G.11(C)·Geometry·Similarity & MeasurementReadiness

Triangle Similarity and Trigonometry

Use trigonometric ratios and similar-triangle proportions to determine unknown side lengths.

Students use triangle similarity and side-splitter proportions to find lengths, then use sine, cosine, and tangent to express or calculate right-triangle sides from either complementary acute angle.

Lesson plan and worked examples by TEKSForge · not a TEA-issued lesson. Videos by independent educators.

Archived course: TEKS code and readiness label from TEA's 2013 Geometry key. Geometry is not one of Texas's current required STAAR EOC assessments, and this lesson does not change the Biology and Algebra I Study Plans.

Practice the released item Ask the tutor All Geometry released practice55 minutes

Learn the concept

Label from the angle

For a chosen acute angle, the hypotenuse is opposite the right angle, the opposite side faces the chosen angle, and the adjacent side touches the chosen angle without being the hypotenuse.

Choose the ratio

$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ , and $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ . Pick the ratio that includes the side you know and the side you need.

Why the ratios are constant

Right triangles with the same acute angle are similar by AA. Their corresponding sides are proportional, so each trigonometric ratio has one value for that angle, regardless of the triangle's size.

A parallel segment splits sides proportionally

In triangle ABC, let D lie on AB and E lie on AC, with $DE\parallel BC$ . Corresponding angles show $\triangle ADE\sim\triangle ABC$ by AA. Thus $\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$ , and the side-splitter relationship is $\frac{AD}{DB}=\frac{AE}{EC}$ . Compare part to whole on both sides, or corresponding parts on both sides; do not mix the two.

Express a side from complementary angles

The acute angles of a right triangle sum to 90°. Changing the reference angle swaps opposite and adjacent, so $\sin\theta=\cos(90^\circ-\theta)$ . If a leg of length 17 is adjacent to 25° and the opposite leg is x, then $x=17\tan25^\circ=\frac{17}{\tan65^\circ}$ . If the hypotenuse is c, that same opposite leg is $x=c\sin25^\circ=c\cos65^\circ$ . An exact trig expression is a valid answer without a decimal.

Study released STAAR visuals

Use the original graphs, tables, and diagrams from released STAAR questions to practice reading labels, values, and relationships before answering the items below.

Archived released STAAR Geometry EOC · 2013 · Item 7 · 1 point

Highway 1 and 10th Street meet at a right angle. Main Street is the hypotenuse and makes a 40-degree angle with 10th Street. — Main Street right-triangle mapOriginal visual from the released STAAR test.

Archived released STAAR Geometry EOC · 2013 · Item 23 · 1 point

Right triangle QRS with the right angle at R. Q is above R, and S is to the right of R. Side RS is 19 centimeters, and angle S measures 20 degrees. — Right triangle QRSOriginal visual from the released STAAR test.

Archived released STAAR Geometry EOC · 2013 · Item 52 · 1 point

Triangle DAC with A at the top. E is on side AD and B is on side AC, and segment EB is parallel to DC. AE is 18 centimeters and ED is 4.8 centimeters. — Triangle DAC with segment EBOriginal visual from the released STAAR test.

Worked example

A right triangle has a 35° angle and a hypotenuse of 12. Find the side adjacent to the 35° angle.

The known side is the hypotenuse, and the unknown side is adjacent.
Use cosine: $\cos 35^\circ=\frac{x}{12}$ .
Multiply: $x=12\cos 35^\circ\approx 12(0.8192)\approx 9.83$ .

About 9.83 units

Watch the lesson

Texas public-school, Texas educator-created, and TEKS-aligned lessons, hosted by their creators. Each card names its source.

Texas public school video

Geometry: Trigonometry

UT IPSI · Kid2Kid / Texas Education Agency

Use sine, cosine, and tangent to find unknown sides in right triangles.

Watch on YouTube

Try it yourself

TEKSForge practice · unscored

1. A right triangle has a 50° angle with an adjacent side of 8. Find the opposite side.
Show answer and reasoning
About 9.53 units
Use tangent: $x=8\tan 50^\circ\approx 8(1.1918)\approx 9.53$ .
2. A right triangle has legs of 6 and 8. Find the hypotenuse.
Show answer and reasoning
10 units
This is a multiple of the 3-4-5 Pythagorean triple, so the hypotenuse is 10.
3. In triangle ABC, D lies on AB and E lies on AC, with $DE\parallel BC$ . If AD = 4, DB = 6, and AE = 10, find EC.
Show answer and reasoning
15 units
Use corresponding split segments: $\frac{4}{6}=\frac{10}{EC}$ , so $EC=\frac{6(10)}{4}=15$ .
4. A right triangle has a 25° angle and an adjacent leg of 17 cm. Express the opposite leg using tangent of each acute angle, then approximate its length.
Show answer and reasoning
$17\tan25^\circ=\frac{17}{\tan65^\circ}\approx7.93$ cm
$\tan25^\circ=\frac{x}{17}$ , while $\tan65^\circ=\frac{17}{x}$ . Solve each equation for x.

Released STAAR practice

Texas Education Agency · archived 2013 released EOC items · scored with TEA's answer key; step-by-step explanations written by TEKSForge because this release has no rationales. Original graphs, tables, and diagrams are shown where the released item includes one.

Archived released STAAR Geometry EOC · 2013 · Item 7 · 1 point

On the map, Main Street, 10th Street, and Highway 1 intersect to form a right triangle. The distance between 10th Street and Main Street along Highway 1 is 5.6 miles. Which measure is closest to the length of Main Street from Highway 1 to 10th Street?

8.7 mi
3.6 mi
4.7 mi
7.3 mi

Show STAAR answer and explanation

8.7 mi

Highway 1 is opposite the 40° angle and Main Street is the hypotenuse. Therefore $5.6/\sin40°\approx8.7$ miles.

3.6 mi: This is approximately $5.6\sin40°$ , but dividing by sine is required to find the hypotenuse.

4.7 mi: This is approximately $5.6\tan40°$ and would not be the hypotenuse; the answer must exceed 5.6 miles.

7.3 mi: This uses cosine instead of sine. The 5.6-mile Highway 1 side is opposite the 40° angle.

Ask the tutor about this question

Archived released STAAR Geometry EOC · 2013 · Item 23 · 1 point

Which expression can be used to find the length of $\overline{QR}$ in centimeters?

$19(\cos 20°)$
$19(\tan 20°)$
$19(\sin 70°)$
$19(\tan 70°)$

Show STAAR answer and explanation

$19(\tan 20°)$

Relative to the 20° angle, QR is opposite and RS = 19 is adjacent. Therefore $\tan20°=QR/19$ , so $QR=19\tan20°$ .

$19(\cos 20°)$ : Cosine compares adjacent leg to hypotenuse; QR is opposite the 20° angle.

$19(\sin 70°)$ : Relative to 70°, the 19-cm side is opposite, not the hypotenuse required by this sine multiplication.

$19(\tan 70°)$ : At 70°, tangent is $19/QR$ , so QR would be $19/\tan70°$ , not the product.

Ask the tutor about this question

Archived released STAAR Geometry EOC · 2013 · Item 52 · 1 point

In $\triangle DAC$ shown, $\overline{EB} \parallel \overline{DC}$ . If $AC = 28.5$ cm, what is the length of $\overline{AB}$ ?

22.5 cm
14.4 cm
36.1 cm
23.7 cm

Show STAAR answer and explanation

22.5 cm

Parallel EB and DC make triangles AEB and ADC similar. Since $AD=18+4.8=22.8$ , $AB/28.5=18/22.8$ . Thus $AB=28.5(18/22.8)=22.5$ cm.

14.4 cm: This does not apply the matching-side ratio $18/(18+4.8)$ .

36.1 cm: This reverses the similarity ratio and gives AB longer than the entire side AC.

23.7 cm: This subtracts 4.8 directly from AC instead of scaling corresponding sides.

Ask the tutor about this question

Start scored STAAR practice

Learning objectives

By the end of the lesson, students can:

Label opposite, adjacent, and hypotenuse relative to a given acute angle.
Choose sine, cosine, or tangent to express a side using either complementary acute angle.
Solve for unknown sides and connect trigonometric ratios to similar right triangles.
Use a segment parallel to a triangle's side to write proportional side lengths.

Instructional sequence

1
Engage
5 min
Show two similar right triangles with the same acute angle. Ask what stays the same about their side ratios.
2
Direct instruction
10 min
Define sine, cosine, and tangent relative to an acute angle and label sides on several orientations.
3
Guided practice
15 min
Solve multiplication and division cases, deciding whether the unknown side is in the numerator or denominator.
4
Map application
12 min
Solve a street-map right triangle and interpret the answer in miles.
5
Parallel segments and complementary angles
5 min
Set up a side-splitter proportion and compare equivalent side expressions using complementary acute angles.
6
Closure
8 min
Students complete the exit ticket and check their calculator mode.

Misconceptions to address

““The opposite side is always the vertical side.””

Teacher move: Opposite and adjacent depend on the chosen angle. Point to the angle before labeling the sides.

““Always multiply by the trig value.””

Teacher move: Write the ratio first. If the unknown is in the denominator, solving requires division.

““A part-to-part ratio can equal any part-to-whole ratio.””

Teacher move: Match corresponding segments consistently. Label whole sides as sums before writing a proportion.

Materials

•Right-triangle diagrams
•Calculators set to degree mode
•SOH-CAH-TOA reference card

Teaching tips

Have students mark the reference angle with a star before labeling sides.
Use the same two similar triangles to show the ratio does not depend on size.
Estimate first: a hypotenuse must be longer than either leg.

Exit ticket · 3–5 min

A ramp 10 feet long rises at an angle of 6°. How high is the top of the ramp?

Show expected answer

$\sin 6^\circ=\frac{h}{10}$ , so $h=10\sin 6^\circ\approx 1.05$ feet.

Sources & alignment

TEA 2013 Geometry released test TEA 2013 Geometry answer key and standards Texas Gateway, home of the UT IPSI Kid2Kid lessons

Alignment follows the standard TEA assigned to each 2013 released item. Video playback depends on YouTube access; written instruction remains available here.

Learn the concept

Label from the angle

Choose the ratio

Why the ratios are constant

A parallel segment splits sides proportionally

Express a side from complementary angles

Study released STAAR visuals

Archived released STAAR Geometry EOC · 2013 · Item 7 · 1 point

Archived released STAAR Geometry EOC · 2013 · Item 23 · 1 point

Archived released STAAR Geometry EOC · 2013 · Item 52 · 1 point

Worked example

Watch the lesson

Geometry: Trigonometry

Try it yourself

1. A right triangle has a 50° angle with an adjacent side of 8. Find the opposite side.

2. A right triangle has legs of 6 and 8. Find the hypotenuse.

3. In triangle ABC, D lies on AB and E lies on AC, with DE∥BCDE\parallel BCDE∥BC. If AD = 4, DB = 6, and AE = 10, find EC.

4. A right triangle has a 25° angle and an adjacent leg of 17 cm. Express the opposite leg using tangent of each acute angle, then approximate its length.

Released STAAR practice

Archived released STAAR Geometry EOC · 2013 · Item 7 · 1 point

Archived released STAAR Geometry EOC · 2013 · Item 23 · 1 point

Archived released STAAR Geometry EOC · 2013 · Item 52 · 1 point

Learning objectives

Instructional sequence

Engage

Direct instruction

Guided practice

Map application

Parallel segments and complementary angles

Closure

Misconceptions to address

3. In triangle ABC, D lies on AB and E lies on AC, with $DE\parallel BC$ . If AD = 4, DB = 6, and AE = 10, find EC.