Core practice ← Back to lesson

Trigonometry in Measurement: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

0 of 10 done

Progress saved in this browser.

Problem 1 of 10

Problem 1 A field observation

An observer and a vertical tree stand on the same horizontal ground in a common vertical sight plane. The eye is $1.55$ m above ground.

The horizontal separation from the observer to the tree base is $24.0$ m. The elevation angle to the top, measured from the eye's horizontal, is $41.0^\circ$ .

Estimate the whole tree height to the nearest tenth of a meter.

Only the stated measurements and the unknown height are labeled.

Text description of this figure

Schematic, not for measuring. E is the eye above ground point F. The object BT is vertical. F and B have the same ground elevation. The dashed horizontal through E meets BT and forms a right angle. The line ET rises at 41° above that horizontal. The horizontal run is 24 m, eye height is 1.55 m, and object height is labeled H.
1. Hint 1
  Draw a horizontal reference through the eye and identify the vertical part above it.
2. Hint 2
  Tangent connects that rise with the horizontal run. The whole height also contains the eye height.
Answer

The modeled tree height is approximately $22.4$ m.

Full solution

The right triangle has horizontal run $24$ m and vertical rise $H-1.55$ from eye level to the top.
The same ground elevation at both bases justifies that subtraction.

Use tangent.
$\tan41^\circ=\frac{H-1.55}{24}.$

Solve before rounding.
$H=1.55+24\tan41^\circ.$
The whole modeled height is approximately $22.4$ m.

As a check, subtracting the eye height from the unrounded whole height and dividing by $24$ returns $\tan41^\circ$ .

The completed height label is rounded after the calculation.

Answer

The modeled tree height is approximately $22.4$ m.

Key idea

A sight-line rise starts at the eye, so account for the eye’s elevation before reporting a whole height.
Problem 2 A straight ramp

A straight ramp is $6.80$ m long along its surface and makes $7.00^\circ$ with the horizontal.

Find its vertical rise and horizontal run to the nearest hundredth of a meter. Find its percent gradient to the nearest tenth of a percent.

The given surface length is the sloping side.

Text description of this figure

Schematic, not for measuring. A straight ramp is the sloping hypotenuse, labeled 6.80 m. The horizontal run is d and the perpendicular vertical rise is r. The angle with the horizontal is labeled 7°.
1. Hint 1
  The ramp surface is the hypotenuse of the profile triangle.
2. Hint 2
  Use sine for the rise and cosine for the horizontal run. Form the gradient from the unrounded rise and run.
Answer

Rise approximately $0.83$ m; horizontal run approximately $6.75$ m; percent gradient approximately $12.3\%$ .

Full solution

The rise is opposite the inclination, and the sloped surface is the hypotenuse.
$r=6.80\sin7^\circ.$
Thus the rise is approximately $0.83$ m.

The horizontal run is adjacent to the inclination.
$d=6.80\cos7^\circ.$
The run is approximately $6.75$ m.

Use the unrounded values in rise/run.
$G=100\left(\frac rd\right).$
The gradient is approximately $12.3\%$ .

The squared unrounded rise and run sum to $6.80^2$ , checking their roles as legs.
The run is shorter than the surface, as a leg must be.

The profile keeps one geometric scale; computed labels are rounded.

Answer

Rise approximately $0.83$ m; horizontal run approximately $6.75$ m; percent gradient approximately $12.3\%$ .

Key idea

The gradient uses the horizontal run, even when the supplied length follows the surface.
Problem 3 Certifying a reference

A surveyor draws sight line $EP$ from higher point $E$ to lower point $P$ in one vertical plane. Reference line $h$ passes through $E$ , and reference line $r$ passes through $P$ . The two acute alternate interior angles made by $EP$ with $h$ and $r$ are equal.

(a) Line $h$ has been independently checked perpendicular to a true vertical line $v$ . May the surveyor treat $r$ as horizontal? Explain which conclusions follow from the equal readings and from the vertical check.

(b) A different survey records the same equality of alternate interior angles but checks neither reference against a true vertical or a true horizontal. Do those equal readings certify an elevation or depression angle? Explain.

Use the equal readings and the verified vertical separately.

Text description of this figure

E is higher and left of P. Reference h through E is marked perpendicular to a true vertical line v. Reference r passes through P. Sight line EP crosses the two references. The two acute alternate interior angles carry matching single arcs and the same label alpha, as given. No parallel arrows or perpendicular mark for r are supplied.
1. Hint 1
  Distinguish showing that two references are parallel from showing that they are horizontal.
2. Hint 2
  Use the converse of the alternate interior angle theorem, then ask what the true-vertical check adds.
Answer

(a) Yes, $r$ is horizontal. (b) No. Equal alternate interior readings alone certify parallel references, not horizontal ones.

Full solution

(a) The marked angles are alternate interior angles made by transversal $EP$ .
Their equality gives
$h\parallel r.$

Line $h$ is perpendicular to true vertical $v$ .
A line parallel to $h$ is also perpendicular to $v$ , so $r$ is horizontal in this vertical plane.
The equal readings establish parallelism; the independent vertical check establishes horizontality.

(b) Equal alternate interior readings still make the references parallel.
Both references could nevertheless be tilted.
Turning both references and the sight line together preserves the equal angles while a true vertical stays fixed.

Without an independent horizontal or true-vertical check in part (b), neither reading is certified as an elevation or depression angle.
Those names require an actual horizontal reference.

In part (b), parallel tilted references can keep the two readings equal.

Answer

(a) Yes, $r$ is horizontal. (b) No. Equal alternate interior readings alone certify parallel references, not horizontal ones.

Key idea

Parallel reference lines still need an independent horizontal or true-vertical check.
Problem 4 A recorded percentage

A straight ramp has vertical rise $1.40$ m and surface length $10.0$ m. Jo writes, "The gradient is 14 percent because 1.40 divided by 10.0 is 0.14."

Identify the mismatched length. Find the horizontal run to the nearest hundredth of a meter and the correct percent gradient to the nearest tenth of a percent. Explain why the corrected percentage is slightly greater than Jo's.

The two known lengths have different side roles.

Text description of this figure

Schematic, not for measuring. A straight ramp is the sloping hypotenuse, labeled 10.0 m. The horizontal run is d and the perpendicular vertical rise is 1.40 m. The angle with the horizontal is labeled inclination.
1. Hint 1
  Which length belongs in the denominator of a rise/run ratio?
2. Hint 2
  The surface length is the hypotenuse. Recover the horizontal leg from the two known lengths.
3. Hint 3
  A smaller positive denominator produces a larger ratio with the same rise.
Answer

Jo used the surface length instead of the horizontal run. The run is approximately $9.90$ m and the gradient approximately $14.1\%$ , slightly greater than $14\%$ .

Full solution

Percent gradient uses rise divided by horizontal run.
Jo's denominator $10.0$ m follows the surface, so it is the hypotenuse rather than the required leg.

Find the run by Pythagoras.
$d=\sqrt{10.0^2-1.40^2}.$
The horizontal run is approximately $9.90$ m.

Use the unrounded run.
$G=100\left(\frac{1.40}{d}\right).$
The correct gradient is approximately $14.1\%$ .

The horizontal run is less than the sloped $10.0$ m surface.
Dividing the same positive rise by that smaller length makes the gradient greater than Jo's $14\%$ .

The recovered run is used before rounding the gradient.

Answer

Jo used the surface length instead of the horizontal run. The run is approximately $9.90$ m and the gradient approximately $14.1\%$ , slightly greater than $14\%$ .

Key idea

Rise divided by surface length is a sine ratio, whereas gradient uses rise divided by horizontal run.
Problem 5 An incomplete field record

A report concerns a vertical mast and an observer whose eye is $1.6$ m above the observer's ground. The sight line to the top is $33^\circ$ above a true horizontal.

The only distance recorded is $40$ m along sloping ground from the observer's foot to the mast base. No ground inclination or vertical difference between the bases is known.

Decide whether $1.6+40\tan33^\circ$ is a supported mast-height estimate. Identify the missing geometric information and describe a measurement plan that would supply it. Do not calculate an unsupported height.
1. Hint 1
  The tangent model needs a horizontal run and a vertical rise from the eye.
2. Hint 2
  Eye height above local ground need not be eye elevation above the mast base.
Answer

The estimate is unsupported. Measure the horizontal separation and the vertical difference between the two base levels, and verify a common vertical sight plane. Then use the eye elevation relative to the mast base in the tangent model.

Full solution

The $40$ m tape measurement follows sloping ground.
It is not the horizontal adjacent leg required by the tangent equation, and it does not establish the vertical difference between the observer's ground and mast base.

Record a horizontal separation $d$ and a signed vertical offset $g$ , positive when the observer ground is higher than the mast base.
Verify that the sight line, vertical mast, and observer lie in the selected vertical plane.

The eye elevation relative to the mast base is $1.6+g$ .
With that recorded geometry, a supported model is
$H=1.6+g+d\tan33^\circ.$

The original record leaves both $d$ and $g$ unspecified.
Different sloping-ground configurations can have the same tape length and sight angle but yield different whole mast heights.

Answer

The estimate is unsupported. Measure the horizontal separation and the vertical difference between the two base levels, and verify a common vertical sight plane. Then use the eye elevation relative to the mast base in the tangent model.

Key idea

A numerical formula needs the horizontal distance and the correct vertical reference, not just any recorded distance.
Problem 6 A diagram for a sighting

A vertical tower and a survey station stand on the same horizontal ground. Instrument eye $E$ is $2.0$ m above station foot $F$ . The horizontal distance from $F$ to tower base $B$ is $18.0$ m. The elevation angle from $E$ to top $T$ is $48.0^\circ$ . All points lie in one vertical plane.

Draw and annotate a model on paper. Include the horizontal through $E$ , its intersection $C$ with the tower, a right-angle mark, the given angle and lengths, the full tower height $H$ , and the part of that height above eye level. Write an equation for $H$ ; a decimal height is not requested.
1. Hint 1
  The horizontal through the eye meets the vertical tower at a right angle.
2. Hint 2
  Label the opposite leg as the whole height minus the eye elevation.
Answer

The diagram has $EC$ horizontal and $CT$ vertical, with a right angle at $C$ and elevation $48^\circ$ at $E$ . Label $EC$ as $18$ m, $BC$ as $2$ m, and $CT$ as $H-2$ . The model is
$H=2+18\tan48^\circ.$

Full solution

Place $F$ and $B$ on a horizontal ground line.
Draw $EF$ and $BT$ vertical.
Put $E$ two meters above $F$ , join $E$ to $T$ , and draw the horizontal $EC$ to the tower.

Since $EC$ is horizontal and $BT$ vertical, mark a right angle at $C$ .
Mark the $48^\circ$ elevation between $EC$ and $ET$ , not between the sight line and the vertical.

Equal horizontal levels give $EC$ the given horizontal run $18$ m, and $BC$ the eye elevation $2$ m.
The upper part $CT$ is $H-2$ .

Use tangent in triangle $ECT$ .
$\tan48^\circ=\frac{H-2}{18}.$
Hence
$H=2+18\tan48^\circ.$

The constructed eye-level horizontal identifies the right triangle.

Answer

The diagram has $EC$ horizontal and $CT$ vertical, with a right angle at $C$ and elevation $48^\circ$ at $E$ . Label $EC$ as $18$ m, $BC$ as $2$ m, and $CT$ as $H-2$ . The model is
$H=2+18\tan48^\circ.$

Key idea

A labeled drawing distinguishes the whole height from the right triangle’s rise above the eye.
Problem 7 A station record

Stations $F$ , $N$ , and the base $B$ of a vertical tower lie in that order on one straight horizontal line. Station $N$ is nearer the tower, and $FN$ is $14.0$ m.

Both instrument eyes are $1.60$ m above the ground. They sight the same tower top above eye level in the common vertical plane. Elevation is $28.0^\circ$ at $F$ and $46.0^\circ$ at $N$ .

Find $NB$ and the whole tower height, each to the nearest tenth of a meter. Check both observation equations using unrounded values.

The order F, N, B determines the two ground runs.

Text description of this figure

Schematic, not for measuring. All bases lie on the same horizontal line and both eyes are 1.6 meters above it. The station order is F, N, B, with N nearer the tower. Blue and dashed red sight lines reach the same tower top T. Elevation angles are 28 and 46 degrees. The marked separation is 14 m. Object height is H.
1. Hint 1
  Let the near run be x and use the same rise above both instruments.
2. Hint 2
  The far run includes the baseline plus the near run. Equate two expressions for the shared rise.
3. Hint 3
  Recover the whole tower height by adding the common instrument elevation after solving.
Answer

$\begin{aligned}NB&\approx14.8\text{ m}.\\H&\approx16.9\text{ m}.\end{aligned}$

Full solution

Let $x$ be $NB$ and $u$ be the rise from instrument level to the tower top.
The near observation gives
$u=x\tan46^\circ.$

The station order makes the far run $x+14$ .
Thus
$u=(x+14)\tan28^\circ.$

Equate the rises and solve.
$x=\frac{14\tan28^\circ}{\tan46^\circ-\tan28^\circ}.$
The near horizontal run is approximately $14.8$ m.

Use unrounded $x$ to find the whole height.
$H=1.60+x\tan46^\circ.$
The tower height is approximately $16.9$ m.

As checks with unrounded values, $u/x$ equals $\tan46^\circ$ and $u/(x+14)$ equals $\tan28^\circ$ .
Both ratios describe the same positive rise.
The near angle is larger, consistent with the station order.

The checked measurements share the same target and eye level.

Answer

$\begin{aligned}NB&\approx14.8\text{ m}.\\H&\approx16.9\text{ m}.\end{aligned}$

Key idea

Two stations give useful equations only after their distances and common eye elevation have been fixed.
Problem 8 An angle interval

A vertical mast and an observer share level ground and one vertical sight plane. Treat horizontal run $40.0$ m and eye height $1.60$ m as fixed in the model.

The elevation reading is uncertain: the actual modeled angle may be anywhere from $31.5^\circ$ to $32.5^\circ$ , including the endpoints.

Find the modeled height at each endpoint to the nearest tenth of a meter. Can the record support one guaranteed whole-height value to that precision? Explain why using more calculator digits does not remove the angle uncertainty.

A schematic nominal view; the text supplies the full allowed angle interval.

Text description of this figure

Schematic, not for measuring. E is the eye above ground point F. The object BT is vertical. F and B have the same ground elevation. The dashed horizontal through E meets BT and forms a right angle. The line ET rises at θ above that horizontal. The horizontal run is 40 m, eye height is 1.6 m, and object height is labeled H.
1. Hint 1
  For the same horizontal run, a larger acute elevation gives a larger rise.
2. Hint 2
  Apply the whole-height model at both supplied angle endpoints. Compare their final rounded heights.
Answer

At $31.5^\circ$ , height is approximately $26.1$ m. At $32.5^\circ$ , height is approximately $27.1$ m. These differ to the nearest tenth, so no single rounded height is guaranteed by the record.

Full solution

With fixed horizontal run and eye elevation, raising the sight direction in this acute range raises the intersection with the vertical mast.
The endpoint angles therefore give the smallest and largest modeled heights.

At the lower angle,
$H_{\rm low}=1.60+40\tan31.5^\circ.$
Its nearest-tenth estimate is $26.1$ m.

At the upper angle,
$H_{\rm high}=1.60+40\tan32.5^\circ.$
Its nearest-tenth estimate is $27.1$ m.

The two endpoint estimates already disagree at the requested precision.
More digits in evaluating tangent improve arithmetic accuracy, but they do not tell which angle within the recorded interval was observed.

Answer

At $31.5^\circ$ , height is approximately $26.1$ m. At $32.5^\circ$ , height is approximately $27.1$ m. These differ to the nearest tenth, so no single rounded height is guaranteed by the record.

Key idea

The precision supported by a measurement depends on its input uncertainty, not the calculator display.
Problem 9 Stations on two sides

Stations $A$ and $C$ and vertical mast base $B$ lie on one horizontal line in the order $A$ , $B$ , $C$ . The whole separation $AC$ is $57.0$ m.

Both instrument eyes are $1.40$ m above this ground and sight the same top above eye level in a common vertical plane. Elevation is $39.0^\circ$ at $A$ and $54.0^\circ$ at $C$ .

Find $AB$ , $BC$ , and whole mast height to the nearest tenth of a meter. State which horizontal lengths add to the supplied separation.

The base lies between the two stations.

Text description of this figure

Schematic, not for measuring. All bases lie on the same horizontal line and both eyes are 1.4 meters above it. The tower base B lies between stations A and C. Blue and dashed red sight lines reach the same tower top T. Elevation angles are 39 and 54 degrees. The marked separation is 57 m. Object height is H.
1. Hint 1
  The mast base lies between the two stations, so the two runs add rather than differ by the station separation.
2. Hint 2
  Let u be the shared rise above instrument level. Express each run as u divided by the tangent of its own angle.
3. Hint 3
  Solve the sum equation for u, then recover both runs and add the eye height for the whole mast.
Answer

$\begin{aligned}AB&\approx35.9\text{ m}.\\BC&\approx21.1\text{ m}.\\H&\approx30.5\text{ m}.\end{aligned}$
The two unrounded runs add to $57.0$ m.

Full solution

Let $u$ be the top's rise above the shared instrument level.
The left run is
$AB=\frac{u}{\tan39^\circ}.$

The right run is
$BC=\frac{u}{\tan54^\circ}.$

Because $B$ is between the stations,
$AB+BC=57.$
Substitution gives
$\frac{u}{\tan39^\circ}+\frac{u}{\tan54^\circ}=57.$

Solve for the rise.
$u=\frac{57}{1/\tan39^\circ+1/\tan54^\circ}.$
Use the unrounded rise in both run formulas.
They give approximately $35.9$ m and $21.1$ m.

Add the instrument height.
$H=u+1.40.$
The whole mast height is approximately $30.5$ m.
The two unrounded runs sum to $57.0$ m, and the shorter run has the larger elevation angle.

Both observations use one shared rise above instrument level.

Answer

$\begin{aligned}AB&\approx35.9\text{ m}.\\BC&\approx21.1\text{ m}.\\H&\approx30.5\text{ m}.\end{aligned}$
The two unrounded runs add to $57.0$ m.

Key idea

A tower between stations gives a sum of runs; a tower beyond both stations gives a difference.
Problem 10 A second field check

A vertical mast base is $1.20$ m below the observer's ground. The observer's instrument eye is $1.70$ m above that ground. The horizontal run to the mast is $26.0$ m, and elevation to the top is $34.0^\circ$ . All points lie in one vertical sight plane.

(a) Estimate the whole mast height to the nearest tenth of a meter, accounting for the ground offset.

(b) A second instrument is also $1.70$ m above ground at the same elevation as the first observer's ground, so the eyes have equal elevations. It lies on the same side of the mast and in the same vertical sight plane. Its horizontal run is $40.0$ m. Predict its elevation to the nearest tenth of a degree. Its actual reading, to that precision, is $23.7^\circ$ . Is it consistent with the first sighting?

(c) Does this agreement independently verify the recorded $1.20$ m ground offset? Explain.

The observer ground and mast base have different elevations.

Text description of this figure

Schematic, not for measuring. E is the eye above ground point F. The object BT is vertical. The observer ground is 1.2 meters relative to the object base level. The dashed horizontal through E meets BT and forms a right angle. The line ET rises at 34° above that horizontal. The horizontal run is 26 m, eye height is 1.7 m, and object height is labeled H.
1. Hint 1
  The eye elevation relative to the mast base contains both the eye height and the higher ground.
2. Hint 2
  The second instrument shares the same rise to the top, while its horizontal run is different.
3. Hint 3
  Decide whether the second angle checks the shared rise, the reference elevation, or both.
Answer

(a) Approximately $20.4$ m. (b) Approximately $23.7^\circ$ , consistent with the reading. (c) No; the ground offset is not independently verified.

Full solution

(a) The eye is $2.90$ m above the mast base.
The rise above the eye is
$u=26\tan34^\circ.$
The whole height is
$H=1.20+1.70+u.$
It is approximately $20.4$ m.

(b) The second eye has the same elevation, so its rise to the top is still the unrounded $u$ .
With horizontal run $40$ m,
$\beta=\tan^{-1}\left(\frac u{40}\right).$
The predicted elevation is approximately $23.7^\circ$ , matching the reported $23.7^\circ$ at that precision.

(c) Both angle calculations test the height difference between the common eye level and mast top.
The ground offset is added only when turning that rise into a whole height.
An incorrect shared offset would shift both whole-height estimates together, so the matching second angle does not independently verify it.

The second sight checks the shared rise; its displayed result is rounded.

Answer

(a) Approximately $20.4$ m. (b) Approximately $23.7^\circ$ , consistent with the reading. (c) No; the ground offset is not independently verified.

Key idea

A second observation may check a shared rise while leaving a shared vertical reference untested.

← Back to lesson