Question

Question #81ed9

Answer 1

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`

Answer 2

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