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Question
In a triangle ABC,
∠A=60° and b:c=√3+1:2,then the value of ∠B−∠C=
In a triangle ABC,
∠A=60° and b:c=√3+1:2,then the value of ∠B−∠C=
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Solution
The correct option is D 30°
bc=√3+12⇒b−cb+c=√3+1−2√3+1+2=√3−1(√3+1)√3Using Napier's rule, we get−tan B−C2=b−cb+ccot A2⇒(√3−1(√3+1)√3)×√3=2−√3⇒B−C2=15°Hence, B−C=30°
bc=√3+12⇒b−cb+c=√3+1−2√3+1+2=√3−1(√3+1)√3Using Napier's rule, we get−tan B−C2=b−cb+ccot A2⇒(√3−1(√3+1)√3)×√3=2−√3⇒B−C2=15°Hence, B−C=30°
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