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So have to measure the height of an object $c$ given two known projections ($a$ and $b$). The angle $\phi$ between the two projections is known (it is 85.5°)

What is the length of $c$? (in terms of $\phi$, $a$ and $b$)

Before I was using the Pythagorean theorem, since the angle was thought to be 90°. Upon inspection it is not exactly 90° and thus the height of c is being underestimated.

My gut feeling is I need to use the Law of Cosines, but I can't seem to figure out how.

Notes:

  • The object $c$ can be rotated like in the picture. The angle of rotation is unknown.
  • $\alpha$ and $\beta$ angles are not given, but their sum equals $\phi$

i

Manuel Ruiz
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3 Answers3

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The projection of an object is equal to its length multiplied by the cosine of the angle between the projection and the object. Note that from the diagram, $90 - \alpha$ is the cosine angle for projection $a$. So we have, \begin{align} a &= c \cos \left(90 - \alpha\right) = c \sin \alpha &&\rightarrow \sin \alpha = \frac{a}{c} \\ b &= c \sin \beta &&\rightarrow \sin \beta = \frac{b}{c} \end{align}

Now we are also given the sum of these angles, $\phi$. So we can write one angle in terms of the other; i.e. $\alpha = \phi - \beta$. Applying the sine formula and expanding, \begin{align} \sin \alpha &= \sin (\phi - \beta) = \sin \phi \cos \beta - \cos \phi \sin \beta \\ \frac{a}{c} &= \sin \phi \frac{\sqrt{c^2 - b^2}}{c} - \cos \phi \frac{b}{c} \end{align}

For real problems, and if either $a$ or $b$ are non-zero, then $c$ will also be non-zero. So we can multiple throughout by $c$, and rearrange the rest to get, \begin{gather} c = \sqrt{\left(\frac{a + b \cos \phi}{\sin \phi}\right)^2 + b^2} = \frac{\sqrt{a^2 + b^2 + 2ab \cos \phi}}{\sin \phi} \end{gather}

And we can see that if $\phi = 90$, then the above reduces to the Pythagorean formula, as you were initially thinking.

Tony Mathew
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  • Accepting this answer becaus it was the first, and the most elegant in my opinion, since it uses the more simple logic. – Manuel Ruiz Jun 07 '23 at 15:14
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Consider the blue cyclic quadrilateral below around the required length $c$, with angles (clockwise from the top) $180^\circ - \phi$, $90^\circ$, $\phi$, and $90^\circ$:

Diagram with additional annotation

The required $c$ is the length of one diagonal. The length of the other diagonal can be found using the cosine law:

$$\begin{align*} \text{The other diagonal} &= \sqrt{a^2+b^2-2ab\cos(180^\circ-\phi)}\\ &= \sqrt{a^2+b^2+2ab\cos\phi} \end{align*}$$

$c$ is also the diameter of the circumcircle of the cyclic quad., which can be found by the sine law:

$$\begin{align*} c = 2R &= \frac{\text{The other diagonal}}{\sin(180^\circ-\phi)}\\ &= \frac{\sqrt{a^2+b^2+2ab\cos\phi}}{\sin\phi} \end{align*}$$

peterwhy
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If we move the line segments $a$ and $b$ towards $c,$ we get something like the following enter image description here

Due to the right angles at $A$ and $B$ and because of Thales's theorem, we know that $c$ is the diameter of the circle that goes through $A$, $B$ and $P.$

The length of $d$ can be calculated by means of the law of cosines: $$ d=\sqrt{a^2+b^2-2ab\cos(\pi-\phi)}=\sqrt{a^2+b^2+2ab\cos\phi} $$ and the diameter of the circumcircle of the triangle $ABP$ can be computed as the length of any side of the triangle divided by the sine of the opposite angle, see circumscribed circle. Therefore $$ c = \frac {d}{\sin(\pi-\phi)}=\frac {d}{\sin\phi} =\frac {\sqrt{a^2+b^2+2ab\cos\phi}}{\sin\phi} $$

Reinhard Meier
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