Question

If 3 angleA= 4 angleB = 6angleC, then angleA: angleB : angleC = ? where A, B, C are angles of a triangle
(A) 3:4:6
(B) 4:3:2
(C) 2:3:4
(D) 6:4:3​

Answer 1

Given :

• 3× angle(A) = 4 ×angle(B) = 6× angle (C)
• ABC are angle of t triangle

To find :

angle(A) : angle(B) : angle(C)

Solution :

➝ 3× angle(A) = 4 ×angle(B) = 6× angle (C) = k { let }

Therefore

1) 3× angle(A) = k

➝ angle(A) = k/3

2) 4 × angle(B) = k

➝ angle(B) = k/4

3) 6 × angle(C) = k

➝ angle(C) = k/6

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Also , angle(A) , angle(B) and angle(C) are angle of triangle , this means

$\sf \implies angle(A) + angle(B) + angle(C) = 180^{\circ} \\ \\ \sf \implies \: \dfrac{k}{3} + \dfrac{k}{4} + \dfrac{k}{6} = \: 180^{\circ} \\ \\ \: \sf taking \: lcm \\ \sf \implies \: \: \dfrac{(k \times 8) + (k \times 6) + (k \times 4)}{24} \: = \: 180^{\circ} \\ \\ \sf \implies \: \dfrac{8k + 6k + 4k}{24} = 180^{\circ} \\ \\ \sf \implies \: \: 18k \: = \:24 \times 180^{\circ} \\ \\ \sf \implies \:k \: = \dfrac{24 \times 180}{18} \\ \\ \sf \implies \:k \: = 240$

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1) angle(A) = k/3

➝ angle(A) = 240/3

➝ angle(A) = 80°

2) angle(B) = k/4

➝ angle(B) = 240/4

➝ angle(B) = 60°

3) angle(C) = k/6

➝ angle(C) = 240/6

➝ angle(C) = 40°

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➝ angle(A) : angle(B) : angle(C) = 80° : 60° : 40°

➝ angle(A) : angle(B) : angle(C) = 4 : 3 : 2

ANSWER :

option (B) 4 : 3 : 2