Question

In angle ABC AC=10 and angle C =30°. Find AB and BC by trigonometry equation?

Answer 1

${\huge{\red{\sf{Given}}}}\begin{cases}\leadsto \bf{A\triangle ABC} \\\leadsto \bf{Measure\:of\:angle\:C\:is\:30\degree}\\\leadsto\bf{Measure\:of\:AC\:is\:10cm}\end{cases}$

${\huge{\red{\sf{To\:Find}}}}\begin{cases}\leadsto \bf{Measure\:of\:AB}\\\leadsto\bf{Measure\:of\:BC}\end{cases}$

$\huge\red{\underline{\bf{\green{Answer}}}}$

$\sf{\red{Let\:the\:right\:angled\:triangle\:be\:\triangle ABC. }}$

$\sf{\blue{\underline{So,\:in\:\triangle ABC}}}$

$\sf{\purple{sinC =\dfrac{perpendicular}{hypotenuse}=\dfrac{AB}{AC}}}$$\sf{\pink{cosC =\dfrac{base}{hypotenuse}=\dfrac{BC}{AC}}}$

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$\bf{\underline{\underline{\orange{So,\:using\:above\:equations}}}}$

$\sf{\implies sinC=\dfrac{AB}{AC}=\dfrac{AB}{10cm}}$

$\sf{\implies sin30\degree =\dfrac{AB}{10cm}}$

$\sf{\implies \dfrac{1}{2}=\dfrac{AB}{10cm}}$

$\large\red{\boxed{\green{\bf{sin30\degree=\dfrac{1}{2}}}}}$

$\sf{\implies AB=\dfrac{10cm}{2}}$

$\sf{\implies AB=\dfrac{\cancel{10}^{5}cm}{\cancel{2}}}$

${\underline{\boxed{\pink{\sf{\longmapsto AB=5cm}}}}}$

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$\sf{\underline{\underline{\green{Also,}}}}$

$\sf{\implies cosC=\dfrac{BC}{AC}}$

$\sf{\implies cos30\degree =\dfrac{BC}{10cm}}$

$\sf{\implies \dfrac{\sqrt{3}}{2}=\dfrac{BC}{10cm}}$

$\large\red{\boxed{\green{\bf{cos30\degree=\dfrac{\sqrt{3}}{2}}}}}$

$\sf{\implies BC=\dfrac{10cm\times \sqrt{3}}{2}}$

$\sf{\implies BC=\dfrac{\cancel{10}^{5}cm\times \sqrt{3}}{\cancel{2}}}$

${\underline{\boxed{\pink{\sf{\longmapsto BC=5\sqrt{3}}}}}}$

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${\sf{\underline{\underline{\red{So,}}}}}$

${\huge{\red{\sf{Answer:}}}}\begin{cases}\leadsto \bf{AB=5cm}\\\leadsto \bf{BC=5\sqrt{3}cm }\end{cases}$