Inverse Trigonometric Ratios Color by Number - Free Printable

Problem Analysis:

The image contains several right triangles, and the task is to solve for missing sides or angles using trigonometric principles. Additionally, there is a problem involving finding an angle where sine equals cosine.

Let's solve each part step by step.

---

Problem 7 (Sky Blue)

#### Given:
- Hypotenuse = 45
- One leg = 26
- Find the other leg (?).

#### Solution:
We use the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where \( c \) is the hypotenuse, and \( a \) and \( b \) are the legs of the triangle.

Here:
\[
a = 26, \quad c = 45, \quad b = ?
\]

Substitute the known values:
\[
26^2 + b^2 = 45^2
\]
\[
676 + b^2 = 2025
\]
\[
b^2 = 2025 - 676
\]
\[
b^2 = 1349
\]
\[
b = \sqrt{1349}
\]

Thus, the length of the other leg is:
\[
\boxed{\sqrt{1349}}
\]

---

Problem 8 (Navy Blue)

#### Given:
- One leg = 13
- Hypotenuse = 20
- Find the other leg (?).

#### Solution:
Again, use the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where \( c = 20 \), \( a = 13 \), and \( b = ? \).

Substitute the known values:
\[
13^2 + b^2 = 20^2
\]
\[
169 + b^2 = 400
\]
\[
b^2 = 400 - 169
\]
\[
b^2 = 231
\]
\[
b = \sqrt{231}
\]

Thus, the length of the other leg is:
\[
\boxed{\sqrt{231}}
\]

---

Problem 9 (Yellow)

#### Given:
- One leg = 43
- Hypotenuse = 37
- Find the other leg (?).

#### Solution:
Use the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where \( c = 37 \), \( a = 43 \), and \( b = ? \).

Substitute the known values:
\[
43^2 + b^2 = 37^2
\]
\[
1849 + b^2 = 1369
\]

Notice that \( 43 \) is greater than \( 37 \), which is not possible for a right triangle because the hypotenuse must be the longest side. Therefore, this problem has no solution under the given conditions.

Thus, the answer is:
\[
\boxed{\text{No solution}}
\]

---

Problem 10 (Brown)

#### Given:
- One leg = 72
- Other leg = 54
- Find the hypotenuse (?).

#### Solution:
Use the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 72 \), \( b = 54 \), and \( c = ? \).

Substitute the known values:
\[
72^2 + 54^2 = c^2
\]
\[
5184 + 2916 = c^2
\]
\[
c^2 = 8100
\]
\[
c = \sqrt{8100}
\]
\[
c = 90
\]

Thus, the length of the hypotenuse is:
\[
\boxed{90}
\]

---

Problem 11 (Royal Blue)

#### Given:
- One leg = 48
- Other leg = 14
- Find the hypotenuse (?).

#### Solution:
Use the Pythagorean theorem:
\[
a^2 + b^2 = c^2
\]
where \( a = 48 \), \( b = 14 \), and \( c = ? \).

Substitute the known values:
\[
48^2 + 14^2 = c^2
\]
\[
2304 + 196 = c^2
\]
\[
c^2 = 2500
\]
\[
c = \sqrt{2500}
\]
\[
c = 50
\]

Thus, the length of the hypotenuse is:
\[
\boxed{50}
\]

---

Problem 12 (Red-Orange)

#### Given:
- Find an angle \( x \) where \( \sin x = \cos x \).

#### Solution:
Recall the trigonometric identity:
\[
\sin x = \cos x \implies \tan x = 1
\]
This occurs when:
\[
x = 45^\circ \quad \text{(or } \frac{\pi}{4} \text{ radians)}
\]

Thus, the angle \( x \) is:
\[
\boxed{45^\circ}
\]

---

Final Answers:

1. Problem 7 (Sky Blue): \(\boxed{\sqrt{1349}}\)
2. Problem 8 (Navy Blue): \(\boxed{\sqrt{231}}\)
3. Problem 9 (Yellow): \(\boxed{\text{No solution}}\)
4. Problem 10 (Brown): \(\boxed{90}\)
5. Problem 11 (Royal Blue): \(\boxed{50}\)
6. Problem 12 (Red-Orange): \(\boxed{45^\circ}\)