HSG.SRT.D.10: Proving the Laws of Sines and Cosines
In plain English: HSG.SRT.D.10 is the Common Core geometry standard that asks students to prove the Law of Sines and the Law of Cosines and then use them to solve triangles that are not right triangles. The proofs rest on an altitude, right-triangle trigonometry, the distance formula and sin² + cos² = 1. It is an advanced (+) standard, usually taught in Precalculus or honors Geometry.
(+) Prove the Laws of Sines and Cosines and use them to solve problems.
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.10 or G-SRT.10 · Official standard
Right-triangle trigonometry stops working once a triangle has no right angle. This lesson gives students two tools for any triangle and, as the standard requires, proves each one before using it. The Law of Sines comes from writing one altitude in two ways. The Law of Cosines comes from placing the triangle on a coordinate grid and applying the distance formula together with the Pythagorean identity.
After the proofs, students solve triangles from each kind of given information: two angles and a side, two sides and the included angle, three sides, and two sides with an angle that is not between them. The last case can produce zero, one or two triangles, and students learn to check each candidate angle against the 180° angle sum.
Learning Objectives
By the end of this lesson, students will be able to:
Prove the Law of Sines by drawing an altitude and expressing its length in two ways, for acute and obtuse triangles
Prove the Law of Cosines with coordinates, the distance formula and the identity sin²C + cos²C = 1
Choose the Law of Sines or the Law of Cosines based on the given parts (AAS, ASA, SAS, SSS or SSA) and solve the triangle
Decide whether an SSA situation gives zero, one or two triangles and justify the decision
Prior Knowledge Required
Students should already be comfortable with:
Solving right triangles with sine, cosine and tangent HSG.SRT.C.8
The area formula (1/2)ab sin C and its altitude derivation HSG.SRT.D.9
The Pythagorean identity sin²θ + cos²θ = 1 HSF.TF.C.8
The distance formula on the coordinate plane 8.G.B.8
Draw a triangle with sides 6 cm and 8 cm and a 60° angle between them.
Warm-Up Prompt
"A classmate says the third side is 10 cm because 6² + 8² = 10². Is that right? Measure it on your drawing. Then explain what the Pythagorean theorem would need that this triangle does not have."
Measurements should come out near 7.2 cm, well short of 10 cm. The Pythagorean theorem needs a right angle opposite the unknown side, and 60° is smaller than 90°, so the side is shorter. Record the question on the board and promise an exact answer once students have proved a rule that works for any angle.
Direct Instruction25 minutes
Proof 1: the Law of Sines (Diagram 1). Use the usual labels: side a is opposite angle A, and so on.
Altitude. Draw CD perpendicular to line AB and call its length h.
Two right triangles. In △ADC, sin A = h/b, so h = b sin A. In △BDC, sin B = h/a, so h = a sin B.
Set equal. Both expressions equal h, so b sin A = a sin B. Dividing by sin A sin B (neither is zero in a triangle) gives a/sin A = b/sin B.
Third ratio. Repeat with the altitude from A to line BC to get b/sin B = c/sin C. So a/sin A = b/sin B = c/sin C.
Obtuse case. If A is obtuse, D lands on the extension of BA beyond A, and △ADC has the angle 180° - A. Since sin(180° - A) = sin A, the step h = b sin A still holds.
Proof 2: the Law of Cosines (Diagram 2). Place C at the origin and B at (a, 0). Point A is b units from C in the direction of angle C, so A = (b cos C, b sin C). This works for acute, right and obtuse C: when C is obtuse, cos C is negative and A sits left of the y-axis. The distance formula gives c² = (b cos C - a)² + (b sin C)². Expanding and grouping the b² terms, c² = a² + b²(cos²C + sin²C) - 2ab cos C, and the Pythagorean identity reduces this to c² = a² + b² - 2ab cos C. The same argument with other vertices at the origin gives the versions for a² and b².
Now solve the warm-up and four more triangles. Before each example, ask which parts are given and which law can start the solution.
Law of Cosines, SAS (the warm-up triangle)
Sides 6 and 8 with a 60° included angle. Find the third side c.
Equation: c² = 36 + 64 - 2(6)(8) cos 60° = 100 - 48 = 52, so c = 2√13 ≈ 7.21 cm
Law of Sines, AAS
A = 40°, B = 65°, a = 12. Find C, b and c.
Equation: C = 75°; b = 12 sin 65°/sin 40° ≈ 16.92; c = 12 sin 75°/sin 40° ≈ 18.03
Law of Cosines, SSS
Sides a = 7, b = 9, c = 12. Find the largest angle, C.
Equation: cos C = (49 + 81 - 144)/(2 · 7 · 9) = -14/126 = -1/9, so C ≈ 96.38°
Law of Sines, SSA with two triangles
A = 35°, a = 7, b = 11. Find every possible angle B.
Equation: sin B = 11 sin 35°/7 ≈ 0.9013, so B ≈ 64.33° or B ≈ 115.67°; both leave room for C (80.67° or 29.33°), so there are two triangles
Law of Cosines, right-angle check
Sides a = 5 and b = 12 with C = 90°.
Equation: c² = 25 + 144 - 2(5)(12)(0) = 169, so c = 13: the Pythagorean theorem is the case cos C = 0
Guided Practice15 minutes
Pairs complete two tasks. First, one partner writes the altitude-from-A argument that finishes the Law of Sines (b sin C = c sin B, so b/sin B = c/sin C), and the other checks every reason. Second, the pair solves two triangles and names the law they started with:
Guided practice triangles
Given
Start with
Results
A = 55°, B = 45°, c = 9
Law of Sines (ASA, find C = 80° first)
a ≈ 7.49, b ≈ 6.46
a = 6, b = 7, c = 8
Law of Cosines (SSS)
Largest angle C ≈ 75.52°
Circulate and listen for two errors: dividing by sin A sin B before checking that both sides of the equation contain h, and applying the Law of Sines to an SAS or SSS triangle, where no complete ratio pair is known.
Independent Practice15 minutes
Students work alone:
Redraw Diagram 2 with C = 120°. Find the coordinates of A when b = 6 and explain why the proof still works. (A = (-3, 3√3); cos 120° is negative, and the distance formula does not care about signs.)
a = 10, b = 4, C = 25°: find c. (c ≈ 6.60.)
A = 40°, a = 12, b = 9: find B and explain why only one triangle exists. (B ≈ 28.82°; the supplement 151.18° plus 40° exceeds 180°.)
A = 50°, a = 5, b = 9: show that no triangle exists. (sin B would be about 1.38.)
Closure5 minutes
Exit ticket: (1) For each set of given parts, name the law that starts the solution: SAS, SSS, ASA, SSA. (2) Point to the line of the Law of Cosines proof where sin²C + cos²C = 1 is used, and say what would be left over without it.
Differentiation Strategies
For Struggling Students
Give a proof frame with blanks for each right-triangle ratio, so students fill in h = __ sin __ twice before setting the expressions equal
Provide a decision chart: a known angle with its opposite side means the Law of Sines; otherwise use the Law of Cosines
Have students write the given parts as a row of six boxes (A, B, C, a, b, c) and shade what is known before choosing a law
For Advanced Students
Ask for a second proof of the Law of Cosines using the altitude from B and the Pythagorean theorem in two right triangles, and compare it with the coordinate proof
Show that each ratio a/sin A equals the diameter of the circumscribed circle (an extension beyond this standard)
Ask students to find, for A = 30° and b = 10, every value of a that gives exactly one triangle, and explain the boundary cases
Assessment Guidance
What to Look For
In the proofs, look for a named auxiliary line or coordinate setup, a reason for every equation, and an explicit use of the identity sin²C + cos²C = 1. In the Law of Sines proof, students should explain why dividing by sin A sin B is allowed. In problem solving, check the choice of law, degree mode, rounding only at the end, and, for SSA, a test of the supplementary angle against the angle sum.
02
Classroom Activities
3 Activities
1
Proof Card Shuffle
15 minPairs
Each pair receives 12 cards: six statements and six reasons from the Law of Sines proof. Pairs match every statement with its reason and put the steps in order, then add the obtuse-case adjustment in their own words.
Statement Cards
CD ⊥ AB with length h
sin A = h/b
sin B = h/a
b sin A = a sin B
a/sin A = b/sin B
b/sin B = c/sin C
Reason Cards
Construct the altitude from C
Definition of sine in right △ADC
Definition of sine in right △BDC
Both expressions equal h
Divide both sides by sin A sin B, which is not zero
Same argument with the altitude from A
Modification for Distance Learning
Put the cards on a shared slide as draggable text boxes. Pairs drag each statement next to its reason and record a short voice or text note for the obtuse case.
2
Coordinate Proof on Graph Paper
20 minPairs
Partners draw the same triangle on graph paper with C at the origin and B on the positive x-axis, compute the length of AB with the distance formula, and compare it with a² + b² - 2ab cos C. Then they repeat the calculation with letters to rebuild the proof.
Procedure
Partner A uses a = 8, b = 5, C = 40°; Partner B uses a = 8, b = 5, C = 130°
Each plots A = (5 cos C, 5 sin C), rounded to hundredths, and computes AB with the distance formula
Each then computes √(89 - 80 cos C) and compares (about 5.26 for 40° and about 11.85 for 130°)
Together, they replace the numbers with a, b and C and write the algebra from the distance formula to c² = a² + b² - 2ab cos C
Discussion Questions
Why is A to the left of the y-axis in the 130° drawing, and does the algebra change?
At which step did you need sin² + cos² = 1?
What does the formula give when C = 90°, and why is that expected?
3
Triangle-Solving Stations
20 minGroups of 3-4
Four stations each hold one triangle with a different set of given parts. Groups rotate every 5 minutes, name the case, choose a law, solve the triangle and leave their answer on a sticky note for the next group to check.
Station Cards
Station 1 (ASA): A = 62°, C = 43°, b = 20. (B = 75°, a ≈ 18.28, c ≈ 14.12)
Station 2 (SAS): a = 14, c = 10, B = 72°. (b ≈ 14.47)
Station 3 (SSS): a = 9, b = 10, c = 17. (C ≈ 126.87°)
Station 4 (SSA): B = 42°, b = 8, a = 11. (A ≈ 66.94° or A ≈ 113.06°: two triangles)
Procedure
At each station, one student names the case, one sets up the equation, one computes, and one checks that the angles add to 180° and the largest side faces the largest angle
Groups compare their result with the previous group's sticky note and resolve any difference before moving on
Challenge Variation
At Station 4, ask groups to change b so that the SSA case gives exactly one triangle, and to find the value of b that gives a right triangle (b = 11 sin 42° ≈ 7.36).
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Diagrams & Visual Aids
2 diagrams
Diagram 1: One Altitude, Two Expressions
The altitude CD splits △ABC into two right triangles. Writing its length h once with angle A and once with angle B gives b sin A = a sin B, the heart of the Law of Sines. The triangle is drawn to scale with A = 55°, B = 65° and c = 10.
Diagram 2: A Coordinate Proof of the Law of Cosines
With C at the origin and B on the x-axis, vertex A has coordinates (b cos C, b sin C). The distance formula for AB, together with the Pythagorean identity, gives c² = a² + b² - 2ab cos C. The grid is drawn to scale with a = 8, b = 6 and C = 60°.
04
Homework Assignment
~30 min
HSG.SRT.D.10 Homework: Proving and Using the Laws of Sines and Cosines
Directions: Write each proof with a labeled diagram and a reason for every step. For each triangle, name the case (AAS, ASA, SAS, SSS or SSA) and the law you used. Round sides to two decimal places and angles to the nearest hundredth of a degree.
Part 1: Proofs (Problems 1-2)
Prove that a/sin A = b/sin B for a triangle in which angle A is obtuse. Draw the altitude from C, say where its foot lands, and explain how sin(180° - A) = sin A is used.
Prove the Law of Cosines without coordinates. In an acute △ABC, draw the altitude from B to side AC with foot D. Show that BD = a sin C and AD = b - a cos C, then use the Pythagorean theorem in △ABD to show c² = a² + b² - 2ab cos C.
Part 2: Solving Triangles (Problems 3-4)
In △ABC, A = 48°, C = 71° and b = 15. Find B, a and c.
In △ABC, b = 13, c = 9 and A = 110°. Find a, then find the smallest angle.
Part 3: Special Cases (Problems 5-6)
B = 30° and a = 10. (a) If b = 6, find every possible triangle (all angles and side c). (b) If b = 4, show that no triangle exists and explain the result in terms of the height a sin B.
A triangle has sides 5, 8 and 11. Find all three angles. Explain why you should find the largest angle first.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Proof Structure
Labeled diagram, auxiliary line or coordinates, a reason for every step
Correct steps with missing reasons or diagram
Formula stated without proof
Choice of Law
Case named and correct law chosen for every triangle
One case misnamed or a less efficient choice
Law chosen incorrectly
Accuracy
All sides and angles correct and consistent with the angle sum
One calculation or rounding error
Several errors
SSA Reasoning
Both candidate angles tested; zero, one or two triangles justified
One candidate tested
No test of the supplement
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Set your calculator to degree mode before you start. 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
In △PQR, the altitude from R to side PQ has length h. Which pair of expressions for h leads to the Law of Sines?
Answer: A
The altitude from R makes two right triangles with hypotenuses PR = q and QR = p. So h = q sin P and h = p sin Q, which gives p/sin P = q/sin Q. Choice C pairs each side with its own opposite angle, which is not a right-triangle ratio. Choice B uses cosine, which gives the pieces of PQ, not the height.
Question 2 of 20 · Multiple Choice
After showing x sin M = m sin X in a proof, what is the next step to reach the Law of Sines?
Answer: D
Dividing by sin M sin X gives x/sin X = m/sin M. This is allowed because angles of a triangle are between 0° and 180°, so their sines are not zero. Choice C multiplies instead of dividing, so the sines pile up on each side and no side-to-sine ratio appears. Choices A and B do not move toward a ratio at all.
Question 3 of 20 · Multiple Choice
In a coordinate proof of the Law of Cosines, R is at the origin and S is at (t, 0). Point T is s units from R, and angle R is at the origin. What are the coordinates of T?
Answer: B
T lies s units from the origin in the direction of angle R, so T = (s cos R, s sin R). Choice A uses the wrong side length. Choice C swaps sine and cosine. Choice D would be correct only if s = 1.
Question 4 of 20 · Multiple Choice
A proof step contains the expression 49 cos²θ + 49 sin²θ. What does it simplify to, and why?
Answer: A
Factor out 49 and use the Pythagorean identity: 49(cos²θ + sin²θ) = 49(1) = 49. This is the same move that removes the angle from the b² terms in the coordinate proof of the Law of Cosines. Choice C is wrong because (cos θ + sin θ)² is not the same as cos²θ + sin²θ. Choice D takes a square root that nothing in the expression calls for.
Question 5 of 20 · Multiple Choice
A student's proof ends with c² = a² + b² + 2ab cos C. Test the formula on an equilateral triangle with every side equal to 1. What does the test show?
Answer: C
With a = b = 1 and C = 60°, the student's formula gives 1 + 1 + 2(0.5) = 3, but c² must be 1. The correct law, with -2ab cos C, gives 1 + 1 - 1 = 1. Choice A is what the correct law gives, not the student's version. Choice D is false: a proof must work for every triangle, so any triangle can test it.
Question 6 of 20 · Multiple Choice
In △ABC, B = 50°, C = 60° and b = 10. Find a.
Answer: D
A = 180° - 50° - 60° = 70°. Then a = b sin A/sin B = 10 sin 70°/sin 50° ≈ 12.27. Choice A inverts the ratio (10 sin 50°/sin 70°). Choice B is side c, found with sin 60° instead of sin 70°.
Question 7 of 20 · Multiple Choice
In △ABC, a = 5, b = 7 and C = 60°. Find c.
Answer: B
c² = 25 + 49 - 2(5)(7)(0.5) = 74 - 35 = 39, so c = √39 ≈ 6.24. Choice A adds 2ab cos C instead of subtracting it. Choice C leaves out the cosine term, as if C were 90°. Choice D forgets the square root.
Question 8 of 20 · Multiple Choice
A triangle has sides 4, 5 and 6. What is its largest angle, to the nearest hundredth of a degree?
Answer: C
The largest angle is opposite the side of length 6: cos C = (16 + 25 - 36)/(2 · 4 · 5) = 5/40 = 0.125, so C ≈ 82.82°. Choice A is the smallest angle, opposite the side of length 4. Choice D is 180° minus the correct angle, which comes from a sign error in the cosine.
Question 9 of 20 · Multiple Choice
You know two sides of a triangle and the angle between them. Which law must you use first, and why?
Answer: A
Every Law of Sines equation needs one complete pair (an angle and its opposite side). With SAS, the known angle is opposite the unknown side, so no pair is complete. The Law of Cosines gives the third side directly. Choice D is false: SAS determines a unique triangle.
Question 10 of 20 · Multiple Choice
In △ABC, A = 40°, a = 6 and b = 10. How many triangles are possible?
Answer: D
sin B = 10 sin 40°/6 ≈ 1.07, but a sine cannot exceed 1. Geometrically, the height b sin A ≈ 6.43 is longer than a = 6, so side a cannot reach the base. Choice B would be true only if a were exactly b sin A ≈ 6.43 (or at least 10).
Question 11 of 20 · Multiple Choice
In △ABC, A = 45°, a = 8 and b = 10. How many triangles are possible?
Answer: C
sin B = 10 sin 45°/8 ≈ 0.8839, so B ≈ 62.11° or B ≈ 117.89°. Both work, since 45° + 117.89° = 162.89° is less than 180°. So there are two triangles. Choice B keeps only the acute angle, a common error in SSA problems.
Question 12 of 20 · Multiple Choice
In △ABC, A = 130°. The altitude from C meets line AB outside the triangle, and the right triangle it forms has a 50° angle at A. A proof needs h = b sin 130°. Which fact justifies that step?
Answer: B
In the outside right triangle, h = b sin 50°. The angles 50° and 130° are supplementary, and supplementary angles have equal sines, so h = b sin 130° and the Law of Sines proof goes through unchanged. Choice A is false: cos 130° = -cos 50°. Choice D is false for the same reason: tan 130° = -tan 50°.
Question 13 of 20 · Multiple Choice
The Law of Cosines gives cos C = -0.2 for a triangle. What can you conclude?
Answer: A
cos C < 0 means C is between 90° and 180°. From c² = a² + b² - 2ab cos C, the term -2ab cos C is positive, so c² > a² + b². Choice D is wrong: a cosine of -0.2 is a valid angle, about 101.54°.
Question 14 of 20 · Multiple Choice
In △ABC, A = 58°, B = 72° and c = 20. Find a.
Answer: C
C = 180° - 58° - 72° = 50°. Then a = c sin A/sin C = 20 sin 58°/sin 50° ≈ 22.14. Choice A inverts the ratio. Choice B is side b. Choice D divides by sin 72° instead of sin 50°.
Question 15 of 20 · Short Answer
Prove that a/sin A = c/sin C by drawing the altitude from vertex B. Name the foot of the altitude and give a reason for each step.
Draw BE perpendicular to line AC, with length k. In right triangle ABE, the hypotenuse is AB = c, so sin A = k/c and k = c sin A. In right triangle CBE, the hypotenuse is CB = a, so sin C = k/a and k = a sin C. Both equal k, so c sin A = a sin C. Dividing by sin A sin C, which is not zero, gives a/sin A = c/sin C. If A or C is obtuse, E lies on an extension of AC and the supplementary angle has the same sine.
Question 16 of 20 · Short Answer
Put C at (0, 0) and A at (b, 0), and let B be a units from C so that angle C is at the origin. Write the coordinates of B and use them to prove c² = a² + b² - 2ab cos C.
B = (a cos C, a sin C). By the distance formula, c² = (a cos C - b)² + (a sin C)² = a²cos²C - 2ab cos C + b² + a²sin²C = a²(cos²C + sin²C) + b² - 2ab cos C = a² + b² - 2ab cos C, using cos²C + sin²C = 1. The result does not depend on which side is placed on the x-axis.
Question 17 of 20 · Short Answer
In △ABC, a = 9, b = 12 and C = 35°. Find c.
SAS, so use the Law of Cosines: c² = 81 + 144 - 2(9)(12) cos 35° = 225 - 216 cos 35° ≈ 225 - 176.94 = 48.06, so c ≈ 6.93.
Question 18 of 20 · Short Answer
In △ABC, A = 100°, a = 15 and b = 9. Solve the triangle.
sin B = 9 sin 100°/15 ≈ 0.5909, so B ≈ 36.22°. The supplement, about 143.78°, cannot be used because A is already 100°. So there is one triangle: B ≈ 36.22°, C ≈ 43.78°, and c = 15 sin 43.78°/sin 100° ≈ 10.54.
Question 19 of 20 · Short Answer
An isosceles triangle has sides 10, 10 and 16. Find its vertex angle and base angles.
The vertex angle is opposite the side of 16: cos V = (100 + 100 - 256)/(2 · 10 · 10) = -56/200 = -0.28, so V ≈ 106.26°. The base angles are equal: (180° - 106.26°)/2 ≈ 36.87° each.
Question 20 of 20 · Short Answer
Use the Law of Cosines to show that a triangle with sides 3, 4 and 6 is obtuse, and find its largest angle.
For the angle opposite 6: cos C = (9 + 16 - 36)/(2 · 3 · 4) = -11/24. The cosine is negative, so C is obtuse; equivalently 36 > 9 + 16. C ≈ 117.28°.
0 of 20 answered · 0 correct
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Frequently Asked Questions
10 Questions
What does HSG.SRT.D.10 mean?
It asks students to prove the Law of Sines and the Law of Cosines, not just memorize them, and then use both laws to solve problems. A lesson aligned to it should include a written proof of each law, usually one with an altitude and one with coordinates or the Pythagorean theorem, followed by triangle-solving practice.
Is HSG.SRT.D.10 in Geometry or Precalculus?
Both are possible. HSG.SRT.D.10 is a (+) standard, which Common Core describes as additional mathematics for students who take advanced courses. It usually appears in Precalculus or trigonometry, and some honors Geometry courses include it after right-triangle trigonometry.
How do you prove the Law of Sines?
Draw an altitude from one vertex to the opposite side. It creates two right triangles, and in each one the altitude equals a side times the sine of an angle, for example h = b sin A and h = a sin B. Setting these equal and dividing by sin A sin B gives a/sin A = b/sin B. A second altitude brings in c/sin C.
How do you prove the Law of Cosines?
One common proof places angle C at the origin with one side along the x-axis, so the far vertex is (b cos C, b sin C). The distance formula and the identity sin²C + cos²C = 1 then give c² = a² + b² - 2ab cos C. Another proof uses an altitude and the Pythagorean theorem in two right triangles.
When do I use the Law of Sines versus the Law of Cosines?
Use the Law of Sines when you know an angle and the side opposite it (AAS, ASA after finding the third angle, or SSA). Use the Law of Cosines when no such pair is known: two sides and the included angle (SAS) or three sides (SSS). After the first step, either law may finish the job.
What is the ambiguous case?
It is the SSA situation: two sides and an angle that is not between them. The Law of Sines gives a value of sin B, and two angles between 0° and 180° share that sine. Depending on the numbers, both, one or neither of them fit, so there can be two, one or no triangles. Always test the supplement against the angle sum.
What is a common mistake with the Law of Cosines?
Computing a² + b² - 2ab first and then multiplying by cos C, instead of subtracting the whole product 2ab cos C. Another frequent error is taking the square root too early or rounding cos C to one decimal place, which can move the final side length noticeably.
Why does the Law of Cosines look like the Pythagorean theorem?
Because it is a generalization of it. The term -2ab cos C corrects for the angle not being 90°. When C = 90°, cos C = 0 and the formula becomes c² = a² + b². When C is acute the third side is shorter than in the right-angle case, and when C is obtuse it is longer.
Do students need to know both proofs for tests?
The standard says "prove," so students should be able to reproduce a proof of each law with reasons. Tests aligned to this standard can ask for a full proof, a missing step or a reason for a step, in addition to triangle-solving problems.
Are the Laws of Sines and Cosines on the SAT?
The digital SAT Geometry and Trigonometry domain focuses on right triangles, and the SAT does not require the Laws of Sines and Cosines. Students taking Precalculus or later courses such as calculus and physics will use them, especially for vectors and forces.
07
Related Standards
5 standards
These standards connect to HSG.SRT.D.10: 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
