In $Δ A B C$ , $∠ C = 30^{\circ}, ∠ B = 80^{\circ}, ∠ A = 70^{\circ}, A B = 12 c m$ as shown in the figure. Find the length of the side AC.

$⎡ ⎢ ⎣ \begin{matrix} s i n 30^{\circ} = 0.5 s i n 70^{\circ} = 0.93 s i n 80^{\circ} = 0.98 \end{matrix} ⎤ ⎥ ⎦$

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Solution

Draw diameter AD passing through centre O and connect DB

$∠ D = 30^{\circ}$ (Angle subtended by the same chord AB)

$∠ A B D = 90^{\circ}$ (Angle subtended by the diameter on the circumference)

In right - angled triangle ABD

$s i n 30^{\circ} = \frac{A B}{A D}$

$A D = \frac{A B}{s i n 30^{\circ}} = \frac{12}{0.5} = d = 24 c m . . . (1)$

In $Δ A B C,$ applying sine rule

$\frac{A C}{s i n 80^{\circ}} = \frac{A B}{s i n 30^{\circ}} = d = 24 c m$ ...( from (1))

$A C = 24 \times s i n 80^{\circ} = 23.63 c m$

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In $Δ A B C$ , $∠ C = 30^{\circ}, ∠ B = 80^{\circ}, ∠ A = 70^{\circ}, A B = 12 c m$ as shown in the figure. The length of the side AC is equal to ___

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and

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then,

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In ΔABC, ∠ C=30∘, ∠ B=80∘, ∠ A=70∘, AB=12 cm as shown in the figure. Find the length of the side AC. ⎡⎢⎣sin 30∘=0.5sin 70∘=0.93sin 80∘=0.98⎤⎥⎦

In $Δ A B C$ , $∠ C = 30^{\circ}, ∠ B = 80^{\circ}, ∠ A = 70^{\circ}, A B = 12 c m$ as shown in the figure. Find the length of the side AC.

$⎡ ⎢ ⎣ \begin{matrix} s i n 30^{\circ} = 0.5 s i n 70^{\circ} = 0.93 s i n 80^{\circ} = 0.98 \end{matrix} ⎤ ⎥ ⎦$