Question

In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are

Options

• 70°, 70°, 40°

• 60°, 40°, 80°

• 30°, 40°, 110°

• 60°, 70°, 50°

Answer 1

It is given that D, E and F be the mid-points of BC , CA and AB respectively.

Then,

DE || AB,EF || BC and DF || CA.

Now,DE || AB and transversal CB and CA intersect them at D and E respectively.

Therefore,

∠CDE = ∠B

∠CDE = 40° [∠B = 40° (Given)]

and ∠CED = ∠A

∠CED = 30° [∠A = 30°(Given)]

Similarly, EF || BC

Therefore,

∠AEF = ∠C

∠AEF = 110°[∠C = 110° (Given)]

and ∠AFE = ∠B

∠AFE = 40°  [∠B = 40° (Given)]

Similarly, DF || CA

Therefore,

∠BDF = ∠C

∠BDF = 110° [ ∠C = 110°(Given)]

and∠BFD  = ∠A

∠BFD = 30° [∠A = 30° (Given)]

Now BC is a straight line.

∠BDF +∠FDE +∠EDC = 180

      110° + FDE + 40° = 180°

             ∠FDE + 150° = 180°

                        ∠FDE = 30°

Similarly, ∠DEF = 40°

and ∠EFD = 110°

Hence the correct choice is (c).