Find cosec30° and cos60° geometrically
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Sia ? 5 years, 8 months ago
Let a triangle ABC with each side equal to 2a
{tex}\angle{/tex}A = {tex}\angle{/tex}B = {tex}\angle{/tex}C = 60°
Draw AD perpendicular to BC
In {tex}\triangle{/tex}BDA and {tex}\triangle{/tex}CDA
AB = AC, AD = AD and {tex}\angle BDA=\angle CDA{/tex}
{tex}\triangle B D A \cong \triangle C D A{/tex} by RHS
BD = CD
{tex}\angle{/tex}BAD = {tex}\angle{/tex}CAD = 30° by CPCT
In {tex}\triangle{/tex}BDA, {tex}\text{cosec} 30 ^ { \circ } =\frac HP= \frac { A B } { B D } = \frac { 2 a } { a } = 2{/tex} and {tex}\cos 60 ^ { \circ } =\frac BH= \frac { B D } { A B } = \frac { a } { 2 a } = \frac { 1 } { 2 }{/tex}
2Thank You