If displaystyle sin -2-A-sin -2-B-sin -2-C-9-4- thendisplaystyle angle A-angle B-angle Cdisplaystyle angle A-angle B-angle C-0displaystyle angle A-angle B-angle C-60-circ-displaystyle angle A-angle B-90-circ-angle C

Question

Toppr Admin · Accepted Answer

Given equation-xA0-sin2A-sin2B-sin2C-94Put -x2220-A-x2220-B-x2220-C-60-x2218-sin260-sin260-sin260-x27F9-3-xD7-x221A-32-2-x27F9-3-xD7-34-94-x2234-x2220-A-x2220-B-x2220-C-60-x2218-

If ${sin}^{2} A + {sin}^{2} B + {sin}^{2} C = 9 / 4,$ then
$∠ A + ∠ B = ∠ C$
$∠ A + ∠ B + ∠ C = 0$
$∠ A = ∠ B = ∠ C = 60^{\circ}$
$∠ A + ∠ B = 90^{\circ} - ∠ C$

Given equation

${sin}^{2} A + {sin}^{2} B + {sin}^{2} C = \frac{9}{4}$

Put $∠ A = ∠ B = ∠ C = 60^{\circ}$

${sin}^{2} 60 + {sin}^{2} 60 + {sin}^{2} 60$

$⟹ 3 \times {(\frac{\sqrt{3}}{2})}^{2}$

$⟹ 3 \times \frac{3}{4} = \frac{9}{4}$

$∴ ∠ A = ∠ B = ∠ C = 60^{\circ}$

If sin2A+sin2B+sin2C=9/4, then∠A+∠B=∠C∠A+∠B+∠C=0∠A=∠B=∠C=60∘∠A+∠B=90∘−∠C

Given equation sin2A+sin2B+sin2C=94Put ∠A=∠B=∠C=60∘sin260+sin260+sin260⟹3×(√32)2⟹3×34=94∴∠A=∠B=∠C=60∘

If ${sin}^{2} A + {sin}^{2} B + {sin}^{2} C = 9 / 4,$ then
$∠ A + ∠ B = ∠ C$
$∠ A + ∠ B + ∠ C = 0$
$∠ A = ∠ B = ∠ C = 60^{\circ}$
$∠ A + ∠ B = 90^{\circ} - ∠ C$

Given equation

${sin}^{2} A + {sin}^{2} B + {sin}^{2} C = \frac{9}{4}$

Put $∠ A = ∠ B = ∠ C = 60^{\circ}$

${sin}^{2} 60 + {sin}^{2} 60 + {sin}^{2} 60$

$⟹ 3 \times {(\frac{\sqrt{3}}{2})}^{2}$

$⟹ 3 \times \frac{3}{4} = \frac{9}{4}$

$∴ ∠ A = ∠ B = ∠ C = 60^{\circ}$