Chapter 12.7, Problem 22PPS

a.

To determine

Find the estimated angles by using the protractor of each angle of the triangle.

Answer to Problem 22PPS

The angles of the three triangles are $({90}^{\circ },{90}^{\circ },{90}^{\circ })$ , $({90}^{\circ },{90}^{\circ },{45}^{\circ })$ & $({120}^{\circ },{45}^{\circ },{45}^{\circ })$ .

Explanation of Solution

Given:

It is given in the question that the spherical geometrical figure is shown below:

  

Concept Used:

In this, use the concept of using the protractor and measuring the angles.

Calculation:

Since, if you have rubber bands then use the tape rather than band.

Put the protractor in each angles of the triangle and take measures.

So, there will be three triangles and the measure of the all three will be like $({90}^{\circ },{90}^{\circ },{90}^{\circ })$ , $({90}^{\circ },{90}^{\circ },{45}^{\circ })$ & $({120}^{\circ },{45}^{\circ },{45}^{\circ })$ .

Conclusion:

The angles are $({90}^{\circ },{90}^{\circ },{90}^{\circ })$ , $({90}^{\circ },{90}^{\circ },{45}^{\circ })$ & $({120}^{\circ },{45}^{\circ },{45}^{\circ })$ .

b.

To determine

Write the sum of the angles as find above in the tabular form.

Answer to Problem 22PPS

The values are $({90}^{\circ },{90}^{\circ },{90}^{\circ })$ , $({90}^{\circ },{90}^{\circ },{45}^{\circ })$ & $({120}^{\circ },{45}^{\circ },{45}^{\circ })$ .

Explanation of Solution

Given:

It is given in the question that the spherical geometrical figure is shown below:

  

Since, in the first part the angles are measured by the protractor.

The angles are $({90}^{\circ },{90}^{\circ },{90}^{\circ })$ , $({90}^{\circ },{90}^{\circ },{45}^{\circ })$ & $({120}^{\circ },{45}^{\circ },{45}^{\circ })$ .

Make the values in the tabular form:

Triangle 1	Triangle 2	Triangle 3
${90}^{\circ }$	${90}^{\circ }$	${120}^{\circ }$
${90}^{\circ }$	${90}^{\circ }$	${45}^{\circ }$
${90}^{\circ }$	${45}^{\circ }$	${45}^{\circ }$

c.

To determine

Write a conjecture about the sum of the measures of the triangles in the spherical geometry.

Answer to Problem 22PPS

The sum of the angles of the triangles will always be greater than ${180}^{\circ }$ .

Explanation of Solution

Given:

It is given in the question that the angles of the triangles are $({90}^{\circ },{90}^{\circ },{90}^{\circ })$ , $({90}^{\circ },{90}^{\circ },{45}^{\circ })$ & $({120}^{\circ },{45}^{\circ },{45}^{\circ })$ .

Here, the angles will be $({90}^{\circ },{90}^{\circ },{90}^{\circ })$ , $({90}^{\circ },{90}^{\circ },{45}^{\circ })$ & $({120}^{\circ },{45}^{\circ },{45}^{\circ })$ .

Let’s check the sum of each angles of the triangle:

Triangle (i) - ${90}^{\circ }+{90}^{\circ }+{90}^{\circ }={270}^{\circ }>{180}^{\circ }$

Triangle (ii) - ${90}^{\circ }+{90}^{\circ }+{45}^{\circ }={225}^{\circ }>{180}^{\circ }$

Triangle (iii) - ${120}^{\circ }+{45}^{\circ }+{45}^{\circ }={210}^{\circ }>{180}^{\circ }$