Question

In the adjoining figure which is not drawn accurately, angle BAD is equal to 59°, angle DAC=32° and AD=BD. Calculate the value of angle ACB and state, giving Reasons, which is greater BD or DC.

Answer 1

hey your ans is here.abc+bad=90°ad=bd/ac but bd= ad so sin90-ad=1 so dc greater than dc

Answer 2

In ΔABD,

AB=59°          (side of an isosceles triangle in which ∠ABD=∠BAD)

∠ADB=180-(∠ABD+∠BAD)          (angle of a triangle)

      =180°-(59°+59°)

      =62°

In ΔADC,

∠ADC=180°-∠ADB

       =180°-62°

       =118°

∠ACD=180°-(∠ADC+∠DAC)

       =180°-(118°+32°)

       =30°

∠ACB=30°

In order to compare two angles, you first need to know what is Inequalities Theorem-

"Inequalities Theorem states that a greater angle has a greater side opposite to it and vice versa."

Now, in ΔABD,

∠BAD=59°

∠ABD=59°

∠BAD=∠ABD

BD=AD (by Inequalities Theorem)                               ...I

In ΔADC

∠ACD=30°

∠DAC=32°

∠ACD<∠DAC

AD<DC (by Inequalities Theorem)                              ...II

Comparing I and II, we have,

BD<DC

DC is greater than BD.