Question #60967

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Answer 1 · 2017-09-08T18:34:08

# B-C=30^@.#

In fact, #B=75^@, and, C=45^@.#

We use the following property known as Napier's Rule (NR) :

#" NR : In "DeltaABC, tan((B-C)/2)=(b-c)/(b+c)cot(A/2).#

It can be proved very easily using the Sine Rule for #Delta.#

In our Problem, we have, #b/c=(sqrt3+1)/2.#

By the Componedo-Dividendo, we get,

#(b-c)/(b+c)=(sqrt3+1-2)/(sqrt3+1+2)=(sqrt3-1)/(sqrt3+3).#

#:.," by, NR, "tan((B-C)/2)=(sqrt3-1)/(sqrt3+3)*cot(60^@/2),#

#=(sqrt3-1)/(sqrt3+3)*sqrt3=(sqrt3-1)/(sqrt3(1+sqrt3))*sqrt3,#

#=(sqrt3-1)/(sqrt3+1),#

#:. tan((B-C)/2)=tan15^@.#

#:. B-C=30^@.#

Foot Note : In fact, #B, and C# can also be found individually.

#A=60^@ rArr B+C=180^@-60^@=120^@...........(1).#

# B-C=30^@...............................................................(2).#

Solving, #(1) and (2), B=75^@, and, C=45^@.#

Enjoy Maths.!

Answer 2 · 2018-01-23T13:24:17

Given #b/c=(sqrt3+1)/2#

#=>(2RsinB)/(2RsinC)=(sqrt3+1)/2#

#=>sin(pi-(A+C))/(sinC)=(sqrt3+1)/2#

#=>sin((A+C))/(sinC)=(sqrt3+1)/2#

#=>(sinA cosC+cosAsinC)/(sinC)=(sqrt3+1)/2#

#=>sinA cotC+cosA=sqrt3/2+1/2#

#=>sin60^@ cotC+cos60^@=sqrt3/2+1/2#

#=>sqrt3/2cotC+1/2=sqrt3/2+1/2#

#=>cotC=(sqrt3/2)/(sqrt3/2)=1=cot45^@#

#=>C=45^@#

Now #B-C#

#=(A+B+C)-A-2C#

#=180^@-60^@-2xx45^@=30^@#