ABCD is a rhombus in which ∠BAC = 50°, then 3∠D - 2∠DAC is equal to?

Question

ABCD is a rhombus in which ∠BAC = 50°, then 3∠D - 2∠DAC is equal to?

120º
140º
150º
160º

LohaHridoy

4 views

2025-01-04 04:42

1 Answers

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AhmedNazir · Accepted Answer · 2022-09-04T04:29:57+06:00

Option 2 : 140º

Given:

ABCD is a rhombus.

⇒ ∠BAC = 50°

Concept used:

Rhombus property,

Diagonal bisects.

All sides are equal.

Angle sum in quadrilateral = 360º

Calculation:

[ alt="6247ee50f744d65adef771bc 16551960948901" src="//storage.googleapis.com/tb-img/production/22/06/6247ee50f744d65adef771bc_16551960948901.png" style="height: 100px; width: 108px;">

ABCD is rhombus.

AB = BC = CD = DA

Let ∠ ABC = ∠ ADC = y

Let ∠ BAD = ∠ BCD = 50º + x

AB II CD, AD II BC

⇒ ∠DAC = ∠ BCA = 50º

⇒ ∠ ABC + ∠ BCD + ∠ CDA + ∠ DAB =

⇒ 360º

⇒ y + 50º +x +y + 50º + x = 360º

⇒ 2x + 2y = 260º

x + y = 130º

Diagonal bisects the angle in rhombus,

x = 50º , y = 130º- 50º = 80º

⇒ 3 ∠ D - 2 ∠ DAC = 3 × 80º - 2 × 50º

⇒ 140º

∴ Option 2 is correct.