Triangles
Conventional Notation for a Triangle
The conventional notation for a triangle ABC is shown in Fig. 26-1.
| Right Triangle | Acute Triangle | Obtuse Triangle |
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A triangle that contains no right angle is called an oblique triangle. The six parts of the triangle ABC are the three sides a, b, and c, together with the three angles $ $ , $ $ , and $ $ .
Solving a Triangle
Solving a triangle is the process of determining all the parts of the triangle. In general, given three parts of a triangle, including at least one side, the other parts can be determined. (Exceptions are cases where two possible triangles are determined or where no triangle can be shown to be consistent with the given data.)
Right Triangles
Here one part is known from the outset to be an angle of $ 90^{} $ . Given either two sides, or one side and one of the acute angles, the other parts can be determined using the definitions of the trigonometric functions for acute angles, the Pythagorean theorem, and the fact that the sum of the three angles in a plane triangle is $ 180^{} $ .
EXAMPLE 26.1 Given a right triangle ABC with c = 20 and $ = 30^{} $ , solve the triangle.
Here it is assumed that γ = 90°.
Solve for $ $ :
Since $ + + = 180^{} $ , $ = 180^{} - - = 180^{} - 30^{} - 90^{} = 60^{} $ .
Solve for a:
In the right triangle ABC, $ = $ , hence $ a = c = 20 ^{} = 10 $ .
Solve for b:
From the Pythagorean theorem, $ c{2}=a{2}+b^{2} $ ; hence $ b====10 $
Oblique Triangles
Oblique triangles are solved using the law of sines and the law of cosines. Normally five cases are recognized on the basis of which parts are given: AAS (two angles and a nonincluded side are given), ASA (two angles and an included side), SSA (two sides and a nonincluded angle), SAS (two sides and an included angle), and SSS (three sides).
Law of Sines
In any triangle, the ratio of each side to the sine of the angle opposite that side is the same for all three sides:
\[ \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}\qquad\quad\frac{a}{\sin\alpha}=\frac{c}{\sin\gamma}\qquad\quad\frac{b}{\sin\beta}=\frac{c}{\sin\gamma} \]
Law of Cosines
In any triangle, the square of any side is equal to the sum of the squares of the other two sides, diminished by twice the product of the other two sides and the cosine of the angle included between them:
\[ \begin{aligned}&a^{2}=b^{2}+c^{2}-2bc\cos\alpha\\&b^{2}=a^{2}+c^{2}-2ac\cos\beta\\&c^{2}=a^{2}+b^{2}-2ab\cos\gamma\\ \end{aligned} \]
Accuracy in Computations
In working with approximate data, the number of significant digits in a result cannot be greater than the number of significant digits in the given data. In interpreting calculator results for angles, the following table is useful:
| NUMBER OF SIGNIFICANT DIGITS FOR SIDES | DEGREE MEASURE OF ANGLES TO THE NEAREST |
| 2 | 1 $ ^{} $ |
| 3 | 0.1 $ ^{} $ or 10 $ ^{} $ |
| 4 | 0.01 $ ^{} $ or 1 $ ^{} $ |
Bearing and Heading
In applications involving navigation and aviation, as well as some other situations, angles are normally specified with reference to a north-south axis:
- BEARING: A direction is specified in terms of an angle measured east or west of a north-south axis. Thus Fig. 26-2 shows bearings of N30°E and S70°W.
- HEADING: A direction is specified in terms of an angle measured clockwise from north. Thus the same figure shows headings of $ 30^{} $ and $ 250^{} $ (that is, $ 180^{} + 70^{} $ ).
Angles of Elevation and Depression
ANGLE OF ELEVATION is the angle from the horizontal measured upward to the line of sight of the observer.
ANGLE OF DEPRESSION is the angle from the horizontal measured downward to the line of sight of the observer.
SOLVED PROBLEMS
26.1. Given a right triangle ABC with $ = 42.7^{} $ and a = 68.2, solve the triangle. (See Fig. 26-1.)
Here it is assumed that $ = 90^{} $ . In any right triangle, the acute angles are complementary, since $ + + 90^{} = 180^{} $ implies $ + = 90^{} $ .
Solve for $ $ :
\[ \beta=90^{\circ}-\alpha=90^{\circ}-42.7^{\circ}=47.3^{\circ} \]
Solve for b:
In the right triangle ABC, $ = $ ; hence $ b = = = 73.9 $ .
Solve for c:
In the right triangle ABC, $ = $ ; hence $ c = = = 100.6 $ .
Alternatively, use the Pythagorean theorem to find c, since a and b are known. However, it is preferable to use given data rather than calculated data wherever possible, since errors in calculations accumulate.
26.2. Given a right triangle ABC with c = 5.07 and a = 3.34, solve the triangle. Express angles in degrees and minutes. (See Fig. 26-1.)
Here it is assumed that $ = 90^{} $ .
Solve for $ $ :
In the right triangle ABC, $ = $ ; hence $ = ^{-1} = ^{-1} = 41^{}12’ $ .
Solve for $ $ :
In the right triangle $ ABC, + = 90^{} $ ; hence $ = 90^{} - = 90^{} - 41^{}12’ = 48^{}48’ $ .
Solve for b:
From the Pythagorean theorem, $ c{2}=a{2}+b^{2} $ ; hence $ b===3.81 $ .
26.3. When the angle of elevation of the sun is $ 27^{} $ , a pole casts a shadow 14 meters long on level ground. Find the height of the pole.
Sketch a figure (see Fig. 26-4).
In right triangle STB, let h = height of pole. Given that SB = 14 and $ S = 27^{} $ , then $ S = $ , thus
\[ h=SB\tan S=14\tan27^{\circ}=7.1\text{meters} \]
26.4. An airplane leaves an airport and travels at an average speed of 450 kilometers per hour on a heading of $ 250^{} $ . After three hours, how far south and how far west is it from its original position?
Sketch a figure (see Fig. 26-5).
In the figure, the original position is O and the final position is P. Thus $ OP = (450 )(3 ) = 1350 $ . Since the heading is $ 250^{} $ , $ AOP $ must be $ 250^{} - 180^{} = 70^{} $ .
Hence, in right triangle AOP,
\[ \frac{OA}{OP}=\cos AOP,or OA=OP\cos AOP=1350\cos70^{\circ}=462km south \]
and
\[ \frac{AP}{OP}=\sin AOP,or AP=OP\sin AOP=1350\sin70^{\circ}=1269km west \]
26.5. From a point on level ground the angle of elevation of the top of a building is $ 37.3^{} $ . From a point 50 yards closer, the angle of elevation is $ 56.2^{} $ . Find the height of the building.
Sketch a figure (Fig. 26-6).
Introduce the auxiliary variable x. The cotangent function is chosen, as it leads to the simplest algebra in eliminating x. In right triangle DBT,
\[ \cot TDB=\frac{x}{h} \]
In right triangle CBT,
\[ \cot TCB=\frac{50+x}{h} \]
Therefore.
\[ \cot TCB\;-\;\cot TDB=\frac{50\;+\;x}{h}\;-\;\frac{x}{h}=\frac{50}{h} \]
Hence.
\[ h=\frac{50}{\cot TCB-\cot TDB}=\frac{50}{\cot37.3^{\circ}-\cot56.2^{\circ}}=78yd \]
Note: The accuracy of the result is determined by the least accurate input measurement. Also note that in calculating cotangent on a scientific calculator, the identity $ u = 1/(u) $ is used.
26.6. Derive the law of sines
Two typical situations (acute and obtuse triangles) are sketched (Fig. 26-7).
h represents an altitude of the triangle, drawn perpendicular from one vertex (shown as B) to the opposite side. For the obtuse triangle, the altitude lies outside the triangle. In general, however, triangles ADB and CDB are right triangles. In triangle ADB,
\[ \sin\alpha=\frac{h}{c}\quad so h=c\sin\alpha \]
In triangle CDB, $ BCD $ is either $ $ or $ 180^{} - $ . In either case, $ BCD = = (180^{} - ) $ ; hence
\[ \sin BCD=\sin\gamma=\frac{h}{a}\quad\text{so}h=a\sin\gamma \]
Therefore, $ a = c $ , or, dividing both sides by $ $ ,
\[ \frac{a}{\sin\alpha}=\frac{c}{\sin\gamma} \]
Note: Since the letters are assigned arbitrarily, the other cases of the law of sines can be immediately derived by replacement (sometimes called rotation of letters): replace a with b, b with c, c with a, and also $ $ with $ $ , $ $ with $ $ , $ $ with $ $ .
The law of sines also applies to a right triangle; the proof is left to the student.
26.7. Analyze the AAS and ASA cases of solving an oblique triangle.
In either case, two angles are known; hence the third can be found immediately since the sum of the angles of a triangle is $ 180^{} $ . With all three angles known, and one side given, there is enough information present to substitute into the law of sines to find the second and third sides. For example, given a, then b can be found, since
\[ \frac{a}{\sin\alpha}=\frac{b}{\sin\beta},\qquad\mathrm{so}\;b=\frac{a\sin\beta}{\sin\alpha} \]
26.8. Solve triangle ABC, given \(\alpha = 23.9^{\circ}\), \(\beta = 114^{\circ}\), and \(c = 82.8\).
Since two angles and the included side are given, this is the ASA case.
Solve for $ $ .
Since $ + + = 180^{} $ , $ = 180^{} - - = 180^{} - 23.9^{} - 114^{} = 42.1^{} $ .
Solve for a:
From the law of sines, $ = $ ; hence
\[ a=\frac{c\sin\alpha}{\sin\gamma}=\frac{82.8\sin23.9^{\circ}}{\sin42.1^{\circ}}=50.0 \]
Solve for b:
Applying the law of sines again, $ = $ , hence
\[ b=\frac{c\sin\beta}{\sin\gamma}=\frac{82.8\sin114^{\circ}}{\sin42.1^{\circ}}=113 \]
26.9. Fire station B is located 11.0 kilometers due east of fire station A. Smoke is spotted at a bearing of S23°40’E from station A and at S68°40’W from station B. How far is the fire from each fire station?
Sketch a figure (Fig. 26-8).
Given side \(AB = c = 11.0\), \(\angle S_{1}AC = 23^{\circ}40^{\prime}\), and \(\angle S_{2}BC = 68^{\circ}40^{\prime}\), it follows that
\[ \alpha=90^{\circ}-23^{\circ}40^{\prime}=66^{\circ}20^{\prime}and\beta=90^{\circ}-68^{\circ}40^{\prime}=21^{\circ}20^{\prime} \]
Thus two angles and the included side are given and this is the ASA case.
Solve for $ $ :
\[ \gamma=180^{\circ}-\alpha-\beta=180^{\circ}-66^{\circ}20^{\prime}-21^{\circ}20^{\prime}=92^{\circ}20^{\prime} \]
Solve for a:
From the law of sines, $ = $ ; hence
\[ a=\frac{c\sin\alpha}{\sin\gamma}=\frac{11.0\sin66^{\circ}20^{\prime}}{\sin92^{\circ}20^{\prime}}=10.1km from B \]
Solve for b:
Applying the law of sines again, $ = $ ; hence
\[ b=\frac{c\sin\beta}{\sin\gamma}=\frac{11.0\sin21^{\circ}20^{\prime}}{\sin92^{\circ}20^{\prime}}=4.01km from A \]
26.10. Analyze the SSA case of solving an oblique triangle.
There are several possibilities. Assume for consistency that \(a, b\), and \(\alpha\) are given. Draw a line segment of unspecified length to represent \(c\), then draw angle \(\alpha\) and side \(b\). Then the following cases can be distinguished:
α acute (see Fig. 26-9):
| $ b > a \(</td><td>\) b = a \(</td><td>\) b < a < b \(</td><td>\) a b $ | |||
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| No triangle | One triangle | Two triangles | One triangle |
α obtuse (see Fig. 26-10):
| $ a b $ | a > b |
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| No triangle | One triangle |
In every case, start by calculating the possible values of $ $ , using the law of sines:
Since $ = $ , it follows that $ = $ .
If the value of $ $ calculated in this way is greater than 1, there is no solution to this equation and no triangle is possible. If this value of $ = 1 $ , there is one (right) triangle possible. If this value of $ < 1 $ , there are two solutions for $ $ :
\[ \beta=\sin^{-1}\frac{b\sin\alpha}{a}\quad and\quad\beta^{\prime}=180^{\circ}-\sin^{-1}\frac{b\sin\alpha}{a}. \]
If both of these solutions, substituted into $ + + = 180^{} $ , yield a positive value for $ $ , then two triangles are possible; if only the first solution leads to a positive value for $ $ , then there is 1 possible triangle, and if neither solution leads to a positive value for $ $ , then there is no possible triangle.
The SSA case is sometimes referred to as the ambiguous case, both because there are so many possibilities and because there may be two triangles determined by the given information.
26.11. Solve triangle ABC, given $ = 23.9^{} $ , a = 43.7, and b = 35.1.
Sketch a figure, starting with a line segment of unspecified length to represent c: Since two sides and a nonincluded angle are given, this is the SSA case.
From Fig. 26-11, and the fact that a > b, only one triangle is determined by the given data.
Solve for $ $ :
\[ \begin{aligned}\sin\beta&=\frac{b\sin\alpha}{a}=\frac{35.1\sin23.9^{\circ}}{43.7}=0.3254\\\beta&=\sin^{-1}0.3254=19.0^{\circ}\end{aligned} \]
Solve for $ $ :
\[ \gamma=180^{\circ}-\alpha-\beta=180^{\circ}-23.9^{\circ}-19.0^{\circ}=137.1^{\circ} \]
Note that the second solution of $ $ , namely, $ {}-{-1}0.3254=161^{} $ , is too large to fit into the same triangle as $ ^{} $ ; this possibility must be discarded.
Solve for c:
From the law of sines $ = $ , hence
\[ c=\frac{a\sin\gamma}{\sin\alpha}=\frac{43.7\sin137.1^{\circ}}{\sin23.9^{\circ}}=73.4 \]
26.12. Derive the law of cosines
Two typical situations (acute and obtuse angle $ $ ) are sketched (Fig. 26-12).
In either case, since A is a point b units from the origin, on the terminal side of $ $ in standard position, the coordinates of A are given by $ (b , b ) $ . The coordinates of B are given by $ (a, 0) $ . Then the distance from A to B, labeled c, is given by the distance formula as:
\[ c=d(A,B)=\sqrt{(a-b\cos\gamma)^{2}+(0-b\sin\gamma)^{2}} \]
Squaring yields
\[ c^{2}=(a-b\cos\gamma)^{2}+(b\sin\gamma)^{2} \]
Simplifying yields:
\[ \begin{aligned}c^{2}&=a^{2}-2ab\cos\gamma+b^{2}\cos^{2}\gamma+b^{2}\sin^{2}\gamma\\&=a^{2}-2ab\cos\gamma+b^{2}(\cos^{2}\gamma+\sin^{2}\gamma)\end{aligned} \]
Hence by the Pythagorean identity,
\[ c^{2}=a^{2}+b^{2}-2a b\cos\gamma \]
Note: Since the letters are assigned arbitrarily, the other cases of the law of cosines can be immediately derived by replacement (sometimes called rotation of letters): replace a with b, b with c, c with a, and also $ $ with $ $ , $ $ with $ $ , $ $ with $ $ .
The law of cosines also applies to a right triangle; for the side opposite the right angle, it reduces to the Pythagorean theorem; the proof is left to the student.
26.13. Analyze the SAS and SSS cases of solving an oblique triangle
In the SAS case, two sides and the angle included between them are given. Label them a, b, and $ $ . Then none of the three ratios in the law of sines is known at the outset, so no information can be derived from this law. From the law of cosines, however, the third side, c, can be determined. Then with three sides and an angle known, the law of sines can be used to determine a second angle. If the smaller of the two unknown angles (smaller because it is opposite a smaller side) is chosen, this angle must be acute, hence there is only one possibility. The third angle follows immediately since the sum of the three angles must be $ 180^{} $ .
In the SSS case, again, none of the three ratios in the law of sines is known at the outset, so no information can be derived from this law. However, the law of cosines can be solved for the cosine of any unknown angle to yield:
\[ \cos\alpha=\frac{b^{2}+c^{2}-a^{2}}{2bc}\quad\cos\beta=\frac{a^{2}+c^{2}-b^{2}}{2ac}\quad\cos\gamma=\frac{a^{2}+b^{2}-c^{2}}{2ab} \]
If the angle opposite the largest side is calculated first, then if the cosine of this angle is negative, the angle is obtuse; otherwise, the angle is acute. In either case, the other two angles cannot be obtuse and must be acute. Therefore, the second angle can be found from the law of sines without ambiguity, since this angle must be acute. The third angle follows immediately since the sum of the three angles must be $ 180^{} $ .
26.14. Solve triangle ABC, given a = 3.562, c = 8.026, and $ = 14^{}23’ $ .
Since two sides and the angle included between them are given, this is the SAS case.
Solve for b:
From the law of cosines:
\[ b^{2}=a^{2}+c^{2}-2ac\cos\beta=(3.562)^{2}+(8.026)^{2}-2(3.562)(8.026)\cos14^{\circ}23^{\prime}=21.7194 \]
Hence
\[ b=\sqrt{21.7194}=4.660 \]
Solve for the smaller of the two unknown angles, which must be $ $ :
From the law of sines, $ = $ hence $ = = = 0.18986 $ . The only acceptable solution of this equation must be an acute angle; hence $ = ^{-1}0.18986 $ , or, expressed in degrees and minutes, $ = 10^{}56’ $ .
Solve for $ $ :
\[ \gamma=180^{\circ}-\alpha-\beta=180^{\circ}-10^{\circ}56^{\prime}-14^{\circ}23^{\prime}=154^{\circ}41^{\prime} \]
26.15. Solve triangle ABC, given a = 29.4, b = 47.5, and c = 22.0.
Since three sides are given, this is the SSS case. Start by solving for the largest angle, $ $ , largest because it is opposite the largest side, b.
Solve for $ $ :
From the law of cosines,
\[ \cos\beta=\frac{a^{2}+c^{2}-b^{2}}{2ac}=\frac{(29.4)^{2}+(22.0)^{2}-(47.5)^{2}}{2(29.4)(22.0)}=-0.70183 \]
The only acceptable solution of this equation must be an obtuse angle; hence $ = ^{-1}(-0.70183) $ , or, expressed in degrees, $ = 134.6^{} $ .
Solve for $ $ :
From the law of sines, $ = $ ; hence $ ===0.44090 $ . The
only acceptable solution of this equation must be an acute angle, hence $ = ^{-1}0.44090 $ , or, expressed in degrees, $ = 26.2^{} $ .
Solve for $ $ :
\[ \gamma=180^{\circ}-a-b=180^{\circ}-26.2^{\circ}-134.6^{\circ}=19.2^{\circ} \]
26.16. A car leaves an intersection traveling at an average speed of 56 miles per hour. Five minutes later, a second car leaves the same intersection and travels on a road making an angle of $ 112^{} $ with the first, at an average speed of 48 miles per hour. Assuming the roads are straight, how far apart are the cars 15 minutes after the first car has left?
Sketch a figure (see Fig. 26-13).
Let x = the required distance. Since the first car travels 56 mph for $ $ hr, it goes a distance of $ 56() = 14 $ miles. The second car travels 48 mph for $ $ hr, so it goes a distance of $ 48() = 8 $ miles. In the triangle, two sides and the angle included between them are given, hence by the law of cosines,
\[ x^{2}=8^{2}+14^{2}-2(8)(14)\cos112^{\circ}=343.9 \]
Hence $ x = = 19 $ miles to the accuracy of the input data.
26.17. A regular pentagon is inscribed in a circle of radius 10.0 units. Find the length of one side of the pentagon.
Sketch a figure (see Fig. 26-14).
Let x = the length of the side. Since the pentagon is regular, angle $ = $ of a full circle = $ 72^{} $ . Hence, from the law of cosines,
\[ x^{2}=(10.0)^{2}+(10.0)^{2}-2(10.0)(10.0)\cos72^{\circ}=138.2,so x=\sqrt{138.2}=11.8units \]
SUPPLEMENTARY PROBLEMS
26.18. Solve a right triangle given \(a = 350\) and \(\alpha = 73^{\circ}\).
Ans. \(\beta = 17^{\circ}, b = 107, c = 366\)
26.19. Solve a right triangle given \(b = 9.94\) and \(c = 12.7\).
Ans. \(a = 7.90, \beta = 51.5^{\circ}, \alpha = 38.5^{\circ}\)
26.20. A rectangle is 173 meters long and 106 meters high. Find the angle between a diagonal and the longer side.
Ans. $ 31.5^{} $
26.21. From the top of a tower the angle of depression of a point on level ground is $ 56^{}30’ $ . If the height of the tower is 79.4 feet, how far is the point from the base of the tower?
Ans. 52.6 feet
26.22. A radio antenna is attached to the top of a building. From a point 12.5 meters from the base of the building, on level ground, the angle of elevation of the bottom of the antenna is 47.2° and the angle of elevation of the top is 51.8°. Find the height of the antenna.
Ans. 2.39 meters
26.23. Show that the law of sines holds for a right triangle.
26.24. Show that the law of cosines holds for a right triangle, and reduces to the Pythagorean theorem for the side opposite the right angle.
26.25. Show that the area of a triangle can be expressed as one-half the product of any two sides times the sine of the angle included between them. $ (A=bc) $
26.26. How many triangles are possible on the basis of the given data?
\(\alpha = 20^{\circ}, b = 30, \gamma = 40^{\circ};\) (b) \(\alpha = 20^{\circ}, b = 30, a = 5;\) (c) \(a = 30, c = 20, \gamma = 10^{\circ};\)
\(a = 30, c = 30, \gamma = 100^{\circ};\) (e) \(\beta = 20^{\circ}, b = 50, c = 30\)
Ans. (a) 1; (b) 0; (c) 2; (d) 0; (e) 1
26.27. Solve triangle ABC given (a) $ = 35.5^{} $ , $ = 82.6^{} $ , c = 7.88; (b) $ = 65^{}50’ $ , $ = 78^{}20’ $ , c = 15.3.
Ans. (a) $ = 61.9^{} $ , b = 4.61, a = 7.01; (b) $ = 35^{}50’ $ , a = 23.8, b = 25.6
26.28. Solve triangle ABC given (a) a = 12.3, b = 84.5, $ = 71.0^{} $ ; (b) a = 84.5, b = 12.3, $ = 71.0^{} $ ; (c) a = 4.53, c = 6.47, $ = 39.3^{} $ ; (d) a = 934, b = 1420, $ = 108^{} $ .
Ans. (a) No triangle can result from the given data; (b) \(\beta = 7.91^{\circ}, \gamma = 101^{\circ}, c = 87.7\);
- Two triangles can result from the given data; triangle 1: \(\gamma = 64.8^{\circ}\), \(\beta = 75.9^{\circ}\), \(b = 6.94\), triangle 2: \(\gamma' = 115.2^{\circ}\), \(\beta' = 25.5^{\circ}\), \(b' = 3.08\); (d) \(\alpha = 38.7^{\circ}\), \(\gamma = 33.3^{\circ}\), \(c = 819\)
26.29. How many triangles are possible on the basis of the given data?
$ a = 30, = 40^{}, c = 50 $ ; (b) a = 80, b = 120, c = 30; (c) a = 40, b = 50, c = 35;
$ = 75^{}, = 35^{}, = 70^{} $ ; (e) a = 40, b = 40, $ = 130^{} $
Ans. (a) 1; (b) 0; (c) 1; (d) an infinite number; (e) 1
26.30. Solve triangle ABC given (a) b = 78, c = 150, $ = 83^{} $ ; (b) a = 1260, b = 1440, c = 1710.
Ans. (a) \(a = 160, \beta = 29^{\circ}, \gamma = 68^{\circ}\); (b) \(\alpha = 46.2^{\circ}, \beta = 55.5^{\circ}, \gamma = 78.3^{\circ}\)
26.31. Points A and B are on opposite sides of a lake. To find the distance between them, a point C is located 354 meters from B and 286 meters from A. The angle between AB and AC is found to be $ 46^{}20’ $ . Find the distance between A and B.
Ans. 485 meters
26.32. Two sides of a parallelogram are 9 and 15 units in length. The length of the shorter diagonal of the parallelogram is 14 units. Find the length of the long diagonal.
Ans. $ 4 $ units
26.33. A plane travels 175 miles with heading $ 130^{} $ and then travels 85 miles with heading $ 255^{} $ . How far is the plane from its starting point?
Ans. 144 miles
26.34. (a) Use the law of sines to show that in any triangle $ = $ .
- Use the result of part (a) to derive Mollweide’s formula: $ = $
26.35. Because it contains all six parts of a triangle, Mollweide’s formula is sometimes used to check results in solving triangles. Use the formula to check the results in Problem 26.27a.
Ans. $ = 1.4746 $ , $ = 1.4751 $ ; the two sides agree to the accuracy of the given data.