Question-216332

Question Number 216332 by mr W last updated on 04/Feb/25

Answered by cherokeesay last updated on 04/Feb/25

Answered by A5T last updated on 04/Feb/25

Commented by A5T last updated on 04/Feb/25

\frac{1}{2} \times 6 \times 10 \times \sin θ = \frac{6 \times 10 \times \sqrt{6^{2} + 10^{2} - 2 \times 6 \times 10 \cos θ}}{4 R}

\Rightarrow 2 Rsin θ = \sqrt{136 - 120 \cos θ} \Rightarrow R = \frac{\sqrt{34 - 30 \cos θ}}{\sin θ}

\Rightarrow R^{2} = \frac{34 - 30 \cos θ}{1 - \cos^{2} θ} \dots (i)

\frac{\sin (180 - θ)}{AB} = \frac{\sin 90}{10} \Rightarrow AB = 10 \sin θ

{(2 R)}^{2} = {(2 AB)}^{2} + 6^{2}

\Rightarrow R^{2} = {100 \sin}^{2} θ + 9 = 109 - {100 \cos}^{2} θ \dots (ii)

(i) & (ii) \Rightarrow 109 - {100 \cos}^{2} θ = \frac{34 - 30 \cos θ}{1 - \cos^{2} θ}

\Rightarrow \cos θ = \frac{3 \underline{+} \sqrt{209}}{20}

\Rightarrow R^{2} = 109 - \frac{109 \underline{+} 3 \sqrt{209}}{2} = \frac{109 \bar{+} 3 \sqrt{209}}{2}

\Rightarrow R = \sqrt{\frac{109 + 3 \sqrt{209}}{2}} = \frac{3 + \sqrt{209}}{2} \approx 8.728

Commented by mahdipoor last updated on 04/Feb/25

w h y ({(2 R)}^{2} = {(2 A B)}^{2} + 6^{2}) ?

Commented by A5T last updated on 04/Feb/25

The diameter, length 2 R, subtends an angle of

90 ° at the circumference .

Commented by A5T last updated on 04/Feb/25

Commented by A5T last updated on 04/Feb/25

CE = 2 R; AE = 2 AB; AC = 6

∠ CAE = 90 ° \Rightarrow {CE}^{2} = {AE}^{2} = {AC}^{2}

\Rightarrow {(2 R)}^{2} = {(2 AB)}^{2} + 6^{2}

Commented by mr W last updated on 04/Feb/25

��

Answered by mr W last updated on 04/Feb/25

Commented by mr W last updated on 04/Feb/25

R = x + 3

y^{2} = {(x + 3)}^{2} - 3^{2} = 10^{2} - x^{2}

x^{2} + 3 x - 50 = 0

\Rightarrow x = \frac{- 3 + \sqrt{209}}{2}

\Rightarrow R = \frac{- 3 + \sqrt{209}}{2} + 3 = \frac{3 + \sqrt{209}}{2} ✓

Answered by A5T last updated on 05/Feb/25

Commented by A5T last updated on 05/Feb/25

R = 3 + 10 \cos (180 - θ) = 3 - 10 \cos θ

AB = 10 \sin θ = 10 \sqrt{1 - \cos^{2} θ}

{(2 R)}^{2} = 6^{2} + {(2 AB)}^{2} \Rightarrow R^{2} = 3^{2} + {AB}^{2}

\Rightarrow {10 \cos}^{2} θ - 3 \cos θ - 5 = 0 \Rightarrow \cos θ = \frac{3 \underline{+} \sqrt{209}}{20}

\Rightarrow R = \frac{3 + \sqrt{209}}{2}

Facebook Tweet Pin