Question

in figure 7.49 Angle B is is greater than angle A and angle C is greater than angle d show that the angle ad is greater than AC​

Answer 1

Answer:

In △AOB,

∠B<∠A

∴AO<BO

∴ Side opposite to the greater angle is longer.

In △COD,

∠C<∠D⇒OD<OC

⇒AO+OD<OB+OC

∴AD<BC

Answer 2

ANSWER

ab<bc

Step-by-step explanation:

in triangle abo,

in triangle abo,angle b<angle a

in triangle abo,angle b<angle a =ao<bo___1

in triangle abo,angle b<angle a =ao<bo___1in triangle doc,

in triangle abo,angle b<angle a =ao<bo___1in triangle doc,angle c<angle d

in triangle abo,angle b<angle a =ao<bo___1in triangle doc,angle c<angle d=od<oc___2

in triangle abo,angle b<angle a =ao<bo___1in triangle doc,angle c<angle d=od<oc___2add 1 and 2

in triangle abo,angle b<angle a =ao<bo___1in triangle doc,angle c<angle d=od<oc___2add 1 and 2 ao+od<Bo+oc

in triangle abo,angle b<angle a =ao<bo___1in triangle doc,angle c<angle d=od<oc___2add 1 and 2 ao+od<Bo+ocad<bc

in triangle abo,angle b<angle a =ao<bo___1in triangle doc,angle c<angle d=od<oc___2add 1 and 2 ao+od<Bo+ocad<bc hence showed that ad<bc