Selina solutions

Selina solutions

Grade 7

Triangles | Exercise 15(A)

Question 11

Q11) One angle of a triangle is 61° and the other two angles are in the ratio 1\frac{1}{2}:1\frac{1}{3}.. Find these angles.

Solution:

In \triangle ABC

Let ∠A = 61°

But ∠A + ∠B + ∠C = 180°

(Angles of a triangle)

⇒ 61° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° - 61° = 119°

But ∠B : ∠C = 1.1/2 : 1.1/3 = 3/2 : 4/3

= (9 : 8)/6 = 9 : 8

Let ∠B = 9x and ∠C = 8x,

then, 9x + 8x = 119°

⇒ 17x = 119°

⇒ x = 119°/17 = 7°

∠B = 9x = 9 × 7° = 63°

C = 8x = 8 x 7° = 56°

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subject-cta

Question 11

Q11) One angle of a triangle is 61° and the other two angles are in the ratio 1\frac{1}{2}:1\frac{1}{3}.. Find these angles.

Solution:

In \triangle ABC

Let ∠A = 61°

But ∠A + ∠B + ∠C = 180°

(Angles of a triangle)

⇒ 61° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° - 61° = 119°

But ∠B : ∠C = 1.1/2 : 1.1/3 = 3/2 : 4/3

= (9 : 8)/6 = 9 : 8

Let ∠B = 9x and ∠C = 8x,

then, 9x + 8x = 119°

⇒ 17x = 119°

⇒ x = 119°/17 = 7°

∠B = 9x = 9 × 7° = 63°

C = 8x = 8 x 7° = 56°

Still have questions? Our expert teachers can help you out

Book a free class now

Want to top your mathematics exam ?

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Book a free class now
subject-cta
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