HSG.SRT.D.11: Applying the Laws of Sines and Cosines to Real Measurements
In plain English: HSG.SRT.D.11 is the Common Core geometry standard that asks students to understand the Law of Sines and the Law of Cosines and use them to find unknown sides and angles in right and non-right triangles, as in surveying and resultant-force problems. It is an advanced (+) standard, usually taught in Precalculus or honors Geometry.
(+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e.g., surveying problems, resultant forces).
Common Core State Standards for Mathematics · Domain: Similarity, Right Triangles, and Trigonometry (SRT) · Cluster: Apply trigonometry to general triangles Also written as HSG-SRT.D.11 or G-SRT.11 · Official standard
Surveyors, pilots and engineers rarely get a right triangle. A distance across a pond, the height of a hill seen from a road, the path of a plane that turns, and the combined pull of two ropes all lead to triangles with no 90° angle. This lesson treats the Law of Sines and the Law of Cosines as working tools for those situations. The proofs belong to HSG.SRT.D.10; here the goal is to understand what each law says, when it applies, and how to turn a word problem into a labeled triangle.
Students first see that the two laws also hold in right triangles, where they reduce to the familiar ratios and the Pythagorean theorem. Then they solve four families of problems: inaccessible distances, heights from two angles of elevation, navigation with bearings, and the resultant of two forces. Each answer is checked for sense: the longest side must face the largest angle, and a resultant can never be longer than the two forces added together.
Learning Objectives
By the end of this lesson, students will be able to:
Explain what the Law of Sines and the Law of Cosines state, and show that in a right triangle they reduce to sin A = a/c and the Pythagorean theorem
Decide from the known measurements which law, or right-triangle trigonometry, gives the next unknown
Model surveying, height and navigation situations with a labeled triangle and find unknown distances and angles
Find the magnitude and direction of the resultant of two forces with the parallelogram rule and the two laws
Judge whether an answer is reasonable from the angle and side sizes and from the context
Prior Knowledge Required
Students should already be comfortable with:
Solving right triangles with sine, cosine, tangent and the Pythagorean theorem HSG.SRT.C.8
Statements of the Law of Sines and the Law of Cosines HSG.SRT.D.10
Angles of elevation and the triangle angle sum 8.G.A.5
Representing forces or velocities as arrows with a length and a direction HSN.VM.A.1
Project a map sketch: a park ranger stands at point C and can walk to a dock A and to a boathouse B, but a pond lies between the dock and the boathouse (Diagram 1).
Warm-Up Prompt
"The ranger measures CA = 180 m, CB = 240 m and the angle at C, 64°. Can right-triangle trigonometry give the distance AB? If not, what is missing? Estimate AB from a scale drawing."
Students notice that no angle of the triangle is 90°, so SOH-CAH-TOA does not apply directly. Scale drawings should give roughly 225-235 m. Keep the estimates on the board and return to them after Example 1.
Direct Instruction25 minutes
What the two laws say. In any triangle labeled with side a opposite angle A, and so on:
Law of Sines: a/sin A = b/sin B = c/sin C. Each side divided by the sine of the angle across from it gives the same number. It is useful once you know one complete pair: an angle and the side across from it.
Law of Cosines: c² = a² + b² - 2ab cos C. It is the Pythagorean theorem with a correction term. It is useful when you know two sides and the angle between them, or all three sides.
Right triangles are included. If C = 90°, then cos C = 0 and the Law of Cosines becomes c² = a² + b². Since sin 90° = 1, the Law of Sines becomes a/sin A = c, so sin A = a/c, the usual opposite-over-hypotenuse ratio. The laws do not replace right-triangle trigonometry; they extend it.
Modeling routine. Sketch the situation, mark the triangle, label every known side and angle with its value, mark the unknown, then choose the law. For forces, draw the two force arrows from one point and complete the parallelogram: the diagonal is the resultant, and the triangle used has the angle 180° minus the angle between the forces (Diagram 2).
Surveying an inaccessible distance (Law of Cosines, Diagram 1)
From point C, a ranger measures CA = 180 m to a dock, CB = 240 m to a boathouse and ∠ACB = 64°. Find the distance AB across the pond.
Equation: AB² = 180² + 240² - 2(180)(240) cos 64° ≈ 90,000 - 37,876 = 52,124, so AB ≈ 228.3 m
Height from two angles of elevation (Law of Sines, then right-triangle trig)
From point P on a flat road, the angle of elevation to a hilltop T is 18°. After driving 300 m straight toward the hill to point Q, the angle is 27°. Find the height of the hill.
Equation: In △PQT, ∠PQT = 180° - 27° = 153° and ∠T = 180° - 18° - 153° = 9°. QT = 300 sin 18°/sin 9° ≈ 592.6 m, and the height is QT sin 27° ≈ 269.0 m
Resultant force (parallelogram rule, Diagram 2)
Two ropes pull a stuck crate from the same point with forces of 65 N and 40 N, and the angle between the ropes is 50°. Find the size of the combined pull and its angle from the 65 N rope.
Equation: The triangle has 130° between the 65 N and 40 N sides: R² = 65² + 40² - 2(65)(40) cos 130° ≈ 9167.5, so R ≈ 95.7 N. Then sin θ = 40 sin 130°/95.7, θ ≈ 18.7°
A right triangle solved with the laws
A playground slide is 20 ft long and makes an angle A = 35° with the ground, with the right angle at C below the top of the slide. Find the height a and the horizontal run b.
Equation: Law of Sines: a = 20 sin 35°/sin 90° ≈ 11.47 ft, the same as 20 sin 35° from SOH-CAH-TOA. Run b = 20 cos 35° ≈ 16.38 ft, and a² + b² ≈ 400 = 20²
Navigation with bearings (Law of Cosines)
A sailboat travels 15 km on a bearing of 050°, then turns and sails 22 km on a bearing of 160°. How far is it from its starting point?
Equation: At the turn, the bearing back to the start is 230°, so the angle between the two legs is 230° - 160° = 70°. d² = 15² + 22² - 2(15)(22) cos 70° ≈ 483.3, so d ≈ 22.0 km
Guided Practice15 minutes
Pairs sketch and solve each situation, name the law they start with, and write one sentence about why the answer is reasonable.
Guided practice situations
Situation
Start with
Result
Two fire lookouts, A and B, are 10 km apart on a straight road. A fire F is seen at 52° from the road at A and 71° from the road at B. How far is the fire from each lookout?
Law of Sines (∠F = 57°)
AF ≈ 11.27 km, BF ≈ 9.40 km
A triangular garden has sides 14 m, 17 m and 23 m. What is its largest angle?
Law of Cosines (three sides)
About 95.3°, opposite the 23 m side
As pairs work, ask: Which side is longest, and is it across from the largest angle? Watch for pairs who use the 52° and 71° angles as if they were opposite the 10 km side; the angle across from AB is the one at the fire.
Independent Practice20 minutes
Students work alone, with a sketch for each item:
Two people pull a stuck kayak with ropes tied at the same point, with forces of 120 N and 90 N and 35° between the ropes. Find the resultant and its angle from the 120 N force. (R ≈ 200.5 N, about 14.9°.)
A plane flies with an airspeed of 480 km/h, and the wind blows at 60 km/h. The angle between the plane's heading and the wind direction is 70°. Find the ground speed. (About 503.7 km/h.)
Use the Law of Cosines on a right triangle with legs 7 and 24 and the right angle between them. Show why the cosine term disappears, and find the hypotenuse. (cos 90° = 0, so c² = 49 + 576 = 625 and c = 25.)
Closure5 minutes
Exit ticket: (1) A surveyor knows two sides of a lot and the angle between them. Which law finds the third side, and why not the other one? (2) Two forces of 50 N each act on a point. Without computing, what is the largest their resultant can be, and when does that happen?
Differentiation Strategies
For Struggling Students
Give a planning sheet for each problem with three boxes: sketch, known parts (angle-side pairs circled), and the law chosen
Start with the right-triangle version of each problem so students see the laws give the same answers as SOH-CAH-TOA before moving to non-right triangles
For bearings, provide a compass rose template so students can mark the back bearing at the turn point
For Advanced Students
Ask students to solve Example 3 with components (x and y parts of each force) and compare with the Law of Cosines result
Pose a surveying problem with an SSA setup, such as a triangular lot described by two sides and an angle that is not between them, and ask students to decide whether one or two lots fit the description
Have students research how triangulation from known baselines was used to map large regions, then design their own two-station measurement of a distant landmark
Assessment Guidance
What to Look For
Look for a labeled sketch in every problem, with the known angle-side pairs identified before a law is chosen. In force and navigation problems, check that students use the angle inside the triangle (180° minus the angle between the forces, or the angle at the turn found from the back bearing). Final answers should have units, be rounded only at the end, and pass a sense check: the longest side across from the largest angle, and a resultant no longer than the sum of the forces.
02
Classroom Activities
3 Activities
1
Schoolyard Triangulation
25 minGroups of 3-4
Groups find the distance to a landmark they cannot walk to directly, such as a flagpole behind a fence, using a measured baseline and two angles. They then check their answer with a second baseline or, where possible, a tape measure.
Procedure
Mark two stations, A and B, on open ground and tape the baseline AB
At each station, measure the angle between the baseline and the line of sight to the landmark P with a protractor on a clipboard or a phone angle app
Sketch △ABP, find ∠P from the angle sum, and use the Law of Sines to find AP and BP
Sample data from a trial run: AB = 40 m, ∠A = 68°, ∠B = 75°, so ∠P = 37°, AP = 40 sin 75°/sin 37° ≈ 64.2 m and BP = 40 sin 68°/sin 37° ≈ 61.6 m
Reflection Questions
If each angle could be off by 1°, which of your answers changes the most, and why does a small ∠P make the result sensitive?
Why is a longer baseline usually better?
Modification for Distance Learning
Students use a satellite map of a local park, pick two points on a path as the baseline, measure the angles on screen with a protractor tool, and compare their Law of Sines result with the map's distance tool.
2
Force Table with Spring Scales
20 minGroups of 3
Two students pull a ring with spring scales at a set angle while a third holds it still with a third scale. The third scale reads the size of the resultant, which groups predict first with the Law of Cosines.
Procedure
Tape a large protractor drawing under the ring and set the two pulling strings at 60° apart, with readings of 6 N and 8 N
Predict the resultant: the triangle angle is 180° - 60° = 120°, so R² = 36 + 64 - 2(6)(8) cos 120° = 148 and R ≈ 12.2 N
Read the holding scale, which should point opposite the predicted resultant, and compute the percent difference
Repeat with the strings at 90° apart and compare with the Pythagorean prediction of 10 N
Discussion Questions
What happens to the resultant as the angle between the strings grows from 0° to 180°?
Why must the triangle use 120° and not 60°?
3
Which Tool? Scenario Cards
15 minPairs
Each pair receives 8 scenario cards. For each card, they sketch the triangle, circle any known angle-side pair, and decide whether to begin with right-triangle trigonometry, the Law of Sines or the Law of Cosines. Then each pair fully solves two cards of its choice.
Scenario Cards
Card 1: Two fences of a field are 35 m and 50 m with 110° between them; find the third side (Law of Cosines, about 70.2 m)
Card 2: Two survey stations are 12 km apart, with sight angles of 40° and 75° to a mountain peak (Law of Sines, after ∠ = 65°)
Card 3: A roof truss has sides 4 m, 4 m and 6.5 m; find the peak angle (Law of Cosines, about 108.7°)
Card 4: A 5 m loading ramp rises at 8° with a right angle at the ground (right-triangle trigonometry, or the Law of Sines with sin 90° = 1)
Card 5: Forces of 200 N and 150 N act 45° apart (Law of Cosines for the size, about 323.9 N, then the Law of Sines for the direction)
Card 6: A 14 m cable and a 9 m strut, with 32° opposite the strut (Law of Sines; check for a second triangle)
Card 7: Two angles of elevation to a tower from points 80 m apart on level ground (Law of Sines, then right-triangle trigonometry)
Card 8: A hiker walks 6 km on a bearing of 090°, then 8 km on a bearing of 180° (the turn is 90°, so the Pythagorean theorem: 10 km)
Challenge Variation
Pairs rewrite Card 8 so the turn is not 90° and trade with another pair, who must solve it and explain which law the change requires.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Measuring Across a Pond
The ranger can tape CA and CB and measure the angle at C, but not AB. Two sides with the included angle call for the Law of Cosines. Drawn to scale for the warm-up and Example 1.
Diagram 2: The Resultant of Two Forces
Drawing both forces from O and completing the parallelogram gives the resultant R as the diagonal. The triangle with sides 65 N, 40 N and R has a 130° angle, so the Law of Cosines gives the size of R and the Law of Sines gives its direction θ. Drawn to scale for Example 3.
04
Homework Assignment
~30 min
HSG.SRT.D.11 Homework: Surveying, Navigation and Forces
Directions: Draw and label a triangle for each problem, circle any known angle-side pair, and name the law or ratio you use at each step. Round lengths to the nearest tenth and angles to the nearest tenth of a degree, rounding only at the end. Include units and a one-sentence sense check.
Part 1: Distances and Heights (Problems 1-3)
A surveyor at point C cannot cross a ravine between points A and B. She measures CA = 95 m, CB = 130 m and ∠ACB = 72°. Find the width AB.
From a point on a beach, the angle of elevation to the top of a sea cliff is 22°. After walking 200 m straight toward the cliff on level sand, the angle is 35°. Find the height of the cliff.
A triangular building lot has sides of 48 m, 61 m and 70 m. Find all three angles of the lot, starting with the largest.
Part 2: Forces, Navigation and Right Triangles (Problems 4-6)
Two people pull a stalled go-kart with ropes tied at the same point, with forces of 250 N and 180 N and 72° between the ropes. Find the magnitude of the resultant and its angle from the 250 N rope.
A small plane flies 320 km on a bearing of 070°, then 250 km on a bearing of 185°. Find the angle between the two legs at the turn and the plane's distance from its starting airport.
(a) Two triangles both have a = 5 and b = 8. In one, C = 90°; in the other, C = 120°. Use the Law of Cosines to find c in each, and explain how each result compares with the Pythagorean theorem. (b) In right △ABC with C = 90°, hypotenuse c = 26 cm and A = 40°. Use the Law of Sines to find a and b, and check both with sine and cosine ratios.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Model
Labeled sketch with every known value and the unknown marked
Sketch with one missing or misplaced label
No sketch, or a triangle that does not match the situation
Choice of Law
Correct law or ratio named at each step, with the angle inside the triangle used for forces and bearings
Correct law but the wrong angle used once
Law chosen that cannot work with the known parts
Accuracy
All values correct with units, rounded only at the end
One calculation or rounding error
Several errors
Sense Check
Answer checked against the context and the side-angle order
Check stated but incomplete
No check
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Use a calculator in degree mode and sketch each triangle before answering. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which set of known measurements requires the Law of Cosines as the first step?
Answer: B
With three sides, no angle is known, so no angle-side pair exists for the Law of Sines; the Law of Cosines gives an angle directly. Choices A and C contain a complete pair (after the angle sum in A), so the Law of Sines works. Choice D is a right triangle that sine and cosine ratios solve at once.
Question 2 of 20 · Multiple Choice
For which angles C does the Law of Cosines, c² = a² + b² - 2ab cos C, give a value of c² that is greater than a² + b²?
Answer: C
c² is greater than a² + b² exactly when the term -2ab cos C is positive, which needs cos C < 0, so C is obtuse. Choice A has the sign backward: for an acute angle cos C > 0 and the side is shorter than in the right-angle case. Choice B gives c² = a² + b² exactly, with no excess.
Question 3 of 20 · Multiple Choice
In right △ABC with C = 90°, side a = 8 and hypotenuse c = 17. A student uses the Law of Sines, a/sin A = c/sin C, to find angle A. What does she get, to the nearest hundredth of a degree?
Answer: D
sin C = sin 90° = 1, so the equation becomes 8/sin A = 17 and sin A = 8/17 ≈ 0.4706, giving A ≈ 28.07°. This is the same result as opposite over hypotenuse, which shows the law includes right triangles. Choice A is angle B, the complement. Choice B uses the inverse tangent of 8/17. Choice C is the sine value itself, not the angle.
Question 4 of 20 · Multiple Choice
To find the distance AB across a lake, a surveyor at C measures CA = 50 m, CB = 70 m and ∠C = 80°. What is AB, to the nearest hundredth?
Answer: A
AB² = 50² + 70² - 2(50)(70) cos 80° ≈ 7400 - 1215.54 = 6184.46, so AB ≈ 78.64 m. Choice B leaves out the cosine term, as if ∠C were 90°. Choice D adds 2ab cos C instead of subtracting it. Choice C simply adds the two sides.
Question 5 of 20 · Multiple Choice
Two forces of 30 N and 40 N act on an object at a right angle to each other. What is the magnitude of the resultant?
Answer: B
With 90° between the forces, the parallelogram is a rectangle and the triangle is a right triangle, so R = √(30² + 40²) = 50 N. Choice A adds the forces as if they pointed the same way. Choice C subtracts them, which is the case of opposite directions.
Question 6 of 20 · Multiple Choice
Two forces of 30 N and 40 N act on an object with 60° between them. What is the magnitude of the resultant, to the nearest hundredth?
Answer: C
The triangle angle is 180° - 60° = 120°, so R² = 900 + 1600 - 2(30)(40) cos 120° = 2500 + 1200 = 3700 and R ≈ 60.83 N. Choice A puts 60° itself into the Law of Cosines instead of the angle inside the force triangle. Choice B is the 90° answer.
Question 7 of 20 · Multiple Choice
Ranger stations A and B are 8 km apart. A campfire F is sighted at 48° from the line AB at station A and 61° from the line AB at station B. How far is the campfire from station B?
Answer: D
∠F = 180° - 48° - 61° = 71°. BF is across from ∠A, so BF = 8 sin 48°/sin 71° ≈ 6.29 km. Choice A is AF, which is across from ∠B. Choice B divides by sin 61° instead of sin 71°, the sine of the angle across from the 8 km baseline.
Question 8 of 20 · Multiple Choice
A triangular sail has edges of 7 ft, 10 ft and 13 ft. What is its largest angle, to the nearest hundredth of a degree?
Answer: A
The largest angle is across from the 13 ft edge: cos C = (49 + 100 - 169)/(2 · 7 · 10) = -20/140 = -1/7, so C ≈ 98.21°. Choice B is 180° minus the answer, from dropping the negative sign. Choice C is the middle angle, across from the 10 ft edge, and choice D is the smallest angle, across from the 7 ft edge.
Question 9 of 20 · Multiple Choice
A ship sails 12 km on a bearing of 030°, then 12 km on a bearing of 150°. How far is it from its starting point?
Answer: C
At the turn, the bearing back to the start is 210°, so the angle between the legs is 210° - 150° = 60°. Two 12 km sides with 60° between them make an equilateral triangle, so the distance is 12 km (the Law of Cosines gives 144 + 144 - 144 = 144). Choice A uses 120°, the change in bearing, as the triangle angle. Choice D assumes a 90° turn.
Question 10 of 20 · Multiple Choice
A buoy B is 7 km from a lighthouse L. A ship S sails along a straight course from L, and the course makes a 40° angle with LB. How many positions on the course are exactly 5 km from the buoy?
Answer: C
In △LSB, ∠L = 40°, the side across from it is SB = 5 and LB = 7. The Law of Sines gives sin S = 7 sin 40°/5 ≈ 0.900, so ∠S ≈ 64.1° or 115.9°. Both leave a positive third angle (75.9° or 24.1°), so two positions fit. Choice B keeps only the acute angle. Choice A would be right only if the sine were greater than 1.
Question 11 of 20 · Multiple Choice
A student solves a triangle and finds that the side across from a 30° angle is 14 m, while the side across from an 80° angle is 9 m. What should the student conclude?
Answer: D
In any triangle, the longer side is across from the larger angle, which also follows from the Law of Sines because sin 80° > sin 30°. So a 9 m side cannot face 80° while a 14 m side faces 30°. Choice A ignores this relationship.
Question 12 of 20 · Multiple Choice
From point P, the angle of elevation to the top of a radio tower is 30°. From point Q, 100 m closer on level ground, it is 45°. How tall is the tower, to the nearest hundredth of a meter?
Answer: A
In △PQT, ∠PQT = 135° and ∠T = 15°. QT = 100 sin 30°/sin 15° ≈ 193.19 m, and the height is QT sin 45° ≈ 136.60 m. Choice B stops at QT, the slanted line of sight, not the height. Choice C multiplies 100 by sin 45°, as if the 100 m walk were the line of sight from Q to the top. Choice D is 100 tan 45°, as if Q were 100 m from the base of the tower.
Question 13 of 20 · Multiple Choice
Two forces act from the same point with 25° between them. A student puts 25° into the Law of Cosines with the two force magnitudes. What has the student actually found?
Answer: B
The triangle built from the two force arrows with 25° between them has, as its third side, the segment joining the arrow tips: the short diagonal of the parallelogram, which represents the difference of the forces. The resultant is the long diagonal, and its triangle has the angle 180° - 25° = 155°. Choice D is false: the law holds for every triangle angle.
Question 14 of 20 · Multiple Choice
A surveyor measures a triangular lot with ∠A = 47°, ∠B = 68° and AB = 250 m. Find the length of side BC, to the nearest hundredth.
Answer: C
∠C = 180° - 47° - 68° = 65°, and it is across from AB, which completes an angle-side pair. BC is across from ∠A, so BC = 250 sin 47°/sin 65° ≈ 201.74 m. Choice A is AC, across from ∠B. Choice B inverts the ratio. Choice D divides by sin 68° instead of sin 65°.
Question 15 of 20 · Short Answer
Two dock workers pull a small boat with ropes tied at the same point, with forces of 55 N and 75 N and 40° between the ropes. Find the magnitude of the resultant and its angle from the 75 N rope.
The triangle angle is 140°. R² = 55² + 75² - 2(55)(75) cos 140° ≈ 8650 + 6319.9 = 14,969.9, so R ≈ 122.4 N. Then sin θ = 55 sin 140°/122.4 ≈ 0.289, so θ ≈ 16.8° from the 75 N rope.
Question 16 of 20 · Short Answer
A student wants the third side of a triangle with sides 8 cm and 11 cm and a 50° angle between them, and starts with the Law of Sines. Explain why that cannot work, then find the third side.
The Law of Sines needs a known angle together with the side across from it. The 50° angle is across from the unknown side, and the angles across from 8 and 11 are unknown, so no ratio is complete. The Law of Cosines works: c² = 64 + 121 - 176 cos 50° ≈ 71.87, so c ≈ 8.5 cm.
Question 17 of 20 · Short Answer
A park is shaped like a triangle with sides of 25 m, 38 m and 44 m. Find all three angles to the nearest tenth of a degree.
Across from 44 m: cos = (625 + 1444 - 1936)/(2 · 25 · 38) = 133/1900 = 0.07, so the angle is about 86.0°. Across from 25 m: cos = (1444 + 1936 - 625)/(2 · 38 · 44) ≈ 0.824, so about 34.5°. The third angle is 180° - 86.0° - 34.5° = 59.5°, across from 38 m.
Question 18 of 20 · Short Answer
A helicopter flies 400 km on a bearing of 100°, then 300 km on a bearing of 225°. Find the angle at the turn and the distance back to the start.
The bearing back to the start from the turn is 280°, so the angle between the legs is 280° - 225° = 55°. d² = 400² + 300² - 2(400)(300) cos 55° ≈ 250,000 - 137,658 = 112,342, so d ≈ 335.2 km.
Question 19 of 20 · Short Answer
From a point on level ground, the angle of elevation to the top of a water tower is 25°. From a point 50 m closer, it is 38°. Find the height of the tower.
In the triangle formed by the two points and the top, the angle at the closer point is 180° - 38° = 142°, so the angle at the top is 13°. The line of sight from the closer point is 50 sin 25°/sin 13° ≈ 93.9 m, and the height is 93.9 sin 38° ≈ 57.8 m.
Question 20 of 20 · Short Answer
A triangle has sides 9, 12 and 15. Use the Law of Cosines to find the angle across from the 15 side, then use the Law of Sines to find the angle across from the 9 side. What kind of triangle is it?
cos C = (81 + 144 - 225)/(2 · 9 · 12) = 0, so C = 90° and the triangle is a right triangle. Then sin A = 9 sin 90°/15 = 0.6, so A ≈ 36.9°, which matches the right-triangle ratio opposite/hypotenuse = 9/15.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.SRT.D.11 mean?
It asks students to understand what the Law of Sines and the Law of Cosines say and to use them to find unknown sides and angles in any triangle, right or not. The official text names surveying and resultant forces as typical applications, so lessons aligned to it focus on measurement problems from real situations.
How is HSG.SRT.D.11 different from HSG.SRT.D.10?
HSG.SRT.D.10 is about proving the two laws. HSG.SRT.D.11 is about understanding and applying them to find measurements. In practice the two are often taught together, with the proofs first and the applications after.
Is HSG.SRT.D.11 a Geometry or Precalculus standard?
It is an advanced (+) standard in the Geometry category. Common Core describes (+) standards as additional mathematics for students who take advanced courses, and this one usually appears in Precalculus, trigonometry or honors Geometry.
Do the Laws of Sines and Cosines work in right triangles?
Yes. With a 90° angle, the Law of Cosines becomes the Pythagorean theorem because cos 90° = 0, and the Law of Sines gives sin A = opposite/hypotenuse because sin 90° = 1. For right triangles the usual ratios are quicker, but the laws give the same answers.
How do I know which law to use in a word problem?
Sketch the triangle and look for an angle whose opposite side is also known. If you find one, the Law of Sines can start. If not, start with the Law of Cosines. After the first unknown is found, either law may finish.
How are the laws used to find resultant forces?
Draw the two forces from one point and complete the parallelogram. The resultant is the diagonal from that point. In the triangle made by one force, a copy of the other and the resultant, the angle is 180° minus the angle between the forces. The Law of Cosines gives the size of the resultant and the Law of Sines gives its direction.
How do surveyors use the Law of Sines?
A surveyor measures one baseline and the angles from each end of it to a distant point. The angle sum gives the third angle, which sits across from the baseline, and the Law of Sines then gives the distances to the point. This method, called triangulation, finds distances that cannot be taped.
What mistakes do students make with bearings problems?
A frequent one is using the change in bearing as the angle of the triangle. The triangle angle at the turn is found from the back bearing (the first bearing plus 180°) and the new bearing. A quick sketch with a north line at each point prevents this error.
Why is my Law of Sines answer wrong when the angle is obtuse?
The inverse sine key only returns angles up to 90°, but sin θ = sin(180° - θ). If the angle you want is obtuse, or could be, subtract the calculator result from 180° and check which angle fits the triangle and the context. Using the Law of Cosines for the largest angle avoids the issue, because cosine is negative for obtuse angles.
Is HSG.SRT.D.11 tested on the SAT?
The digital SAT Geometry and Trigonometry domain concentrates on right-triangle trigonometry and does not require these laws. They appear in Precalculus and later in physics and engineering courses, where forces and velocities are combined.
07
Related Standards
5 standards
These standards connect to HSG.SRT.D.11: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSG.SRT.C.8Prerequisite
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems
