Tamil Nadu Board of Secondary EducationHSC Arts Class 11th

# In a ∆ABC, if a = 3-1, b = 3+1 and C = 60° find the other side and other two angles - Mathematics

Sum

In a ∆ABC, if a = sqrt(3) - 1, b = sqrt(3) + 1 and C = 60° find the other side and other two angles

#### Solution

In a ∆ABC,

Given A = sqrt(3) - 1

B = sqrt(3) + 1

C = 60°

Using cosine formula

C2 = a2 + b2 – 2 ab cos C

= (sqrt(3) - 1)^2 + (sqrt(3) + 1)^2 - 2(sqrt(3) - 1) xx (sqrt(3) + 1)  cos 60^circ

= 3 -2sqrt(3) + 1 + 3 + 2sqrt(3) + 1 - 2(3 - 1) xx 1/2

C2 = 8 – 2 = 6

⇒ C = √6

Using since formula

"a"/sin"A" = "b"/sin"B" = "c"/sin"C"

(sqrt(3) - 1)/sin"A" = (sqrt(3) + 1)/sin"B" = sqrt(6)/sin60^circ

(sqrt(3) - 1)/sin"A" = sqrt(6)/sin 60^circ

(sqrt(3) - 1)/sin"A" = sqrt(6)/(sqrt(3)/2)

⇒ (sqrt(3) - 1)/sin"A" = (2sqrt(3) * sqrt(2))/sqrt(3)

⇒ (sqrt(3) - 1)/sin"A" = 2sqrt(2)

⇒ sin A = (sqrt(3) - 1)/(2sqrt(2))  ......(1)

sin(45° – 30°) = sin 45° . cos 30° – cos 45° sin 30°

= 1/sqrt(2) * sqrt(3)/2 - 1/sqrt(2) * 1/2

sin 15° = (sqrt(3) - 1)/(2sqrt(2))  ......(2)

From equations (1) and (2), we have

sin A = sin 15°

⇒ A = 15°

In ∆ABC,

We have A + B + C = 180°

15° + B + 60° = 180°

B = 180° – 75°

B = 105°

∴ The required sides and angles are

C = sqrt(6), A = 15°, B = 105°

Concept: Application to Triangle
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