Precalculus · G.SRT.D.11

Applications of oblique triangles

You cannot stretch a tape to the far bank, the hut across the ice or the top of an antenna. How can you still find where it is, how far, and how big?

You need

Heron's formulaPrecalculus

Where you are 6 in Laws of sines and cosines

Trigonometry7

Laws of sines and cosines6

Law of sines Area from two sides and the included angle The ambiguous case Law of cosines Heron's formula

Applications of oblique triangles

Why this exists

A tape measure needs both ends in your hands. The far bank of a river, a hut across a frozen lake, the top of an antenna: you can stand at one end of the distance, never the other. Angles, though, can be measured from where you stand. Triangulation trades the distance you cannot walk for one you can, called a baseline, plus angles sighted from its ends.

The triangle that results almost never has a right angle, so the tools are the law of sines, the law of cosines, the sine area formula and Heron’s formula. This lesson is about using them on real ground: setting up the triangle and choosing which tool the measurements allow.

Definition

Triangulation locates a point R from a known segment PQ, the baseline, by measuring the angles \angle QPR and \angle PQR at its two ends. When \angle P + \angle Q < 180^\circ, two angles and the side between them is ASA, so exactly one triangle fits on each side of the line PQ, and R is fixed once you know which side it is on.

An application of oblique triangles is any measurement whose triangle has no right angle: locating a point, correcting a course, measuring land, finding a height you cannot reach. Each is solved by asking which three parts of a triangle the measurements give, and choosing the tool from that.

Intuition

Figure 1Example 2. Two sight lines from the ends of a known baseline cross at exactly one point.

Stand at P and sight the boulder: it lies somewhere on that ray, but you do not know how far along. Walk to Q and sight it again: a second ray. Two rays from different points cross at most once, and the crossing is the boulder. The baseline turns two directions into a position.

The same picture runs through every application. A course correction is the triangle formed by the start, the destination and where you actually are. A plot with more than three corners cuts into triangles along diagonals. A height you cannot reach becomes a flat triangle on the ground, to find how far away the foot is, then one right triangle to climb it.

Results

Choosing the tool. What the measurements give, not what the question asks, decides the first step.

Given	Case	First tool
two angles and any side, such as a baseline and both its end angles	ASA, AAS	angle sum, then the law of sines
two sides and the angle between them	SAS	law of cosines for the third side; \tfrac12 ab\sin C for the area
three sides	SSS	law of cosines for an angle, largest first; Heron’s formula for the area
two sides and an angle not between them	SSA	law of sines with the height test: zero, one or two triangles

Triangulation. With baseline PQ = d and angles P and Q measured at its ends, R = 180^\circ - P - Q and

PR = \frac{d \sin Q}{\sin R}, \quad QR = \frac{d \sin P}{\sin R}.

Course correction. You meant to travel D on one bearing but traveled t on a bearing \theta away from it. The distance x still to go satisfies

x^2 = t^2 + D^2 - 2tD\cos\theta,

and the turn you must make is \theta + H, where H is the triangle’s angle at the destination.

Area of a plot. Diagonals from one corner cut a convex plot with n corners into n - 2 triangles, and the plot’s area is the sum of theirs.

Why it is true

Triangulation. The triangle PQR has two angles and the included side, so the angle sum gives R and the law of sines gives the rest. Each distance from a stake is across from the angle measured at the other stake: PR is across from Q.

Course correction. Call the start S, your position A and the destination H. The triangle has SA = t, SH = D and the angle \theta between them, which is SAS, so the law of cosines gives AH = x. You were heading along SA extended past A. The turn is the angle between that extension and AH, which is the exterior angle of the triangle at A, and an exterior angle equals the sum of the two interior angles not next to it: \theta + H. When t < D, the angle H is across from SA, which is not the longest side, so H is acute and the inverse sine returns it correctly.

Area. In a convex plot every diagonal from one corner stays inside, so the triangles do not overlap and together fill the plot. Area adds over pieces that do not overlap.

Heights. If the foot T of an object is level with the observer at A and the object stands vertical, the triangle from A to the foot and the top has a right angle at T, so the height is AT \tan(\text{elevation}). The flat triangle supplies AT.

Worked examples

Example 1

Choosing the tool

For each, name the case and the first tool. Do not solve (a) to (c).

(a) A triangular lot has sides 120 and 95 meters meeting at a 38^\circ corner; find the third side. (b) A triangular lot has sides 70, 85 and 110 meters, with no angles measured; find its area. (c) A point is sighted from both ends of a 300-meter baseline; find its distance from one end. (d) A triangle has b = 30, a = 22 and A = 40^\circ; find B.

Solution

(a) Two sides and the angle between them: SAS, so the law of cosines. The law of sines cannot start, since no side is paired with the angle across from it.

(b) Three sides: SSS. For the area alone, Heron’s formula needs no angle at all.

(d) Two sides and an angle not between them: SSA, so the height test first. The height from C is h = b \sin A = 30 \sin 40^\circ \approx 19.28. Since h < a < b, the side a reaches the base line in two places and two triangles fit. These measurements do not fix the triangle; a surveyor would need one more.

Example 2

Locating a boulder

Halona wants to locate a boulder R on the far bank of a river. She drives stakes P and Q 240 meters apart along the straight near bank, Q east of P. At P the angle between the bank (toward Q) and the sight line to the boulder is 48^\circ; at Q the matching angle is 63^\circ. Using coordinates with P at the origin, the x-axis east along the bank and the y-axis across the river, find the boulder’s position and its distance from each stake.

Solution

The angles and the baseline are ASA. The third angle is R = 180^\circ - 48^\circ - 63^\circ = 69^\circ. By the law of sines, each distance sits across from the angle at the other stake:

\begin{aligned} PR &= \frac{240 \sin 63^\circ}{\sin 69^\circ} \approx 229.06, \\ QR &= \frac{240 \sin 48^\circ}{\sin 69^\circ} \approx 191.04. \end{aligned}

The boulder lies on the ray from P at 48^\circ above the x-axis, at distance PR, so by the definition of cosine and sine its coordinates are

\begin{aligned} &\big(229.06\cos 48^\circ,\ 229.06\sin 48^\circ\big) \\ &\approx (153.27,\ 170.22). \end{aligned}

Check from the other stake: 240 - QR\cos 63^\circ \approx 153.27 and QR \sin 63^\circ \approx 170.22. The boulder is 170.22 meters straight out from the bank, 153.27 meters east of P.

Example 3

A snowmobile off course

Figure 2Example 3. Where you meant to go, where you went, and where you started make a triangle.

Aleksi rides a snowmobile from a landing S toward an ice-fishing hut H, 18 kilometers away on a bearing of 060^\circ. After 7.0 kilometers he finds he has been riding on 072^\circ. How far is he from the hut, and on what bearing must he ride now?

Solution

Call his position A. At S the two directions differ by 72^\circ - 60^\circ = 12^\circ, and the sides around that angle are SA = 7.0 and SH = 18: SAS. By the law of cosines,

\begin{aligned} x^2 &= 7^2 + 18^2 - 2(7)(18)\cos 12^\circ \\ &\approx 126.51, \quad x \approx 11.25. \end{aligned}

Now the triangle has a known side-and-angle pair, x across from 12^\circ, so the law of sines gives the angle at the hut, across from SA = 7:

\sin H = \frac{7 \sin 12^\circ}{11.25}, \qquad H \approx 7.43^\circ.

H is acute because 7 is not the longest side. The turn is the exterior angle at A, 12^\circ + 7.43^\circ = 19.43^\circ. The hut’s bearing, 060^\circ, is to the left of his heading, so he turns left: 72^\circ - 19.43^\circ \approx 052.6^\circ.

He has 11.25 kilometers to go on 052.6^\circ. On course he would have had 11, so the mistake costs about 0.25 kilometer. Check by components as in Bearings and navigation: the hut minus his position points on 052.6^\circ.

Example 4

A cranberry bog

Figure 3Example 4. One diagonal cuts the plot into a two-sides-and-angle triangle and a three-sides triangle.

Eveliina surveys a four-sided cranberry bog ABCD. She measures AB = 62, BC = 48, CD = 55 and DA = 70 meters, and the corner angle B = 104^\circ. Find the area.

Solution

Four sides alone do not fix a four-sided shape: the corners can hinge, changing the area. The angle at B is what pins it. The diagonal AC splits the bog into two triangles.

Triangle ABC is SAS. Its area is \tfrac12(62)(48)\sin 104^\circ \approx 1443.80, and the law of cosines gives the diagonal:

\begin{aligned} AC^2 &= 62^2 + 48^2 - 2(62)(48)\cos 104^\circ, \\ AC &\approx 87.11. \end{aligned}

Triangle ACD now has three known sides, 87.11, 55 and 70: SSS, so Heron’s formula. Keeping AC unrounded, s \approx 106.05 and

\sqrt{s(s - AC)(s - 55)(s - 70)} \approx 1923.15.

The bog covers 1443.80 + 1923.15 \approx 3366.95 square meters.

Example 5

An antenna across a quarry

Figure 4Example 5. The horizontal triangle finds how far away the antenna’s foot is; one right triangle then climbs it.

An antenna stands on the far rim of a quarry, level with the near rim, and Halona cannot reach its foot T. On the near rim she lays out AB = 150 meters. At A the horizontal angle between B and T is 64^\circ; at B the angle between A and T is 71^\circ. From A the angle of elevation to the top of the antenna is 14^\circ. How tall is it?

Solution

The flat triangle ABT is ASA, with T = 180^\circ - 64^\circ - 71^\circ = 45^\circ. The distance AT is across from the angle at B:

AT = \frac{150 \sin 71^\circ}{\sin 45^\circ} \approx 200.57.

The antenna is vertical and its foot is level with A, so A, the foot and the top form a right triangle with the 14^\circ angle at A:

\text{height} = 200.57 \tan 14^\circ \approx 50.0 \text{ meters}.

Check: BT = \dfrac{150 \sin 64^\circ}{\sin 45^\circ} \approx 190.66, so from B the top should appear at \tan^{-1}(50.0/190.66) \approx 14.7^\circ, and a second sighting there can confirm it. In Angles of elevation and depression the two sightings lie on a line toward the object; here the quarry blocks that line, so the baseline runs sideways.

Example 6

Two cairns across a lake

Two stone cairns X and Y stand on the far shore of Aleksi’s lake. He lays out AB = 400 meters on the near shore. At A he measures \angle BAX = 112^\circ and \angle BAY = 50^\circ; at B, \angle ABX = 38^\circ and \angle ABY = 98^\circ. How far apart are the cairns?

Solution

Neither cairn can be reached, but each is located from the baseline, and then a third triangle joins them.

In triangle ABX, X = 180^\circ - 112^\circ - 38^\circ = 30^\circ, so AX = \dfrac{400 \sin 38^\circ}{\sin 30^\circ} \approx 492.53.

In triangle ABY, Y = 180^\circ - 50^\circ - 98^\circ = 32^\circ, so AY = \dfrac{400 \sin 98^\circ}{\sin 32^\circ} \approx 747.49.

Both cairns are on the same side of AB, so the angle between the sight lines at A is 112^\circ - 50^\circ = 62^\circ. Triangle AXY is SAS:

\begin{aligned} XY^2 &= 492.53^2 + 747.49^2 \\ &\quad - 2(492.53)(747.49)\cos 62^\circ, \end{aligned}

so XY \approx 675.01 meters. Check from B: BX \approx 741.75, BY \approx 578.23, the angle between them is 98^\circ - 38^\circ = 60^\circ, and the law of cosines again gives 675.01.

Boundaries

Thin triangles magnify errors. When the angle at the far point is small, its sine is small and changes fast compared with its size, so a tiny error in a sighted angle moves the answer a lot (Exercise 15). A baseline too short for the distance gives exactly this triangle.
The ground is treated as a plane. Over a few miles that is accurate. Over hundreds of miles the surface is a sphere, a triangle’s angles add to more than 180^\circ, and these formulas no longer hold exactly.
Splitting into triangles needs the diagonals inside. In a plot with a dented corner, a diagonal from the wrong corner runs outside the plot; choose diagonals that stay in.
The height method needs a level, vertical object. If the foot is higher or lower than the observer, that difference must be measured and added.
SSA data may not fix a triangle. The height test can return two triangles, and then one more measurement is needed.

Where this goes next

A course correction is a difference of two displacements, and vectors in component form handle it without choosing a law at all, which is the check used in Example 3. Real surveys chain triangles: one triangle’s side, once computed, is the baseline for the next, so a single measured baseline can map a whole region.

Exercises

Choosing the tool

Name the case and the first tool, then find the quantity. (a) a = 14, b = 9, C = 52^\circ; find c. (b) Sides 11, 13, 20; find the area. (c) A = 30^\circ, B = 45^\circ, a = 10; find b. (d) a = 8, b = 5, C = 70^\circ; find the area.
A = 35^\circ, a = 12, b = 18. How many triangles fit? Find B in each.
A triangle has sides 7, 10 and 12. Find its largest angle and its area.

Locating a point

Points A and B are 500 meters apart on a straight shore. A buoy is sighted at 57^\circ from the shore at A and 66^\circ at B. Find its distance from A, from B, and from the shore.
A baseline runs from (0, 0) to (300, 0). A point above the x-axis is sighted at 72^\circ from the baseline at the origin and 44^\circ at (300, 0). Find its coordinates.
A baseline AB is 180 meters. A point R is sighted with \angle BAR = 118^\circ and \angle ABR = 27^\circ. Find AR, BR, and the distance from R to the line AB.

Course corrections

Eveliina plans to fly 240 miles on 135^\circ. After 80 miles she finds she has been flying 125^\circ. How far is the destination, on what bearing, and which way does she turn?
Halona plans to hike 9 kilometers on 300^\circ, but walks 4 kilometers on 315^\circ. How far is she from her goal, on what bearing, and through what angle does she turn?
Aleksi rides 5 kilometers on a heading 15^\circ off the one he meant. How far is he from where 5 kilometers on course would end?

Land area

A triangular lot has sides 45, 60 and 80 meters. Find its area.
A plot ABCD has AB = 40, BC = 55, B = 120^\circ, CD = 50 and DA = 45 meters. Find AC and the area.
From a point O inside a triangular field with corners A, B, C, Eveliina measures OA = 30, OB = 42, OC = 36 meters, \angle AOB = 110^\circ, \angle BOC = 130^\circ and \angle COA = 120^\circ. Find the field’s area.

Heights out of reach

A flagpole stands on an island, its foot level with the shore. A baseline AB = 120 meters runs along the shore. The horizontal angles to the foot are 70^\circ at A and 58^\circ at B, and the elevation of the top from A is 18^\circ. How tall is the flagpole?
Same setup with AB = 90 meters, horizontal angles 81^\circ at A and 64^\circ at B, and elevation 23^\circ measured from B. Find the height.

Reasoning

A baseline is 1.00 kilometer, and both end angles are 88^\circ. Find the distance from A to the point. Then find it if the angle at B is really 88.5^\circ. Explain the difference and what it says about choosing a baseline.
Explain why four measured sides do not give the area of a four-sided plot, with two shapes as evidence.
In Example 3 the angle at the hut was acute. Describe a course error in which the angle at the destination is obtuse, and explain why the inverse sine would then mislead.

Answers

(a) SAS, law of cosines: \displaystyle c = \sqrt{196 + 81 - 252\cos 52^\circ} \approx 11.04. (b) SSS, Heron: s = 22, area \displaystyle = \sqrt{22 \cdot 11 \cdot 9 \cdot 2} = \sqrt{4356} = 66. (c) AAS, law of sines: \displaystyle b = \dfrac{10\sin 45^\circ}{\sin 30^\circ} = 10\sqrt{2} \approx 14.14. (d) SAS, sine area: \tfrac12(8)(5)\sin 70^\circ \approx 18.79.
Two. h = 18\sin 35^\circ \approx 10.32 and h < 12 < 18. \sin B = \dfrac{18\sin 35^\circ}{12}, so B \approx 59.36^\circ or B \approx 120.64^\circ.
The largest angle is across from 12: \cos C = \dfrac{49 + 100 - 144}{140} = \dfrac{5}{140}, so C \approx 87.95^\circ. Heron: s = 14.5, area \displaystyle = \sqrt{14.5 \cdot 7.5 \cdot 4.5 \cdot 2.5} \approx 34.98.
The angle at the buoy is 57^\circ. From A: \dfrac{500\sin 66^\circ}{\sin 57^\circ} \approx 544.64 meters. From B: exactly 500 meters, since the angles at A and at the buoy are equal, so the triangle is isosceles. From the shore: 544.64\sin 57^\circ \approx 456.77 meters.
The third angle is 64^\circ, and the distance from the origin is \dfrac{300\sin 44^\circ}{\sin 64^\circ} \approx 231.86. Its coordinates are x = 231.86\cos 72^\circ and y = 231.86\sin 72^\circ, so the point is (71.65,\ 220.52).
R = 35^\circ. AR = \dfrac{180\sin 27^\circ}{\sin 35^\circ} \approx 142.47 meters, BR = \dfrac{180\sin 118^\circ}{\sin 35^\circ} \approx 277.09 meters. Distance to the line: BR\sin 27^\circ \approx 125.79 meters. The foot of that perpendicular lies past A, outside the baseline, because A is obtuse.
The angle at the start is 10^\circ: x^2 = 80^2 + 240^2 - 2(80)(240)\cos 10^\circ, x \approx 161.81 miles. Angle at the destination: \sin H = \dfrac{80\sin 10^\circ}{161.81}, H \approx 4.92^\circ. She turns right 14.92^\circ, to 139.9^\circ.
The angle at the start is 15^\circ: x^2 = 16 + 81 - 72\cos 15^\circ, x \approx 5.24 kilometers. \sin H = \dfrac{4\sin 15^\circ}{5.24}, H \approx 11.40^\circ. She turns left 26.40^\circ, to 288.6^\circ.
Both points are 5 kilometers from the start, 15^\circ apart: x = 2(5)\sin 7.5^\circ \approx 1.31 kilometers, the same as \displaystyle \sqrt{50 - 50\cos 15^\circ}.
s = 92.5, so the area is \displaystyle \sqrt{92.5(47.5)(32.5)(12.5)}, about 1336.02 square meters.
AC^2 = 1600 + 3025 - 4400\cos 120^\circ = 6825, so \displaystyle AC = 5\sqrt{273} \approx 82.61. Triangle ABC: \tfrac12(40)(55)\sin 120^\circ \approx 952.63. Triangle ACD by Heron, s \approx 88.81: about 966.95. Total \approx 1919.58 square meters.
Three SAS triangles around O: \tfrac12(30)(42)\sin 110^\circ + \tfrac12(42)(36)\sin 130^\circ + \tfrac12(36)(30)\sin 120^\circ \approx 592.01 + 579.13 + 467.65 \approx 1638.79 square meters.
The angle at the foot is 52^\circ. AT = \dfrac{120\sin 58^\circ}{\sin 52^\circ} \approx 129.14, height \approx 129.14\tan 18^\circ \approx 41.96 meters.
The angle at the foot is 35^\circ. The sighting is from B, so use BT = \dfrac{90\sin 81^\circ}{\sin 35^\circ} \approx 154.98; height \approx 154.98\tan 23^\circ \approx 65.78 meters.
The point’s angle is 4^\circ: \dfrac{\sin 88^\circ}{\sin 4^\circ} \approx 14.33 kilometers. With 88.5^\circ at B it is 3.5^\circ: \dfrac{\sin 88.5^\circ}{\sin 3.5^\circ} \approx 16.37 kilometers. Half a degree moved the answer about 2 kilometers, because the half degree is an eighth of the small angle at the point, and its sine drops in proportion. A baseline long enough to make that angle large keeps errors small.
The sides can hinge at the corners. A square with sides 10 has area 100; a rhombus with sides 10 and a 30^\circ corner has area 10 \cdot 10\sin 30^\circ = 50. Same four sides, different areas, so a diagonal or an angle must also be measured.
If the traveler overshoots, going farther than the destination’s distance (t > D), the side SA can be the longest, and the angle at the destination, across from it, can be obtuse. The inverse sine returns only angles up to 90^\circ, so it would give the acute supplement; find that angle with the law of cosines instead, or find the angle at A first.