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Circular Measurement of Angle

Find :i BCii ...

Question

Find :

(i) BC
(ii) AD
(iii) AC

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Solution

(i) In right ∆ABC, we have
tan $30^{\circ} = \frac{A B}{B C} \frac{1}{\sqrt{3}} = \frac{12}{B C}$
BC = 12 $\sqrt{3}$ cm
(ii) In right ∆ABD, we have
cos A = $\frac{A D}{A B}$
cos $60^{\circ} = \frac{A D}{A B} \frac{1}{2} = \frac{A D}{12} A D = \frac{12}{2}$
AD = 6 cm
(iii) In right ∆ABC, we have
$sin C = \frac{A B}{A C} sin 30^{\circ} = \frac{A B}{A C} \frac{1}{2} = \frac{12}{A C} A C = 12 \times 2 A C = 24 c m$

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(i) {AC}^{2} = {AD}^{2} + BC . DL + {(\frac{BC}{2})}^{2} (ii) {AB}^{2} = {AD}^{2} - BC . DL + {(\frac{BC}{2})}^{2} (iii) {AB}^{2} + {AC}^{2} = 2 {AD}^{2} - \frac{1}{4} {BC}^{2}

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