Free Printable WorksheetMathTrigonometryMath CurriculumUTRI.TA.1

TRIG.111 Set up a right triangle model from a word problem Worksheet

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Worksheet Overview & Skills Covered

Core academic competencies and curriculum alignment

This Trigonometry Math worksheet provides structured practice on TRIG.111 Set up a right triangle model from a word problem. Students are guided through targeted exercises that build foundational understanding, computational accuracy, and critical thinking skills.

TRIG.111 Set up a right-triangle model from a word problem - Primary standard selected for prompt fields in this prompt cell: UTRI.TA.1. Skill statement in this prompt cell: Select the diagram and equation that correctly represent a word problem as a right triangle. Associated standards listed with this skill in this prompt cell: UTRI.TA.1; CCSS HSG.SRT.C.6-8; TEKS §111.42(c)(4)(E)-(F); FL MA.912.T.1.1-1.2; VA T.TT.1a,c-d. Use this prompt-provided wording as the supporting standard description for this workflow.

Whether you are an educator preparing classroom handouts, a tutor reinforcing weekly topics, or a parent supporting homeschool learning, this printable resource gives students the deliberate practice required to achieve subject fluency and master state standards.

Grade

Trigonometry

Subject

Math

Domain

Math Curriculum

Formats

PDF & DOCX

Sample Practice Problems

Preview the types of problems and questions covered in this worksheet

Below are sample questions representative of the exercises included on this worksheet and related TRIG.111 Set up a right triangle model from a word problem skill assessments:

Problem 1Difficulty: Mastery

A Mars rover camera 1 m above the surface sees the top of a crater rim 31 m above the same surface at an angle of elevation of 19°. Let d be the horizontal distance to the rim. Which diagram description and equation correctly model the situation as a right triangle?

ADiagram: the horizontal distance d is opposite the 19° angle and the height difference 31 - 1 is adjacent; Equation: tan(19°) = d/(31 - 1)

BDiagram: the height difference 31 - 1 is opposite the 19° angle and the horizontal distance d is the hypotenuse; Equation: sin(19°) = (31 - 1)/d

CDiagram: the horizontal distance d is adjacent to the 19° angle and the height difference 31 - 1 is the hypotenuse; Equation: cos(19°) = d/(31 - 1)

DDiagram: the height difference 31 - 1 is opposite the 19° angle, the horizontal distance d is adjacent, and the camera line of sight is the hypotenuse; Equation: tan(19°) = (31 - 1)/d

Solution & Explanation:The opposite side is the rim height above the camera, not above the surface. Relative to the 19° angle, the height difference 31 - 1 is opposite and the horizontal distance d is adjacent. Tangent compares opposite over adjacent. So the correct setup is tan(19°) = (31 - 1)/d.

Problem 2Difficulty: Mastery

A hiker's straight trail segment climbs 680 ft over a 0.9 mi trail path. Use 0.9 mi = 4752 ft. Let theta be the trail angle with the horizontal. Which diagram description and equation correctly model the situation as a right triangle?

ADiagram: the 4752 ft trail path is opposite angle theta and the 680 ft climb is the hypotenuse; Equation: sin(theta) = 4752/680

BDiagram: the 680 ft climb is opposite angle theta, the horizontal map distance is adjacent, and the 4752 ft trail path is the hypotenuse; Equation: sin(theta) = 680/4752

CDiagram: the 680 ft climb is adjacent to angle theta and the 4752 ft trail path is the hypotenuse; Equation: cos(theta) = 680/4752

DDiagram: the 4752 ft trail path is adjacent to angle theta and the 680 ft climb is opposite; Equation: tan(theta) = 680/4752

Solution & Explanation:The trail path is the hypotenuse, so sine uses climb over trail length. Relative to angle theta, the 680 ft climb is opposite and the 4752 ft trail path is the hypotenuse. Sine compares opposite over hypotenuse. So the correct setup is sin(theta) = 680/4752.

Problem 3Difficulty: Mastery

A ferry passenger standing 3 m above the water sees the top of a bridge tower 72 m above the water at an angle of elevation of 7°. Let d be the horizontal distance to the tower. Which diagram description and equation correctly model the situation as a right triangle?

ADiagram: the height difference 72 - 3 is opposite the 7° angle, the horizontal distance d is adjacent, and the line of sight to the bridge tower is the hypotenuse; Equation: tan(7°) = (72 - 3)/d

BDiagram: the horizontal distance d is opposite the 7° angle and the height difference 72 - 3 is adjacent; Equation: tan(7°) = d/(72 - 3)

CDiagram: the height difference 72 - 3 is opposite the 7° angle and the horizontal distance d is the hypotenuse; Equation: sin(7°) = (72 - 3)/d

DDiagram: the horizontal distance d is adjacent to the 7° angle and the height difference 72 - 3 is the hypotenuse; Equation: cos(7°) = d/(72 - 3)

Solution & Explanation:Subtract the passenger height from the tower height to find the opposite side. Relative to the 7° angle, the height difference 72 - 3 is opposite and the horizontal distance d is adjacent. Tangent compares opposite over adjacent. So the correct setup is tan(7°) = (72 - 3)/d.

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Frequently Asked Questions

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QWhat grade level and standards does this worksheet align with?

This worksheet is designed specifically for Trigonometry Math students. It aligns with educational standard UTRI.TA.1 (Primary standard selected for prompt fields in this prompt cell: UTRI.TA.1. Skill statement in this prompt cell: Select the diagram and equation that correctly represent a word problem as a right triangle. Associated standards listed with this skill in this prompt cell: UTRI.TA.1; CCSS HSG.SRT.C.6-8; TEKS §111.42(c)(4)(E)-(F); FL MA.912.T.1.1-1.2; VA T.TT.1a,c-d. Use this prompt-provided wording as the supporting standard description for this workflow.).