In a triangle A BC , if A = 3 0 ∘ , B = 6 0 ∘ and side a = 10 , then the length of side b is:

Using the Sine Rule, we have s i n A a = s i n B b . Substituting the given values, s i n 3 0 ∘ 10 = s i n 6 0 ∘ b . This gives 1/2 10 = 3 /2 b , which simplifies to 20 = 3 2 b . Therefore, b = 10 3 .

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Mathematics - Properties of Triangle Question with Solution | TestHub

MathematicsProperties of TriangleGeneral (basic formula based)Easy2 minai-gemini

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In a triangle $A BC$ , if $A = 3 0^{\circ}$ , $B = 6 0^{\circ}$ and side $a = 10$ , then the length of side $b$ is:

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Solution:

Using the Sine Rule, we have $\frac{a}{s i n A} = \frac{b}{s i n B}$ . Substituting the given values, $\frac{10}{s i n 3 0 ^{\circ}} = \frac{b}{s i n 6 0 ^{\circ}}$ . This gives $\frac{10}{1/2} = \frac{b}{3 /2}$ , which simplifies to $20 = \frac{2 b}{3}$ . Therefore, $b = 103$ .

Stream:JEESubject:MathematicsTopic:Properties of TriangleSubtopic:General (basic formula based)

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