Chapter 6.5, Problem 44E

Soap Bubbles When two bubbles cling together in midair, their common surface is part of a sphere whose center D lies on the line passing through the centers of the bubbles (see the figure). Also, ∠ACB and ∠ACD each have measure 60°.

1. (a) Show that the radius r of the common face is given by

$r=\frac{ab}{a-b}$

[Hint: Use the Law of Sines together with the fact that an angle θ and its supplement 180° − θ have the same sine.]

1. (b) Find the radius of the common face if the radii of the bubbles are 4 cm and 3 cm.
2. (c) What shape does the common face take if the two bubbles have equal radii?

To determine

To show: The radius of common face formed, when two bubbles cling together is given as, $r=\frac{ab}{a-b}$.

Explanation of Solution

To prove the radius of common face between two bubbles, assume one unknown angle as a variable and evaluate the angles with respect to variable.

The given figure is,

Figure (1)

It is given that $\angle ACB=60°$ and $\angle ACD=60°$.

Let assume the unknown angle $\angle CAD$ be $\theta$.

Angles $\angle CAD$ and $\angle CAB$ are in straight line. Thus,

$\begin{array}{c}\angle CAD+\angle CAB=180°\\ \theta +\angle CAB=180°\\ \angle CAB=180°-\theta \end{array}$

That is $\angle CAB=180°-\theta$. (1)

And angle $\angle CAD$ is an exterior angle of $\Delta CAB$.

$\angle CAD=\angle CAB+\angle ACB$

Substitute $\theta$ for $\angle CAD$ and $60°$ for $\angle ACB$.

$\begin{array}{c}\theta =\angle CBA+60°\\ \angle CBA=\theta -60°\end{array}$

That is, $\angle CBA=\theta -60°$. (2)

Similarly angle $\angle CAB$ is an exterior angle of $\Delta CAD$.

$\angle CAB=\angle ACD+\angle CDA$

Substitute the value of $\angle CAB$ from equation (1) and $60°$ for $\angle ACD$.

$\begin{array}{c}180°-\theta =60°+\angle CDA\\ \angle CDA=180°-\theta -60°\\ \angle CDA=120°-\theta \end{array}$

That is, $\angle CDA=120°-\theta$. (3)

Draw the figure of triangle and show above angles.

Figure (1)

The law of sine for triangle,

$\frac{\sin X}{x}=\frac{\sin Y}{y}=\frac{\sin Z}{z}$ (4)

Where,

• X, Y and Z are the angles of triangle
• x, y and z are the sides with respective to angles.

Apply law of sine in $\Delta ABC$. Substitute $\angle CAB$ for X, $\angle CBA$ for Y, $\angle ACB$ for Z, a for x, b for y and AB for z in equation (4).

$\frac{\sin \left(\angle CAB\right)}{a}=\frac{\sin \left(\angle CBA\right)}{b}=\frac{\sin \left(\angle ACB\right)}{AB}$

Substitute $180°-\theta$ for $\angle CAB$ and $\theta -60°$ $\angle CBA$ and $60°$ for $\angle ACB$ in the above formula.

$\begin{array}{c}\frac{\sin \left(180°-\theta \right)}{a}=\frac{\sin \left(\theta -60°\right)}{b}=\frac{\sin \left(60°\right)}{AB}\\ \frac{\sin \left(180-\theta \right)°}{a}=\frac{\sin \left(\theta -60\right)°}{b}\end{array}$

That is, $\frac{\sin \left(180-\theta \right)°}{a}=\frac{\sin \left(\theta -60\right)°}{b}$. (5)

Similarly apply law of sine in $\Delta ABC$ and gets an equation as,

$\frac{\sin \left(\theta \right)°}{r}=\frac{\sin \left(120-\theta \right)°}{b}$ (6)

Add both the equations (5) and (6).

$\begin{array}{c}\frac{\sin \left(180-\theta \right)°}{a}+\frac{\sin \left(\theta \right)°}{r}=\frac{\sin \left(\theta -60\right)°}{b}+\frac{\sin \left(120-\theta \right)°}{b}\\ \frac{\sin \left(\theta \right)°}{a}+\frac{\sin \left(\theta \right)°}{r}=\frac{\sin \left(\theta -60\right)°+\sin \left(120-\theta \right)°}{b}\left[\because \sin \left(180-x\right)°=\sin \left(x\right)°\right]\\ \sin \theta °\left(\frac{1}{a}+\frac{1}{r}\right)=\frac{\sin \left(\theta -60\right)°+\sin \left(120-\theta \right)°}{b}\end{array}$

Use the trigonometric identity $\mathrm{Sin}x+\mathrm{Sin}y=2\sin \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)$ to simplify the equation further.

$\begin{array}{c}\sin \theta °\left(\frac{1}{a}+\frac{1}{r}\right)=\frac{2\sin {\left(\frac{\theta -60+120-\theta }{2}\right)}^{°}\cos {\left(\frac{\theta -60-120+\theta }{2}\right)}^{°}}{b}\\ \sin \theta °\left(\frac{1}{a}+\frac{1}{r}\right)=\frac{2\sin {\left(30\right)}^{°}\cos {\left(\theta -90\right)}^{°}}{b}\\ \sin \theta °\left(\frac{1}{a}+\frac{1}{r}\right)=\frac{2\left(\frac{1}{2}\right)\cos {\left(-\left(90-\theta \right)\right)}^{°}}{b}\left[\because \sin {30}^{°}=\frac{1}{2}\right]\\ \sin \theta °\left(\frac{1}{a}+\frac{1}{r}\right)=\frac{\cos {\left(90-\theta \right)}^{°}}{b}\left[\because \cos {\left(-x\right)}^{°}=\cos x^{°}\right]\end{array}$

Further simplify the equation as,

$\begin{array}{c}\sin \theta °\left(\frac{1}{a}+\frac{1}{r}\right)=\frac{\sin {\theta }^{°}}{b}\left[\because \cos {\left(90-\theta \right)}^{°}=\sin x^{°}\right]\\ \frac{1}{a}+\frac{1}{r}=\frac{1}{b}\\ \frac{1}{r}=\frac{1}{b}-\frac{1}{a}\\ r=\frac{ab}{b-a}\end{array}$

Hence, the radius of common face formed is, $r=\frac{ab}{a-b}$.

To determine

To find: The radius of common face when two bubbles cling together.

Answer to Problem 44E

The radius of common face is 12 cm.

Explanation of Solution

Given:

The radii of bubbles that cling together to form the common face are given as 4 and 3 cm respectively.

Calculation:

From part (a), the radius of common face formed, when two bubbles cling together is given as,

$r=\frac{ab}{a-b}$

Substitute 4 for a and 3 for b.

$\begin{array}{c}r=\frac{4\times 3}{4-3}\\ =\frac{12}{1}\\ =12\text{cm}\end{array}$

Thus, the radius of common face is 12 cm.

To determine

The shape of common face formed by two bubbles of same radii.

Explanation of Solution

From part (a), the radius of common face formed, when two bubbles cling together is given as,

$r=\frac{ab}{a-b}$

It is given that bubbles have same radii. Assume the radii of bubbles as $k$. Substitute $k$ for a and $k$ for b in the above formula.

$\begin{array}{c}r=\frac{k\times k}{k-k}\\ =\frac{k^2}{0}\left[\text{undefined value}\right]\end{array}$

The radius is not defined.

Thus, the shape of common face is a straight line.