Solved 9.6 - Solving Right Triangles Worksheet Name: Date: | Chegg.com - Free Printable

Let’s solve each problem one by one. We’re finding missing angles in right triangles using trigonometry — specifically, the tangent, sine, or cosine ratios. Remember:

- Tangent (tan) = opposite / adjacent
- Sine (sin) = opposite / hypotenuse
- Cosine (cos) = adjacent / hypotenuse

We’ll use inverse functions (like tan⁻¹, sin⁻¹, cos⁻¹) on our calculator to find the angle after setting up the ratio.

---

Problem 1:

Triangle SRT, right angle at R.
Side SR = 28 (opposite to angle T)
Side RT = 43 (adjacent to angle T)
Find angle T.

→ Use tan(T) = opposite / adjacent = 28 / 43
→ T = tan⁻¹(28/43)
→ Calculate: 28 ÷ 43 ≈ 0.6512
→ tan⁻¹(0.6512) ≈ 33.1°

✔ Angle T ≈ 33.1°

---

Problem 2:

Triangle MNL, right angle at N.
Side MN = 40 (adjacent to angle M)
Side NL = 13 (opposite to angle M)
Find angle M.

→ Use tan(M) = opposite / adjacent = 13 / 40
→ M = tan⁻¹(13/40)
→ 13 ÷ 40 = 0.325
→ tan⁻¹(0.325) ≈ 18.0°

✔ Angle M ≈ 18.0°

---

Problem 3:

Triangle ABC, right angle at B.
Side AB = ? (we don’t need it directly)
Side BC = 11 (opposite to angle A)
Side AC = 12 (hypotenuse)
Wait — actually, side AC is labeled 12, and it’s the hypotenuse? But angle B is the right angle, so yes, AC is hypotenuse.

But we have side BC = 11 (opposite to angle A), and hypotenuse AC = 12.

→ So use sin(A) = opposite / hypotenuse = 11 / 12
→ A = sin⁻¹(11/12)
→ 11 ÷ 12 ≈ 0.9167
→ sin⁻¹(0.9167) ≈ 66.4°

✔ Angle A ≈ 66.4°

*(Note: The side labeled “12” is between A and C — that’s the hypotenuse since right angle is at B. Correct.)*

---

Problem 4:

Triangle PQR, right angle at Q.
Side PQ = 20 (adjacent to angle P)
Side PR = 29 (hypotenuse)
Find angle P.

→ Use cos(P) = adjacent / hypotenuse = 20 / 29
→ P = cos⁻¹(20/29)
→ 20 ÷ 29 ≈ 0.6897
→ cos⁻¹(0.6897) ≈ 46.4°

✔ Angle P ≈ 46.4°

---

Problem 5:

Triangle XYZ, right angle at Y.
Side XY = 72 (opposite to angle Z)
Side YZ = 65 (adjacent to angle Z)
Hypotenuse XZ = 97 (check: 65² + 72² = 4225 + 5184 = 9409; √9409 = 97 → correct)

Find angle Z.

→ Use tan(Z) = opposite / adjacent = 72 / 65
→ Z = tan⁻¹(72/65)
→ 72 ÷ 65 ≈ 1.1077
→ tan⁻¹(1.1077) ≈ 47.9°

✔ Angle Z ≈ 47.9°

---

Problem 6:

Triangle FED, right angle at E.
Side FE = 54 (adjacent to angle F)
Side ED = 72 (opposite to angle F)
Hypotenuse FD = 90 (check: 54² + 72² = 2916 + 5184 = 8100; √8100 = 90 → correct)

Find angle F.

→ Use tan(F) = opposite / adjacent = 72 / 54
→ Simplify: 72 ÷ 54 = 4/3 ≈ 1.3333
→ F = tan⁻¹(1.3333) ≈ 53.1°

✔ Angle F ≈ 53.1°

---

Final Answer:
1) 33.1°
2) 18.0°
3) 66.4°
4) 46.4°
5) 47.9°
6) 53.1°