Question #e8728

Part a) :

Let `AC=1` unit
`=> AB=cos25, BC=sin25, CD=tana=tanalpha`
Given Area `DeltaABC` = Area `DeltaACD`
`=> 1/2*AB*BC=1/2*AC*CD`
`=>cancel(1/2)*cos25*sin25=cancel(1/2)*1*tanalpha`
`=> tanalpha=cos25sin25`
`=> alpha=tan^-1(cos25sin25)=20.958^@`
`=> beta=90-20.958=69.042^@`

Part(b)

Let hypotenuse of the lower triangle be h . Then the opposite of `25^@` will be `=hsin25^@` and the adjacent will be `=hcos25^@`

For upper triangle
The hypotenuse `=hsecalpha`
and opposite to `alpha` is `=htanalpha`

By the condition of part(b) of the given problem the perimeter of both the triangle (upper and lower) are same. So we can say that the sum of other two sides excluding common side will be same for both the triangles.

Hence

`hsecalpha+htanalpha=hcos25^@+hsin25^@`

`=>secalpha+tanalpha=cos25^@+sin25^@=1.3289...(1)`

Now

`1.3289(secalpha-tanalpha)=sec^2alpha-tan^2alpha=1`

`=>secalpha-tanalpha=1/1.3289`

`=>secalpha-tanalpha=0.7525...(2)`

Adding (1) and (2) we get

`2secalpha=2.0814`

`=>secalpha=1.0407`

`=cosalpha=0.9609`

`=>alpha=cos^-1(0.9609)= 16.077^@`

And

`beta=90^@-alpha=73.923^@`

Part-(a)

By the given condition of part (a) of the question

Area of the upper triangle = Area of the lower triangle

`=>1/2xxhxxhtanalpha=1/2hcos25^@xxhsin25^@`

`tanalpha=1/2xx2cos25^@xxsin25^@=1/2xxsin50^@=0.3830`

`=>alpha =tan^-1(0.3830)=20.958^@`

Then `beta=90^@-20.95^@=69.042^@`