Chapter 11

Oblique Triangles

Oblique Triangles

Tell all the truth but tell it slant.

• Emily Dickinson

Oblique triangles are non-right triangles. In principle, we can solve any determined oblique triangle by cutting it into right triangles, and then applying right-triangle trigonometry; in practice, this strategy leads to messy computations. A better strategy is to prove – and then use – two theorems that apply directly to oblique triangles: the law of sines and the law of cosines.

The Law of Sines

So then with the mind, I myself serve the law of God; but with the flesh, the law of sin.

• Romans 7:25

The law of sines links four parts of a triangle: any two sides you like, and the angles opposite them. So in the triangle at right, the law of sines will link a, b, $ $ , and $ $ . We can discover the law by simply dropping a perpendicular. Its length is, on the one hand, a $ $ , and on the other, b $ $ (by basic right-angle trigonometry). Thus, a $ = b $ . We can polish this into a more symmetric form by dividing both sides by $ $ , thereby obtaining the law of sines:

The Law of Sines.

If \(a\) and \(b\) are any two sides of a triangle, and \(\alpha\) and \(\beta\) are the angles opposite them, then

\[ \frac{a}{sin\alpha}=\frac{b}{sin\beta}. \]

When a determined triangle has two known angles (ASA or AAS), we can solve it with the law of sines, which I’ll often abbreviate as l.o.s.

Example. Solve the triangle shown at right.

Solution. The missing angle is, of course, $ 41^{} $ .

If we let x be the side opposite the $ 63^{} $ angle, the l.o.s. yields $ = $ .

Equivalently, $ x = (4 ^{}) / ^{} $

Calling the remaining side y and applying the l.o.s. again yields $ = $

Equivalently, $ y = (4 ^{}) / ^{} $ .

The law of sines is a foolproof tool for finding sides. Using it to find angles, however, can be dangerous, because knowing an angle’s sine is not quite enough to determine the angle itself. (For example, if we know that $ /2 $ , then $ $ could be either $ 30^{} $ or $ 150^{} $ .) Fortunately, the point is moot: You’ll never have to use the law of sines to find a triangle’s angles. The law of cosines, which you’ll learn in the next section, is eminently suited for that job.

Exercises

1. Solve the following triangles.

a)

b)

c)

1

2. Did you use the law of sines in exercise 1c? If so, then why? Using the law of sines on a right triangle is excessive – nay, uncouth – like using the quadratic formula to solve $ x^{2}=4 $ . If you did commit this faux pas, go back and solve the triangle again with basic right-angle trigonometry. If you didn’t, then verify (but just this once) that the law of sines works in this context too.

3. The best way to appreciate the law of sines is to solve an oblique triangle without it. Redo exercise 1b by dropping a perpendicular and using basic right-angle trigonometry. Then compare your two solutions.

4. If you examine my derivation of the law of sines carefully, you’ll see that it fails if $ $ or $ $ is obtuse, since in that case (as in the figure at right), the perpendicular falls outside the triangle.

Your problem: Prove that the law of sines holds for obtuse triangles, too.

[Hint: Find two different expressions for CD. For one, you’ll need an identity for $ (180^{}-) $ . You found one in Chapter 10, Exercise 42, which you can recover by drawing a picture and thinking about the unit circle.]

5. Solve the following triangles:

a)

b)

c)

6. The verdant slope of Eden Hill makes a gentle $ 7^{} $ angle with the horizontal plain from which it rises. Divinely parallel, the sun’s rays shine forth; those that strike the plain do so at $ 40^{} $ angles. The forbidden Tree of Knowledge of Trigonometry stands on the hill, growing in the direction perpendicular to the plain. Sidling up to you, the serpent motions towards the tree’s shadow. “That shadow,” it says, “is 150 cubits long.” How tall is the tree?

7. (A challenge) Use a computer program (or a compass) to draw a circle with a fixed diameter D of your choice. Now draw a triangle whose vertices lie on the circle. Divide the length of each side of this triangle by the sine of its opposite angle. The result, surprisingly, will always be D! Try to discover why this is so.

The Law of Cosines

The law of cosines relates four different parts of a triangle: the three sides… and any angle of your choice. Let’s call the chosen angle $ $ , and its opposite side c. Thus, the law of cosines will be an equation that links a, b, c, and $ $ (where a and b are the remaining sides). Let’s derive this law.

Our strategy: Drop a perpendicular to make c into a right triangle’s hypotenuse, then try to express this right triangle’s legs in ways that involve a, b, and $ $ . If we succeed, then the Pythagorean Theorem will bind a, b, c, and $ $ into one equation.

So… Let $ ABC $ be an oblique triangle. Drop a perpendicular from B to side AC to create a right triangle $ ABD $ , whose hypotenuse is c. By basic trig, one of its legs is $ a $ , so if we can get b into an expression for the other leg, AD, all will be well. This is easy: $ AD = b - DC = b - a $ .

We can now turn it over to Pythagoras to deliver the coup de grâce:

\[ \begin{aligned}c^{2}&=(a\sin\gamma)^{2}+(b-a\cos\gamma)^{2}\\&=a^{2}\sin^{2}\gamma+b^{2}+a^{2}\cos^{2}\gamma-2ab\cos\gamma\\&=a^{2}(\sin^{2}\gamma+\cos^{2}\gamma)+b^{2}-2ab\cos\gamma\\&=a^{2}+b^{2}-2ab\cos\gamma\end{aligned} \]

(Pythagorean Theorem)

(Pythagorean Identity).

And that’s that. Let’s summarize what we’ve just discovered.

The Law of Cosines

If \(a, b\), and \(c\) are any triangle’s sides, and \(\gamma\) is the angle opposite side \(c\), then

\[ c^{2}=a^{2}+b^{2}-2a b\cos\gamma. \]

I think it’s best to remember the law of cosines rhetorically:

In every triangle, each side’s square is

The sum of the other sides’ squares… minus

twice the product of the other sides and their included angle’s cosine. $ ^{*} $

Note 1. The law of cosines generalizes the Pythagorean Theorem. Algebraically, it appends an extra term $ (2ab ) $ to the Pythagorean Theorem to account for the effects of a triangle’s obliqueness. When $ $ is a right angle, the extra term is zero, so the law of cosines reduces to the Pythagorean Theorem itself.

Note 2. As we’ll see in the examples that follow, the law of cosines lets us solve any determined triangle in which we know at least two sides (i.e. any triangle determined by SAS or SSS).

Note 3. Unlike the law of sines, the law of cosines is well suited for finding angles, because knowing the cosine of an angle in a triangle is enough to determine the angle itself. (For instance, if $ = 1/2 $ , then $ $ can only be $ 60^{} $ .) For this reason, the law of cosines is the best tool for finding oblique triangles’ angles. It will never lead you astray. You can use the law of sines to find angles, but you must take care when doing so. (See exercise 12 for more on this.)

Example 1. Solve the triangle shown at right.

Solution. Let c be the remaining side. By the law of cosines,

\[ c^{2}=9+25-30\cos40^{\circ}. \]

Running this through a calculator, we find that $ c $ .

5

Let $ $ be the angle opposite the side of length 3. By the law of cosines,

\[ 9=25+(3.32\ldots)^{2}-2(5)(3.32\ldots)\cos\alpha.^{*} \]

Solving for $ $ , we find that $ $ . Consequently, $ =^{-1}(.814){} $ .

Finally, since the angle sum is $ 180^{} $ , the remaining angle must be approximately $ 104.5^{} $ .

Example 2. Solve the triangle shown at right.

Solution. Let $ $ be the angle opposite the side whose length is 6.

By the law of cosines,

\[ 36=25+49-2(5)(7)\cos\alpha. \]

Solving this, we find that $ =^{-1}(.543){} $

Let $ $ be the angle opposite the side whose length is 5.

By the law of cosines,

\[ 25=36+49-2(6)(7)\cos\beta, \]

From which we determine that $ ^{} $

The remaining angle is therefore $ 180^{} - (+ ) ^{} $ .

Now that you’ve learned basic right-angle trigonometry and the laws of sines and cosines, you can solve any determined triangle that you’ll ever meet. (You’ll verify this bold statement in exercise 13.)

Exercises

8. Solve the following triangles, selecting the best mathematical tools for each problem.

a)

b)

c)

d)

9. If you examine my derivation of the law of cosines carefully, you’ll see that it doesn’t quite work if $ $ is obtuse, because in that case, the perpendicular will fall outside the triangle.

Your problem: Prove that the law of cosines holds for obtuse triangles, too. [Hint: Adapt Exercise 4’s hint.]

10. In the figure below, find the length of the line segment marked x.

11. Drunk half out of her mind, Nancy attempts to walk home from the local pub, whose door, as she is wont to boast, stands a mere 648 feet due south of her own apartment. After stepping into the brisk midnight air, Nancy locates the polestar (she always was a keen student of astronomy), and manages, for precisely two minutes, to stumble along due north at a constant speed of 3.6 feet per second. Then, distracted by a passing satellite, she drifts from her course, veering 20° east of her original path, yet maintaining her constant speed.

1. Three minutes after she leaves the pub, how far is Nancy from its door?

2. Three minutes after she leaves the pub, how far is Nancy from the door to her apartment?

3. If Nancy maintains her speed and second direction ( $ 20^{} $ East of North), at what time (to the nearest minute) will she be 1 mile from the door to her apartment? (Assume she left the bar at the stroke of midnight.)

12. 1. Draw a picture showing that there are two distinct angles – one acute, one obtuse – whose sines equal 3/4.

2. If $ $ is a triangle’s angle, and $ = 3/4 $ , do we have enough information to determine $ ’ $ ’s value?

3. Suppose $ $ from part (b) is known to be acute. Now do we have enough information to determine $ $ ’s value? If so, what is it (to the nearest tenth of a degree)? If not, why not?

4. Suppose $ $ from part (b) is known to be obtuse. Now do we have enough information to determine $ ’ $ ’s value? If so, what is it (to the nearest tenth of a degree)? If not, why not?

5. If $ $ is an angle in a triangle and we know that $ = 3/4 $ , do we have enough information to determine $ $ ’s value? If so, what is it (to the nearest tenth of a degree)? If not, why not?

6. Please reread “Note 3” in the preceding section.

7. Use the law of sines – carefully! – to find the obtuse angle $ $ in the figure below to the nearest tenth of a degree. Then check your work by finding it by another method.

13. To solve any triangle determined by SSS, one could proceed as follows: Use the law of cosines to find one angle, use it again to find a second angle, and then find the third angle by subtracting the first two angles from $ 180^{} $ . Give similar sorts of directions for solving triangles determined by SAS, ASA, AAS, and RASS. Having done this, you will have demonstrated that all determined triangles can be solved using the tools we’ve developed. Therefore, we’ve truly solved trigonometry’s basic problem, which I first described in Chapter 9’s first paragraph. Huzzah!

14. It is a curious geometrical fact that in any parallelogram, the sum of the two diagonals’ squares is equal to the sum of the four sides’ squares. Prove it.

The Area of a Triangle: 2 Formulas

It has been well said that if triangles had a god, they would give him three sides.

• Montesquieu, Persian Letters, Letter 59.

In this section, we will prove two formulas for the area of a triangle. Everyone knows the first formula (half the base times the height), but not everyone knows why it holds. We will start by explaining that. We’ll then use that first famous formula to prove a second area formula, which isn’t nearly as well known, probably because it involves a trigonometric function.

Formula 1 (The old favorite). A triangle’s area is half the product of its base and height.

Proof. First, observe that we can cut any parallelogram into two pieces and then reassemble them to form a rectangle, as in the figure. This surgery neither creates nor destroys area, so the two shapes must have the same area. Thus, since the rectangle’s area is obvious (bh), we know the parallelogram’s area, too. We’ve discovered, along the way, something important in its own right:

Any parallelogram’s area is the product of its base and height.

This fact about parallelograms can teach us about triangles.

The key is that every triangle is half a parallelogram, as the figure at right illustrates. Since any parallelogram’s area is its base times its height, it follows that any triangle’s area must be half its base times its height, as claimed.

There’s one unsatisfying thing about the famous formula we’ve just proved: It involves a triangle’s height, which is not one of the six “natural” triangle parts (sides and angles). We’ll now prove another formula for a triangle’s area that has no such flaw. It will give us a triangle’s area in terms of two sides and their included angle. For this reason, I call it the “SAS area formula”.

Formula 2 (SAS area formula).

A triangle’s area is half the product of any two sides and their included angle’s sine.

Proof. Let a and b be any two sides and let $ $ be their included angles. If we think of b as the triangle’s base, then relative to it, the triangle’s height is, by basic right-angle trig, $ a $ . Thus, by Formula 1, the triangle’s area is $ (1/2)(b)(a ) $ . Rearranging the factors, this becomes $ (1/2)(ab ) $ , which is half the product of the two sides times the sine of their included angles, as claimed.

To ensure that you’ve digested this proof and can adapt it to other cases, here are a few exercises.

Exercises

15. In the preceding proof, if we had declared the triangle’s base to be the side labelled a, the SAS area formula still holds. Explain why.

16. Explain why the SAS formula for finding areas still holds even if the included angle is obtuse.

17. Use the SAS formula to find, in your head, the exact area of an equilateral triangle whose sides are 1 unit long.

The Area of a Triangle: Heron’s Formula

We’ll now derive a third area formula – one that gives a triangle’s area exclusively in terms of its sides. We’ll do this by finding a way to rewrite the $ $ in the SAS area formula in terms of the triangle’s sides. Our strategy will be a chesslike combination of two “moves”. First, we’ll use the Pythagorean identity $ (^{2}+ ^{2}= 1) $ to get $ $ out of the SAS area formula and $ $ into it. Then we’ll use the law of cosines (which relates the three sides and an angle’s cosine) to get $ $ out, and the three sides in. Checkmate! Now that you understand the strategy, let’s derive the formula. We’ll initially get it in a rough-hewn form. Then we’ll apply some algebraic polish, a task we’ll consider separately.

Problem 1. Derive a formula for a triangle’s area in terms of its three sides, a, b, and c.

Solution. If \(\theta\) is the angle between \(a\) and \(b\), then the SAS area formula tells us that

\[ \begin{aligned}AREA=\frac{1}{2}ab\sin\theta.\end{aligned} \]

Since, as outlined above, we wish to “trade” $ $ for $ $ via the Pythagorean identity (which involves the squares of $ $ and $ $ ), we’ll first need to square both sides of the SAS formula:

\[ A R E A^{2}=\frac{1}{4}a^{2}b^{2}\sin^{2}\theta. \]

By the Pythagorean identity, this is equivalent to

\[ A R E A^{2}=\frac{1}{4}a^{2}b^{2}(1-\cos^{2}\theta). \]

We’ll now eliminate $ $ . By the law of cosines, $ c^{2}=a{2}+b^{2}-2ab$ , or equivalently, $ =(a^{2}+b{2}-c^{2})/2ab $ . Substituting this into our most recent expression for area yields

\[ A R E A^{2}=\frac{1}{4}a^{2}b^{2}\left[1-\frac{(a^{2}+b^{2}-c^{2})^{2}}{4a^{2}b^{2}}\right], \]

which, apart from taking the square root of both sides, is what we wanted – an expression for the triangle’s area exclusively in terms of its sides.

We’ve found a formula, but it is nasty. With some heroic algebra, we can simplify it considerably.

Problem 2. Simplify the formula we derived in Problem 1 as much as possible.

Solution. Putting everything in the brackets over a common denominator and then polishing the result, we obtain, as you should verify,

\[ A R E A^{2}=\frac{1}{16}[4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}], \]

which is better, but still unwieldy. What else might we do? $ ^{*} $ After staring and thinking for a while, an inspired algebraist (whose mind we’ll follow in this problem; please enjoy the ride) will notice a difference of squares lurking within those brackets.

\[ A R E A^{2}=\frac{1}{16}[(2a b)^{2}-(a^{2}+b^{2}-c^{2})^{2}]. \]

Expanding it in the usual way, we obtain

\[ A R E A^{2}=\frac{1}{16}[2a b-(a^{2}+b^{2}-c^{2})][2a b+(a^{2}+b^{2}-c^{2})]. \]

Our inspired algebraist then observes, within each set of brackets, the ingredients of a squared binomial. These will be easier to see if we rearrange the stuff within the brackets as follows –

\[ A R E A^{2}=\frac{1}{16}[-(a^{2}-2a b+b^{2})+c^{2}][(a^{2}+2a b+b^{2})-c^{2}]. \]

Factoring the squared binomials inside the parentheses gives us

\[ A R E A^{2}=\frac{1}{16}[(a-b)^{2}+c^{2}][(a+b)^{2}-c^{2}]. \]

He now notices that there are differences of squares in both sets of brackets. This will be clearer to the rest of us if we write the equation this way:

\[ A R E A^{2}=\frac{1}{16}[c^{2}-(a-b)^{2}][(a+b)^{2}-c^{2}]. \]

Expanding those differences of squares yields

\[ A R E A^{2}=\frac{1}{16}[c-(a-b)][c+(a-b)][(a+b)-c][(a+b)+c]. \]

If we clear out those parentheses (distributing negatives where necessary) and then rearrange the bracketed terms in alphabetical order, this becomes

\[ A R E A^{2}=\frac{1}{16}[-a+b+c][a-b+c][a+b-c][a+b+c]. \]

Notice this expression’s pattern: in each of the first three brackets, the negative changes places; in the fourth, it vanishes. A lesser man would stop here, content with this pleasing symmetry, but our algebraist has one more trick up his sleeve. Watch closely. First, he turns that factor of 1/16 into four factors of 1/2, and shares them out among the brackets so that

\[ A R E A^{2}=\frac{\left[-a+b+c\right]}{2}\frac{\left[a-b+c\right]}{2}\frac{\left[a+b-c\right]}{2}\frac{\left[a+b+c\right]}{2}. \]

The fourth factor is the triangle’s semiperimeter (i.e. half of its perimeter), which we’ll call s. Remarkably (and it does take an inspired algebraist to notice this), all four factors are related to the semiperimeter: in order, they are $ (s - a) $ , $ (s - b) $ , $ (s - c) $ , and s, as you should verify. Thus,

\[ A R E A^{2}=(s-a)(s-b)(s-c)s. \]

Finally, we take square roots of both sides to produce our beautifully compact formula,

\[ \boldsymbol{A}\boldsymbol{R}\boldsymbol{E}\boldsymbol{A}=\sqrt{s(s-a)(s-b)(s-c)}. \]

This is called Heron’s formula, after Heron of Alexandria, the great mathematician and engineer who proved it in the $ 1^{st} $ century A.D. Let us glorify his achievement in a box.

Heron’s formula. If a triangle’s sides are a, b, and c, then its area is

\[ \sqrt{s(s-a)(s-b)(s-c)}, \]

where s is the triangle’s semiperimeter (i.e. half of its perimeter).

Proving Heron’s Formula requires mathematical artistry. Using it requires none.

Example. Find the area of the triangle whose side lengths are 3, 5, and 6.

Solution. This triangle’s semiperimeter is 7, so according to Heron’s Formula, its area is

\[ \sqrt{7(7-3)(7-5)(7-6)}=\sqrt{7\cdot4\cdot2\cdot1}=\sqrt{56}=2\sqrt{14}units^{2}. \]

Exercises

18. Without a calculator, find the following triangles’ exact areas.

19. Find these figures’ areas to the nearest hundredth of a square unit.

20. Since a triangle is completely determined by ASA, there should, in theory, be an ASA formula for a triangle’s area. Derive one. (In your formula, call the given angles $ $ and $ $ , and their included side c.)

[Hint: Start with a known area formula, and re-express each of its variables in terms of $ $ , $ $ , and c.]

21. Heron’s formula gives a triangle’s area in terms of its three sides alone. Is it possible to derive a formula that gives a triangle’s area in terms of its three angles alone? If so, do so. If not, why not?

22. The SAS area formula has a remarkable analogue for convex quadrilaterals. $ ^{*} $ Namely, any convex quadrilateral’s area is half the product of its diagonals times the sine of the angle at which the diagonals cross.

1. Draw a few pictures of convex and non-convex quadrilaterals. Explain why the statement of the formula above wouldn’t make sense for non-convex quadrilaterals.

2. “The sine of the angle at which the diagonals cross” seems ambiguous: When two lines cross, they make four angles – usually a pair of acute angles and a pair of obtuse angles. I didn’t specify which angle among these we must consider. In fact, I did not need to specify this, because all four of the angles will have the same sine. Explain why this is so.

3. Prove that the formula for the area of a convex quadrilateral holds.

4. Earlier in the chapter, we saw that the law of cosines generalizes the Pythagorean Theorem, in the sense that the law of cosines contains the Pythagorean Theorem as a special case. In a slightly more subtle way, the formula we’re considering in this problem generalizes the SAS area formula. To see this, think of a triangle as a “degenerate quadrilateral” where one side has zero length. Thus, we will think of $ ABC $ as a quadrilateral ABCD in which the points C and D coincide. If we think that way, then what are the triangles’ “diagonals”? And what are the angles between them? Ponder this until you understand how the SAS formula is indeed a special case of this formula for convex quadrilaterals.

23. Brahmagupta’s formula gives the area of a cyclic quadrilateral (i.e. a quadrilateral whose vertices lie on a circle). If a cyclic quadrilateral’s sides are a, b, c, and d, then according to Brahmagupta, its area is

\[ \sqrt{(s-a)(s-b)(s-c)(s-d)}. \]

1. Explain why Heron’s formula is a special case of Brahmagupta’s formula.

2. Brahmagupta wrote his mathematics in Sanskrit verse (!), so you almost certainly can’t read his work directly. You can, however, learn a bit about Brahmagupta himself. Do so. While you are at it, learn about the man whose famous result he generalized, Heron of Alexandria.

3. Brahmagupta’s formula, in turn, is a special case of Bretschneider’s formula, which gives the area of any convex quadrilateral – not just cyclic ones – in terms of its sides and any two opposite angles in the quadrilateral. This area turns out to be

\[ \sqrt{(s-a)(s-b)(s-c)(s-d)-a b c d\cos^{2}\left(\frac{\alpha+\gamma}{2}\right)}, \]

where $ $ and $ $ are the opposite angles. What does this formula suggest must be true about opposite angles in every cyclic quadrilateral?

4. The conjecture you hopefully made in part (c) is correct. You can find a 2300-year-old proof of it in Euclid’s Elements, the Bible of classical geometry. Read Euclid 3.22 (i.e. Book 3, Proposition 22). Its proof will refer you back to 3.21, which in turn will send you back to 3.20, both of which are important theorems about the geometry of circles. Read, ponder, and be enlightened. Every educated person should read some Euclid.