### Clarifications

*Clarification 1:*Problems include right triangles, heights inside of a triangle and heights outside of a triangle.

**Subject Area:**Mathematics (B.E.S.T.)

**Grade:**912

**Strand:**Trigonometry

**Date Adopted or Revised:**08/20

**Status:**State Board Approved

## Benchmark Instructional Guide

### Connecting Benchmarks/Horizontal Alignment

### Terms from the K-12 Glossary

- Angle
- Area
- Hypotenuse
- Right Triangle

### Vertical Alignment

Previous Benchmarks

Next Benchmarks

### Purpose and Instructional Strategies

In grade 6, students learned how to find the area of a triangle using a formula and its connection to the area of a rectangle. In Geometry, students find the area of a triangle when given the length of two sides and the included angle. In later courses, students will use further trigonometric relationships and identities to determine information that is needed to find areas of triangles.- 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., θ, α, β, γ).
- Instruction includes making the connection between the formula for the area of a triangle
(
*A*= $\frac{\text{1}}{\text{2}}$$b$$h$) and the trigonometric functions*(MTR.5.1)*.- For example, if an acute triangle is given with two sides lengths, $a$ and $b$, and the
included angle measure, θ, students can use trigonometric functions to determine
the height of the triangle. Students can select side $b$ to be the base and then
construct the height of the triangle, $h$, to determine the area. Students should
realize that when constructing the height, it creates a right triangle and the length
of $h$ can be found by using the sine of the angle θ: $h$ = $a$ sin θ. Students can now use the area formula
*A*= $b$$h$ and substitution to obtain the new formula*A*= $b$($a$ sin θ). Similarly, students can use the side $a$ as the base and obtain the formula*A*= $\frac{\text{1}}{\text{2}}$$a$($b$ sin θ). - For example, if an obtuse triangle is given with side lengths, $a$ and $b$, and the
included obtuse angle measure, θ, students can use trigonometric functions to
determine the height of the triangle. Students can select side $b$ to be the base and
then construct the height of the triangle, $h$, to determine the area. Students should
realize that when constructing the height, it creates a right triangle outside of the
given obtuse triangle and the length of $h$ can be found by using the sine of the
angle 180 − θ: $h$ = $a$ sin(180 − θ). Students can explore using technology or
the unit circle to show that the sin(180 − θ) is equivalent to sin(θ), therefore $h$ = $a$ sin(θ). Students can now use the area formula
*A*= $\frac{\text{1}}{\text{2}}$$b$$h$ and substitution to obtain the new formula*A*= $\frac{\text{1}}{\text{2}}$$b$($a$ sin θ). Similarly, students can use the side $a$ as the base and obtain the formula*A*= $\frac{\text{1}}{\text{2}}$$a$($b$ sin θ).

- For example, if an acute triangle is given with two sides lengths, $a$ and $b$, and the
included angle measure, θ, students can use trigonometric functions to determine
the height of the triangle. Students can select side $b$ to be the base and then
construct the height of the triangle, $h$, to determine the area. Students should
realize that when constructing the height, it creates a right triangle and the length
of $h$ can be found by using the sine of the angle θ: $h$ = $a$ sin θ. Students can now use the area formula
- Instruction includes making the connection to the Law of Sines.
*(MTR.5.1)*- For example, if given a triangle with side lengths a, b and c, and opposite angles
α, β and γ, then there are three formulas students could use to determine the area:
*A*= $\frac{\text{1}}{\text{2}}$$c$$b$ sin α,*A*= $\frac{\text{1}}{\text{2}}$$a$$c$ sin β and*A*= $\frac{\text{1}}{\text{2}}$$b$$a$ sin γ. Students should realize that each of these formulas are equivalent to one another, so that $\frac{\text{1}}{\text{2}}$$c$$b$ sin α = $\frac{\text{1}}{\text{2}}$$a$$c$ sin β = $\frac{\text{1}}{\text{2}}$$b$$a$ sin γ. To make the connection to the Law of Sines, each expression can be divided by $\frac{\text{1}}{\text{2}}$$a$$b$$c$ (Division Property of Equality), to obtain $\frac{\text{sin \alpha}}{\text{a}}$ = $\frac{\text{sin \beta}}{\text{b}}$ = $\frac{\text{sin \gamma}}{\text{c}}$.

- For example, if given a triangle with side lengths a, b and c, and opposite angles
α, β and γ, then there are three formulas students could use to determine the area:

### Common Misconceptions or Errors

- Students may misidentify a side as the height of the triangle and attempt to use
*A*= $\frac{\text{bh}}{\text{2}}$ to find the area. - Students may have difficulty determining the height of an obtuse triangle.

### Instructional Tasks

*Instructional Task 1 (MTR.3.1, MTR.4.1)*

- Triangle
*ABC*is given with*AB*= 22 units,*AC*= 36 units and the measure of angle α = 29°.- Part A. If side
*AB*is the considered the base of the triangle, determine the height of the triangle. - Part B. If side
*AC*is the considered the base of the triangle, determine the height of the triangle. - Part C. Determine the area of the triangle. Compare your method with a partner.
- Part D. Determine the length of side
*BC*using the Law of Cosines. - Part E. Can you use the answers from Part C and Part D to determine the height of the
triangle if
*BC*is considered as the base

- Part A. If side

### Instructional Items

*Instructional Item 1*

- Find the area of triangle
*ABC*.

**The strategies, tasks and items included in the B1G-M are examples and should not be considered comprehensive.*

## Related Courses

## Related Access Points

## Related Resources

## Lesson Plan

## Problem-Solving Tasks

## Student Resources

## Problem-Solving Tasks

Using a chart of diameters of different denominations of coins, students are asked to figure out how many coins fit around a central coin. (For this task, United States coins are used, but the task can be adapted for coins from other countries.)

Type: Problem-Solving Task

This problem solving task asks students to find the area of an equilateral triangle. Various solutions are presented that include the Pythagoren theorem and trigonometric functions.

Type: Problem-Solving Task

This modeling task involves several different types of geometric knowledge and problem-solving: finding areas of sectors of circles, using trigonometric ratios to solve right triangles, and decomposing a complicated figure involving multiple circular arcs into parts whose areas can be found.

Type: Problem-Solving Task

## Parent Resources

## Problem-Solving Tasks

Using a chart of diameters of different denominations of coins, students are asked to figure out how many coins fit around a central coin. (For this task, United States coins are used, but the task can be adapted for coins from other countries.)

Type: Problem-Solving Task

This problem solving task asks students to find the area of an equilateral triangle. Various solutions are presented that include the Pythagoren theorem and trigonometric functions.

Type: Problem-Solving Task

This modeling task involves several different types of geometric knowledge and problem-solving: finding areas of sectors of circles, using trigonometric ratios to solve right triangles, and decomposing a complicated figure involving multiple circular arcs into parts whose areas can be found.

Type: Problem-Solving Task