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प्रश्न
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
पर्याय
70°, 70°, 40°
60°, 40°, 80°
30°, 40°, 110°
60°, 70°, 50°
उत्तर
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).
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