This document discusses how to solve right triangles by finding missing parts. It explains that if an angle and side are given, trig ratios can be used to find the missing side and remaining angle. If two sides are given, the Pythagorean theorem or trig ratios can be used to find the missing side or angle. Examples are provided to demonstrate solving for missing parts in right triangles when given different information. Trig ratios, Pythagorean theorem, subtraction of angles from 90 degrees, and rounding are techniques presented.

1. Solving Right Triangles
The student is able to (I can):
• Find the missing parts of a right triangle
• Use trig ratios to solve problems

2. To “solve” a right triangle means to find all of the missing
parts of the right triangle.
If you are given an angle and a side:
• Subtract the angle from 90° to find the other acute angle
• Use trig ratios to find one of the missing sides
• Use either trig ratios or Pythagorean Theorem to find the
third side

3. If you are given two sides:
• Use Pythagorean Theorem to find the missing side
• Use trig ratios to find an angle
– Unless all of your sides work out to be whole numbers,
be sure to use the two given sides in your trig ratio to
prevent rounding errors.
• Subtract the angle you found from 90° to find the other
angle

4. Examples
Solve the triangles. Round sides to the nearest tenth and
angles to the nearest whole degree.
1. m∠D=_____
ED = ______
BD = ______
6
67°B
E
D

5. Examples
Solve the triangles. Round sides to the nearest tenth and
angles to the nearest whole degree.
1. m∠D=_____
ED = ______
BD = ______
6
67°B
E
D
tan67
6
6tan67
14.1
ED
ED
° =
= °
≈
m 90 67 23D∠ = − = °

6. Examples
Solve the triangles. Round sides to the nearest tenth and
angles to the nearest whole degree.
1. m∠D=_____
ED = ______
BD = ______
6
67°B
E
D
adj
opp
hyp
tan67
6
6tan67
14.1
ED
ED
° =
= °
≈
m 90 67 23D∠ = − = °
23°
14.1

7. Examples
Solve the triangles. Round sides to the nearest tenth and
angles to the nearest whole degree.
1. m∠D=_____
ED = ______
BD = ______
6
67°B
E
D
adj
opp
hyp
tan67
6
6tan67
14.1
ED
ED
° =
= °
≈
m 90 67 23D∠ = − = °
2 2 2
6 14.1
234.81
15.3
BD
BD
= +
=
≈
23°
14.1
15.3

8. 2. TA = _____
m∠A = _____
m∠T = _____
12
18
T
A
M

9. 2. TA = _____
m∠A = _____
m∠T = _____
12
18
T
A
M2 2 2
18 12
468 21.6
TA
TA
= +
= ≈
hyp
21.6

10. 2. TA = _____
m∠A = _____
m∠T = _____
It doesn’t matter whether you use ∠T or ∠A. The
triangle is labeled for ∠T.
12
18
T
A
M2 2 2
18 12
468 21.6
TA
TA
= +
= ≈
hyp
opp
adj
21.6

11. 2. TA = _____
m∠A = _____
m∠T = _____
It doesn’t matter whether you use ∠T or ∠A. The
triangle is labeled for ∠T.
12
18
T
A
M2 2 2
18 12
468 21.6
TA
TA
= +
= ≈
hyp
opp
adj
1
12
tan
18
12
tan
18
34
T
T −
=
 
=  
 
≈ °
90 34
56
m A∠ = −
= °
21.6
56°
34°