Chapter 4.3, Problem 69E

To determine

Find the coordinates of the centre of each hole.

Answer to Problem 69E

  $\boxed{\left(x_2,y_2\right)=\left(28,28\sqrt{3}\right)}$

Explanation of Solution

Given information:

A steel plate has the form of one-fourth of a circle with a radius of $60$ centimetres. Two two-centimeter holes are drilled in the plate, positioned as shown in the figure. Find the coordinates of the centre of each hole.

Calculation:

A steal plate has thew form of one-fourth of a circle with a radius of $60$ centimeters. Two two- centimeter holes are to be drilled in the plate as shown by the figure below.

We have to find the coordinates of the centre of each hole, that is, the positions $\left(x_1,y_1\right),\left(x_2,y_2\right)$

Now, let us consider the triangle, with the angle at the vertex as $30°$ with the $x-axis$ as the base of the triangle. The hypotenuse of the triangle is the radius of the circle, that is, $56$ centimeters.

Now, we know that $\cos \theta =\frac{adjacent-side-of-\theta }{\text{Hypotenuse}}$

Now, the length of adjacent side to $\theta =30°$ is $x_1$ .

Thus, using the above formula, we have,

  $\cos 30°=\frac{x_1}{56}$

Or

  $x_1=56\times \cos 30°$

  $=56\times \frac{\sqrt{3}}{2}$

  $=28\sqrt{3}$

The length of the side opposite to $\theta =30°$ is $y_1$ .

Now, we know that, $\cos \theta =\frac{adjacent-side-of-\theta }{\text{Hypotenuse}}$

Thus, using the above formula, we have,

  $\cos 30°=\frac{y_1}{56}$

Or

  $y_1=56\times \sin 30°$

= $=56\times \frac{1}{2}$

  $=28$

Hence, the coordinates of the centre of the hole are

  $\boxed{\left(x_1,y_1\right)=\left(28\sqrt{3},28\right)}$

Now, let us find the coordinates $\left(x_2,y_2\right)$ .

Now, let us consider the triangle, with the anglew at the vertex as $60°$ with the $x-axis$ as the base of the triangle. The hypotenuse of the triangle is the radius of the circle, that is, $56$ centimeters.

Now, we know that $\cos \theta =\frac{adjacent-side-of-\theta }{\text{Hypotenuse}}$

Now, the length of adjacent side to $\theta =60°$ is $x_2$ .

Thus, using the above formula, we have,

  $\cos 60°=\frac{x_2}{56}$

Or

  $x_2=56\times \cos 60°$

  $=56\times \frac{1}{2}$

  $=28$

The length of the side opposite to $\cos 60°$ is $y_2$ .

Now, we know that, $\sin \theta =\frac{opposite-side-of-\theta }{\text{Hypotenuse}}$

Thus, using the above formula, we have,

  $y_2=56\times \sin 60°$

  $=56\times \frac{\sqrt{3}}{2}$

  $=28\sqrt{3}$

Hence, the coordinates of the centre of the hole are $\boxed{\left(x_2,y_2\right)=\left(28,28\sqrt{3}\right)}$