MA.912.T.1.2

Benchmark Instructional Guide

Connecting Benchmarks/Horizontal Alignment

• MA.912.GR.1.2
• MA.912.GR.1.6
• MA.912.GR.6.3

Terms from the K-12 Glossary

• Angle 
• Equilateral Triangle 
• Hypotenuse 
• Isosceles Triangle 
• Right Triangle

Vertical Alignment

Previous Benchmarks

• MA.8.GR.1.1
• MA.8.GR.1.2

Next Benchmarks

• MA.912.T.1.5
• MA.912.T.1.6
• MA.912.T.1.7
• MA.912.T.1.8
• MA.912.T.2
• MA.912.T.3

Purpose and Instructional Strategies

• Within the Geometry course, the expectation is to use angle measures given in degrees and not in radians. Additionally, it is not the expectation for students to master the trigonometric ratios of secant, cosecant and cotangent within this course. 
• It is customary to use Greek letters to represent angle measures (e.g., Ø, α, β, γ). 
• Problem types include cases where some information about the side lengths or angle measures of a right triangle is missing and one must use trigonometric ratios, inverse of trigonometric ratios or Pythagorean Theorem to determine the unknown length(s) or angle measure(s) within a mathematical or real-world context. 
• Instruction includes the concept of inverse trigonometric ratios to determine unknown angle measures and how to find these values using technology, including a calculator. Students should have practice using both notations for the inverse trigonometric ratios (sin⁻¹ A or arcsin A; cos⁻¹ A or arccos A; and tan⁻¹ A or arctan A). 
• Instruction includes exploring the relationships of the side lengths of special right triangles 45° − 45° − 90° and 30° − 60° − 90°. 
  • For example, students should realize that the special right triangle 45° − 45° − 90° is an isosceles right triangle. Therefore, two of its angle measures and side lengths are equivalent. So, if a side length is $x$ units, then students can use the Pythagorean Theorem to determine that the hypotenuse is $x$$\sqrt{2}$ units. Additionally, students can make the connection to its trigonometric ratios:  sin 45° = $\frac{\text{1}}{\text{<math><msqrt><mn>2</mn></msqrt></math>}}$ (or equivalently $\frac{\text{<math><msqrt><mn>2</mn></msqrt></math>}}{\text{2}}$); cos 45° = $\frac{\text{1}}{\text{<math><msqrt><mn>2</mn></msqrt></math>}}$ (or equivalently $\frac{\text{<math><msqrt><mn>2</mn></msqrt></math>}}{\text{2}}$); and tan 45° = 1. 
  • For example, students should realize that the special right triangle 30° − 60° − 90° is half of an equilateral triangle. Students can use that knowledge to determine that the shorter leg is one-half the length of the hypotenuse. So, if the shorter leg is $x$ units and the hypotenuse is 2$x$ units, then students can use the Pythagorean Theorem to determine that the other leg is $x$$\sqrt{3}$ units. Additionally, students can make the connection to its trigonometric ratios such as, sin 30° = $\frac{\text{1}}{\text{2}}$; cos 30° = $\frac{\text{1}}{\text{<math><msqrt><mn>3</mn></msqrt></math>}}$ and tan 30° = (or equivalently $\frac{\text{<math><msqrt><mn>3</mn></msqrt></math>}}{\text{3}}$).

Common Misconceptions or Errors

• Students may choose the incorrect trigonometric ratio when solving problems. 
• Students may misidentify the sides of triangles. 
  • For example, students may identify the hypotenuse as being the adjacent leg or confuse the adjacent and opposite sides.

Instructional Tasks

• ABCD is a square.
  
  
  
  • Part A. What is the measure of segment BD? 
  • Part B. What is the measure of segment AC? 
  • Part C. If the measure of segment BD is 14 units, what is the measure of segment BC?

• The right triangle ABC is shown. Angle B is the right angle and the length of AB is 1.5 centimeters and the length of BC is 3.1 centimeters.

• Part A. Determine the measure of angles A and C. 
• Part B. Determine the length of AC.

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