Algebra and triangle angles: six triangles with angles expressed as algebraic terms (e.g., 2f, 3f, 5p, 4p, etc.), demonstrating how to use the triangle angle sum property to solve for variables.

We are given six triangles with angles expressed in terms of variables (f, n, k, p, e, d). The key principle we use is:

> The sum of the interior angles of any triangle is always 180 degrees.

We’ll solve each one by setting up an equation where the sum of the three angles equals 180°, then solve for the variable.

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(1) Angles: f, 2f, 3f

Equation:
f + 2f + 3f = 180
→ 6f = 180
→ f = 30

✔ So, the angles are: 30°, 60°, 90°.

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(2) Angles: n, 3n, 5n

Equation:
n + 3n + 5n = 180
→ 9n = 180
→ n = 20

✔ So, the angles are: 20°, 60°, 100°.

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(3) Angles: 9k, 10k, 11k

Equation:
9k + 10k + 11k = 180
→ 30k = 180
→ k = 6

✔ So, the angles are: 54°, 60°, 66°.

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(4) Right triangle — one angle is 90°, others are 5p and 4p

Equation:
5p + 4p + 90 = 180
→ 9p = 90
→ p = 10

✔ So, the angles are: 50°, 40°, 90°.

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(5) Right triangle — one angle is 90°, others are 22e and 23e

Equation:
22e + 23e + 90 = 180
→ 45e = 90
→ e = 2

✔ So, the angles are: 44°, 46°, 90°.

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(6) Right triangle — one angle is 90°, others are 8d and 7d

Equation:
8d + 7d + 90 = 180
→ 15d = 90
→ d = 6

✔ So, the angles are: 48°, 42°, 90°.

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## ✔ Final Answers:

(1) f = 30
(2) n = 20
(3) k = 6
(4) p = 10
(5) e = 2
(6) d = 6

These values make each triangle’s angles sum to 180°, satisfying the triangle angle sum theorem.