Year 11 3U trigonometry question, please help? (1 Viewer)

extmathyo · May 22, 2015

So the question is as follows.( . * - means degrees.)

From a point P an oberserver finds that the angle of elevation of the top of a vertical tower is a*. After walking x meters horizontaly towards the foot of the tower, the observer finds that the angle of elevation is b*. If the height of the twoer is h meters, show that:
x = h(cot(a*)-cot(b*))
Please try show construction if you can, thankyou !

InteGrand · May 22, 2015

extmathyo said:
So the question is as follows.( . * - means degrees.)

From a point P an oberserver finds that the angle of elevation of the top of a vertical tower is a*. After walking x meters horizontaly towards the foot of the tower, the observer finds that the angle of elevation is b*. If the height of the twoer is h meters, show that:
x = h(cot(a*)-cot(b*))
Please try show construction if you can, thankyou !

$The starting point of the man is $P$. Draw a horizontal line $\ell$ through $P$ to represent the ground. Let's say he walks to the right from $P$. Let the base of the tower be $B$ and the top of the tower be $T$. $BT$ is drawn as a vertical line with $B$ lying on $\ell$. $BT$ has length $h$ m (label this in the diagram so you can see all lengths). Let $A$ be the point the man walks to, so label a point $A$ on $\ell$ that is between $P$ and $B$, and label $PA=x$ m. Join $PT$ and $AT$. Based on the given angles of elevation, label the following angles: $\angle TPB = a^\circ$ and $\angle TAB = b^\circ$. Also label $\angle TBP$ as a right angle.$

$Now, using right-angled trigonometry in $\triangle TPB$, we see that $\cot a^\circ = \frac{x+AB}{h}\Rightarrow x = h\cot a^\circ - AB$.$

$Similarly in $\triangle TAB$, we see that $\cot b^\circ = \frac{AB}{h}\Rightarrow AB = h\cot b^\circ$. Substitute this result into the previous equation to obtain: $x =h\cot a^\circ - h\cot b^\circ = h(\cot a^\circ -\cot b^\circ)$. \blacksquare$

extmathyo · May 22, 2015

InteGrand said:
$The starting point of the man is $P$. Draw a horizontal line $\ell$ through $P$ to represent the ground. Let's say he walks to the right from $P$. Let the base of the tower be $B$ and the top of the tower be $T$. $BT$ is drawn as a vertical line with $B$ lying on $\ell$. $BT$ has length $h$ m (label this in the diagram so you can see all lengths). Let $A$ be the point the man walks to, so label a point $A$ on $\ell$ that is between $P$ and $B$, and label $PA=x$ m. Join $PT$ and $AT$. Based on the given angles of elevation, label the following angles: $\angle TPB = a^\circ$ and $\angle TAB = b^\circ$. Also label $\angle TBP$ as a right angle.$

$Now, using right-angled trigonometry in $\triangle TPB$, we see that $\cot a^\circ = \frac{x+AB}{h}\Rightarrow x = h\cot a^\circ - AB$.$

$Similarly in $\triangle TAB$, we see that $\cot b^\circ = \frac{AB}{h}\Rightarrow AB = h\cot b^\circ$. Substitute this result into the previous equation to obtain: $x =h\cot a^\circ - h\cot b^\circ = h(\cot a^\circ -\cot b^\circ)$. \blacksquare$

Thankyou so much!

Feynman · May 22, 2015

Diagram

$cot(a) = \frac{x+hcot(b)}{h}$

$hcot(a) = x + hcot(b)$

$x = hcot(a) - hcot(b)$

$x = h(cot(a) - cot(b))$

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Year 11 3U trigonometry question, please help? (1 Viewer)

extmathyo

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InteGrand

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extmathyo

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Feynman

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