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Aseem Mahajan 4 years, 3 months ago
[tex]\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf \red{\angle A} & \red{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \red{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &\red{ \sf{90}^{ \circ}} \\ \hline \\ \rm \red{sin A} & \green{0} & \green{\dfrac{1}{2}}& \green{\dfrac{1}{ \sqrt{2} }} &\green{ \dfrac{ \sqrt{3}}{2} }&\green{1} \\ \hline \\ \rm \red{cos \: A} & \green{1} &\green{ \dfrac{ \sqrt{3} }{2}}&\green{ \dfrac{1}{ \sqrt{2} }} & \green{\dfrac{1}{2}} &\green{0} \\ \hline \\\rm \red{tan A}& \green{0} &\green{ \dfrac{1}{ \sqrt{3} }}&\green{1} & \green{\sqrt{3}} & \rm \green{\infty} \\ \hline \\ \rm \red{cosec A }& \rm \green{\infty} & \green{2}& \green{\sqrt{2} }&\green{ \dfrac{2}{ \sqrt{3} }}&\green{1} \\ \hline\\ \rm \red{sec A} & \green{1 }&\green{ \dfrac{2}{ \sqrt{3} }}& \green{\sqrt{2}} & \green{2} & \rm \green{\infty} \\ \hline \\ \rm \red{cot A }& \rm \green{\infty} & \green{\sqrt{3}}& \green{1} & \green{\dfrac{1}{ \sqrt{3} }} & \green{0}\end{array}}}}[/tex]
Aseem Mahajan 4 years, 3 months ago
$$secA=y$$
$$\sf \bf :\longmapsto \dfrac{1}{cosA} = y$$
$$\sf \bf :\longmapsto \dfrac{sinA}{cosA} = xy$$
$$\sf \bf :\longmapsto tanA =xy$$
Learn more:
$$\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf {\angle A} & \{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &{ \sf{90}^{ \circ}} \\ \hline \\ \rm {sin A} & \green{0} & {\dfrac{1}{2}}& {\dfrac{1}{ \sqrt{2} }} &{ \dfrac{ \sqrt{3}}{2} }&{1} \\ \hline \\ \rm {cos \: A} & {1} &{ \dfrac{ \sqrt{3} }{2}}&{ \dfrac{1}{ \sqrt{2} }} & {\dfrac{1}{2}} &{0} \\ \hline \\\rm {tan A}& {0} &{ \dfrac{1}{ \sqrt{3} }}&{1} & {\sqrt{3}} & \rm {\infty} \\ \hline \\ \rm {cosec A }& \rm {\infty} & {2}& {\sqrt{2} }&{ \dfrac{2}{ \sqrt{3} }}&{1} \\ \hline\\ \rm {sec A} & {1 }&{ \dfrac{2}{ \sqrt{3} }}& {\sqrt{2}} & {2} & \rm {\infty} \\ \hline \\ \rm {cot A }& \rm {\infty} & {\sqrt{3}}& {1} & {\dfrac{1}{ \sqrt{3} }} &{0}\end{array}}}}$$
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Aseem Mahajan 4 years, 3 months ago
$$\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf \red{\angle A} & \red{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \red{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &\red{ \sf{90}^{ \circ}} \\ \hline \\ \rm \red{sin A} & \green{0} & \green{\dfrac{1}{2}}& \green{\dfrac{1}{ \sqrt{2} }} &\green{ \dfrac{ \sqrt{3}}{2} }&\green{1} \\ \hline \\ \rm \red{cos \: A} & \green{1} &\green{ \dfrac{ \sqrt{3} }{2}}&\green{ \dfrac{1}{ \sqrt{2} }} & \green{\dfrac{1}{2}} &\green{0} \\ \hline \\\rm \red{tan A}& \green{0} &\green{ \dfrac{1}{ \sqrt{3} }}&\green{1} & \green{\sqrt{3}} & \rm \green{\infty} \\ \hline \\ \rm \red{cosec A }& \rm \green{\infty} & \green{2}& \green{\sqrt{2} }&\green{ \dfrac{2}{ \sqrt{3} }}&\green{1} \\ \hline\\ \rm \red{sec A} & \green{1 }&\green{ \dfrac{2}{ \sqrt{3} }}& \green{\sqrt{2}} & \green{2} & \rm \green{\infty} \\ \hline \\ \rm \red{cot A }& \rm \green{\infty} & \green{\sqrt{3}}& \green{1} & \green{\dfrac{1}{ \sqrt{3} }} & \green{0}\end{array}}}}$$
Aseem Mahajan 2 years, 5 months ago
$$\sf \bf :longmapsto cosA=\dfrac{8}{x}$$
Taking second equation,
$$\sf \bf :longmapsto 15cosecA = 8secA$$
$$\sf \bf :longmapsto 15\dfrac{1}{sinA} = 8\dfrac{1}{cosA}$$
$$\sf \bf :longmapsto tanA = \dfrac{15}{8}$$
Thus, you can see using a triangle that sides are respectively, 15k, 8k, 17k
$$\sf \bf :longmapsto cosA = \dfrac{8}{17}$$
Hence , x = 17
More information:
[tex]\sf \color{aqua}{Trigonometry\: Table}\\ \blue{\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \sf \red{\angle A} & \red{\sf{0}^{ \circ} }&\red{ \sf{30}^{ \circ} }& \red{\sf{45}^{ \circ} }& \red{\sf{60}^{ \circ}} &\red{ \sf{90}^{ \circ}} \\ \hline \\ \rm \red{sin A} & \green{0} & \green{\dfrac{1}{2}}& \green{\dfrac{1}{ \sqrt{2} }} &\green{ \dfrac{ \sqrt{3}}{2} }&\green{1} \\ \hline \\ \rm \red{cos \: A} & \green{1} &\green{ \dfrac{ \sqrt{3} }{2}}&\green{ \dfrac{1}{ \sqrt{2} }} & \green{\dfrac{1}{2}} &\green{0} \\ \hline \\\rm \red{tan A}& \green{0} &\green{ \dfrac{1}{ \sqrt{3} }}&\green{1} & \green{\sqrt{3}} & \rm \green{\infty} \\ \hline \\ \rm \red{cosec A }& \rm \green{\infty} & \green{2}& \green{\sqrt{2} }&\green{ \dfrac{2}{ \sqrt{3} }}&\green{1} \\ \hline\\ \rm \red{sec A} & \green{1 }&\green{ \dfrac{2}{ \sqrt{3} }}& \green{\sqrt{2}} & \green{2} & \rm \green{\infty} \\ \hline \\ \rm \red{cot A }& \rm \green{\infty} & \green{\sqrt{3}}& \green{1} & \green{\dfrac{1}{ \sqrt{3} }} & \green{0}\end{array}}}}[/tex]
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$$\sf \bf :\longmapsto P = \dfrac{5}{50} = \dfrac{1}{10}$$
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