In $Δ A B C, ∠ A = 45^{\circ}, ∠ B = 60^{\circ}$ and $∠ C = 75^{\circ}$ . The angles of the triangle formed by joining the mid – points of the sides of this triangles are:

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Solution

The correct option is B

The measures of the angles of the $Δ D E F$ formed by joining the mid-point points of the sides of the $Δ A B C$ is given as:

$∠ D = ∠ A,$ [Since AFDE is a parallelogram]
$∠ E = ∠ B$ [Since BFED is a parallelogram]
and $∠ F = ∠ C$ [Since CDFE is a parallelogram]
Hence,the angles of the triangle formed by joining the mid – points of the sides of this triangles are $45^{\circ}, 60^{\circ}, 75^{\circ} .$

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, and

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In ΔABC,∠A=45∘,∠B=60∘ and ∠C=75∘. The angles of the triangle formed by joining the mid – points of the sides of this triangles are:

In $Δ A B C, ∠ A = 45^{\circ}, ∠ B = 60^{\circ}$ and $∠ C = 75^{\circ}$ . The angles of the triangle formed by joining the mid – points of the sides of this triangles are: