β})^2 - (\frac{h}{tan α})^2 = (5a)^2 - (3a)^2\\\\<br /> \frac{h^2 tan^2 α - h^2 tan^2 β }{(tan^2 β) (tan^2 α)} = 16a^2\\<br /> h = 4a\sqrt{\frac{(tan^2 β) (tan^2 α)}{tan^2 α - tan^2 β }}<br />

At this point I figure that either the question is floored or I've made a mistake. Usually it's the latter.

... or you have to take this a bit further.

What is $\displaystyle\ \left(\frac{(tan^2 β) (tan^2 α)}{tan^2 α - tan^2 β }\right)^{-1}\$ ?

 

[Appleton]

... or you have to take this a bit further.

What is $\displaystyle\ \left(\frac{(tan^2 β) (tan^2 α)}{tan^2 α - tan^2 β }\right)^{-1}\$ ?

Ah yes, I'm kicking myself. Thanks for the nudge.