Trigonometry Demystified, 2nd edition
Stan Gibilisco
Explanations for Quiz Answers in Chapter 12
1. In the spherical triangle of Fig. 12-10, the two vertices on the equator both have interior right angles, because they're points at which a meridian crosses the equator. (On the earth's surface, all meridians cross the equator at right angles.) Right angles, by definition, always measure π/2 rad. The interior angle at the north pole measures 67.5, which equals 3/16 of a full circle. When we convert that value to radians, we get

67.5 = 3/16 x 2π
= 3π/8 rad

When we add up the measures of the three interior angles in radians, we obtain a sum of

π/2 + π/2 + 3π/8
= 4π/8 + 4π/8 + 3π/8
= (4 + 4 + 3) π/8
= 11π/8 rad

The correct choice is C.

2. The chapter text gives us a formula for calculating the distance dlon-deg (in kilometers) between any two meridians separated by exactly one degree of longitude, as measured along any particular parallel at latitude θ on the earth's surface. That formula is

dlon-deg = 111.2 cos θ

In this case, we're told to work at 30.00 north latitude, so θ = 30.00. Therefore

dlon-deg = 111.2 cos 30.00
= 111.2 x 0.8660254
= 96.30202 kilometers per degree

Our meridians are 5.00 apart, so we must multiply the above number by 5.00. When we carry out that arithmetic and round off to the nearest kilometer, we get a distance of 482 kilometers as measured along the parallel corresponding to 30 north latitude. The correct choice is C.

3. This situation resembles the one described in Question 2, except that now we have θ = 60.00, so the distance between two meridians one degree apart works out as

dlon-deg = 111.2 cos 60.00
= 111.2 x 0.5000000
= 55.600000 kilometers per degree

Multiplying by 5.00, and rounding off to the nearest kilometer, we get a distance of 278 kilometers as measured along the parallel corresponding to 60 north latitude. The answer is A.

4. We can use the spherical law of sines to solve this problem. We've been told that the three sides of a spherical triangle have angular extents of

q = 30

r = 25

s = 20

Let's name the interior angles as follows:

  • We use the label ψq for the interior angle opposite side q
  • We use the label ψr for the interior angle opposite side r
  • We use the label ψs for the interior angle opposite side s

The spherical law of sines tells us that

(sin q) / (sin ψq) = (sin r) / (sin ψr) = (sin s) / (sin ψs)

The problem states that the angle opposite side q, which we call ψq, constitutes a right angle, so we have the additional known value

ψq = 90

Now we can put all of our known values into the formula for the spherical law of sines, getting

(sin 30) / (sin 90) = (sin 25) / (sin ψr) = (sin 20) / (sin ψs)

Working out to four significant figures, a calculator can simplify the formula to

0.5000 / 1.000 = 0.4226 / (sin ψr) = 0.3420 / (sin ψs)

Leaving out the last third of the formula and simplifying the first third, we get

0.5000 = 0.4226 / (sin ψr)

A little algebra yields the equation

sin ψr = 0.8452

When we take the Arcsine of both sides, we get

Arcsin (sin ψr) = Arcsin 0.8452

which solves to

ψr = 58

rounded off to the nearest degree. The correct choice is C.

5. We keep the same scenario as we had in Question 4 and its solution, but now we want to know the measure of ψs, the angle opposite side s. Let's go back to the general formula that we got in the middle of our solution to Question 4, which told us that

0.5000 / 1.000 = 0.4226 / (sin ψr) = 0.3420 / (sin ψs)

Leaving out the middle third of the formula and simplifying the first third, we get

0.5000 = 0.3420 / (sin ψs)

which simplifies to

sin ψs = 0.6840

When we take the Arcsine of both sides, we get

Arcsin (sin ψs) = Arcsin 0.6840

which solves to

ψs = 43

rounded off to the nearest degree. The answer is D.

6. We've found the measures of all three interior angles of the spherical triangle described in Question 4. Those measures are

ψq = 90

ψr = 58

ψs = 43

They add up to 191. The answer is B.

7. Each and every individual meridian intersects the equator at a right angle, or π/2 rad. That would of course include the meridian corresponding to 45 east longitude. The correct choice is B.
8. We need the spherical law of cosines to solve this problem. If we know the angular extents of two of the sides, say q and r, of a spherical triangle, and we also know the measure of the spherical angle ψs between those two sides, then we can calculate the cosine of the angular extent of the third side s as

cos s = cos q cos r + sin q sin r cos ψs

In the situation of Fig. 12-11, let's use the following names for the sides of the spherical triangle and their angular extents:

q = 65

r = 55

s = (unknown)

Let's name the interior angles this way:

  • We use the label ψq for the interior angle opposite side q
  • We use the label ψr for the interior angle opposite side r
  • We use the label ψs for the interior angle opposite side s

The diagram below is an enhanced version of Fig. 12-11 with these labels included.

When we plug in the known values to the formula for the spherical law of cosines, we get

cos s = cos 65 cos 55 + sin 65 sin 55 cos 100

When we work out the sines and cosines to four significant figures (a few more than we'll need, to ensure that we don't suffer with cumulative rounding errors), we get

cos s = 0.4226 x 0.5736 + 0.9063 x 0.8192 x (-0.1736)
= 0.1135

Taking the Arccosine of both sides, and rounding off the result to the nearest degree, we get

Arccos (cos s) = Arccos 0.1135
= 83

which tells us that s = 83. The correct choice is A.

9. We can use the spherical law of sines to solve this problem. Once again, the formula is

(sin q) / (sin ψq) = (sin r) / (sin ψr) = (sin s) / (sin ψs)

We want to find the measure of the interior angle at the far-left vertex. That's angle ψr. We found the angular extent of side s (see the drawing above) when we worked out the solution to Question 8. We can therefore set up the ratio equality

(sin r) / (sin ψr) = (sin s) / (sin ψs)

and plug in the known values to get

(sin 55) / (sin ψr) = (sin 83) / (sin 100)

which calculates out to

0.8192 / (sin ψr) = 0.9925 / 0.9848

and simplifies to

sin ψr = 0.8128

When we take the Arcsine of both sides and round off the result to the nearest degree, we obtain

Arcsin (sin ψr) = Arcsin 0.8128
= 54

so ψr = 54. The answer is A.

10. We can use the spherical law of sines to solve this problem, along similar lines to the previous solution. Once again, we start with the original formula for the law of sines:

(sin q) / (sin ψq) = (sin r) / (sin ψr) = (sin s) / (sin ψs)

We want to find the measure of the interior angle at the top-right vertex. That's angle ψq. We found the angular extent of side s (see the drawing above) when we worked out the solution to Question 8. We can therefore set up the ratio equality

(sin q) / (sin ψq) = (sin s) / (sin ψs)

and plug in the known values to get

(sin 65) / (sin ψq) = (sin 83) / (sin 100)

which calculates out to

0.9063 / (sin ψq) = 0.9925 / 0.9848

and simplifies to

sin ψq = 0.8993

When we take the Arcsine of both sides and round off the result to the nearest degree, we have

Arcsin (sin ψq) = Arcsin 0.8993
= 64

so ψq = 64. The answer is B.