Question

pls tell answer of this question​

Answer 1

FHG -70°

FHE-37°

I hope this is the answer

Answer 2

Given:

• EFGH is a rhombus
• Angle OGF = 37°
• Angle EFG = 70°

To find:

• Angle OFG
• Angle FHG

Solution:

The diagonals EG and FH divide the rhombus into four right angle triangles. Let us take ΔFGO.

Here, ∠FGO = 37° (given)

         ∠FOG = 90° (diagonals perpendicularly bisect each other in a rhombus)

The angle sum property of triangles states that the sum of all interior angles of a triangle is 180°. Therefore,

⇒ ∠FGO + ∠FOG + ∠OFG = 180

⇒ 37 + 90 + ∠OFG = 180

⇒ 127 + ∠OFG = 180

⇒ ∠OFG = 180 - 127

⇒ ∠OFG = 53°

Thus, the value of ∠OFG is 53 degrees.

Now, let us find ∠FHG.

We know that in rhombus, a parallelogram, opposite sides are parallel. Therefore, EF and HG, a pair of opposite sides in the rhombus, are parallel. That makes ∠FHG and ∠EFG a pair of alternate interior angles on the transversal FH. Alternate interior angles are equal. So ∠FHG = ∠EFG = 70°.

Thus, the value of ∠FHG = 70°.