Pls, can anyone help me, getting problem solutions? When following details are given?: Let ABC be a triangle with side lengths a=b=1, c=x (unknown) and angles α=A=2π/5, β=B=2π/5, γ=C= π/5

1) Show that cos(2π/5)=x/2
2) Let D be a point on the edge of BC such that AD bisects the angle BAC. Show that AD and CD both have same length x
3) Using cosine rule in the triangle ABD, show that x2+x-1=0
4) Deduce that cos(2π/5) = (√5-1)/4

1 Answer
Barney V.
Apr 29, 2018

See explanation.

Explanation:

Convert radians to degrees:-

#:.(2pi)/5xx180/pi=360/5=72^@#

#:.pi/5xx180/pi=36^@#

#:.x/sinC=b/sinB#

#:.x/(sin36^@)=1/sin72^@#

#:.x=(1xxsin36^@)/sin72^@#

#:.x=0.618033988#

1.) cos#(2pi)/5=x/2#

#:.cos72^@=0.309016994#

#:.x/5=0.618033988/2=0.309016994#

:. L.H.S.=R.H.S.

2.)
in #:.triangle ACD(180^@-(36^@+36^@)=108^@=angleD)#

#:.(AD)/(sin36^@)=(AC)/sin108^@#

#:.AD=(1xxsin36^@)/sin108^@#

#:.AD=0.587785252/0.951056516#

#:.AD=0.618#

#:.(CD)/(sin36^@)=(AD)/(sin36^@)#

#:.CD=(0.618xxsin36^@)/sin36^@#

#:.CD=0.618#

#3.)#
#x^2+x-1=0#

#:.(0.618^2)+0.618-1)#

#:.=0.381924+0.618-1=-0.0=0#

#4.)#

#;.cos((2pi)/5)=(sqrt 5-1)/4#

#:.cos72^@=0.309016994#

#:.(sqrt 5-1)/4=(2.236067978-1)/4=0.309016994#

#:.L,H.S.-R.H.S.#

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