Question #81ed9

2 Answers
Jun 13, 2016

See below the answers.

Explanation:

#i)# #2 x cos(theta)=4->cos(theta)=2/x#
#ii)# #hat C = arctan((sqrt(15)x)/x) = arctan(sqrt(15))#
#iii)# #sin(theta)/x=sin(hat C)/4# but #sin(hat C) = (sqrt(15)x)/(4x)=sqrt(15)/4#

so #sin(theta) = sqrt(15)/16 x#
#iv)# #sin^2(theta)+cos^2(theta)=1->(sqrt(15)/16 x)^2+(2/x)^2=1#

Jun 13, 2016

Steps as below

Explanation:

(i) For a triangle with sides #a,b and c# and angles #A, B and C# the cosine rule states:
#a^2 = b^2 + c^2 - 2bc cos A#
Applying in #DeltaDBC# where #angle D=theta#

#x^2 = x^2 + 4^2 - 2xx x xx4 cos theta#, rearranging and solving for #cos theta#
#=>8x cos theta=16#
#=>cos theta=16/(8x)#
#=>cos theta=2/x#, Proved.
(ii) In #Delta ABC#
#sin AhatCB="perpendicular"/"hypotenuse"=(sqrt15 x)/(4x)=sqrt15/4#
(iii) For a triangle with sides #a,b and c# and angles #A, B and C# the sine rule states:
#sinA/a = sinB/b = sinC/c#
In #DeltaBCD# applying the law of sines for angles #D and C# and using value from (ii) we get

#sintheta/x = (sqrt15/4)/4#
#sintheta = (sqrt15 x)/16#
(iv) LHS #=sin^2theta+cos^2theta# Inserting values as obtained above

#=>LHS=((sqrt15 x)/16 )^2+(2/x)^2#
#=(15 x^2)/256 +4/x^2#, Setting it equal to RHS#=1# we obtain
#(15 x^2)/256 +4/x^2=1#
#=>(15 x^2)/256 +4/x^2-1=0#
Rearranging and multiplying both sides with LCM of all the denominators, we obtain

#256x^2[(15 x^2)/256 -1+4/x^2=0]#
#=>15x^4-256x^2+1024=0#, Proved