Answer to Question #127308 in Geometry for Michael Essel Moses

Question

The angle at the vertex of a cone is measured using a 30mm diameter coin as shown in the figure below. If the coin lies 3.72mm below the top of the cone, determine the value of angle θ.

Answer 1

Let the cone be imagined as shown in the above figure.

Here AB is the diameter of the coin and NB is the radius of the coin.

From the question we have to find $\angle AMB$

Let $\angle NMB = \theta$

In $\triangle MNB$ right angled at N, $\sin{\theta}$ = $NB \div BM$

Using Pythagoras theorem $\lparen MN\rparen ^2 + \lparen BN\rparen ^2 = \lparen BM\rparen ^2$

= $\lparen 3.72\rparen ^2 + \lparen 15\rparen ^2 = \lparen BM\rparen ^2$

= 13.84 + 225

= 238.84

$\therefore BM = \sqrt{238.84}$

= 15.45

So $\sin{\theta} = 15\div15.45$

= 0.97

$\therefore \theta = \sin^{-1}\lparen0.97\rparen$

= 75.9$\degree$

$\therefore$ angle at the vertex = $2\theta$

= 2*75.9

= 151.8$\degree$