Question

help with this math problem​

Answer 1

★GIVEN:-

$\implies \: \sin\alpha = \frac{b}{ \sqrt{a {}^{2} + b {}^{2} } }$

★verified:-

$\implies \: b \: \cot \: \alpha = a$

★solution:-

we know that

$\implies \red{\sin \theta = \frac{b}{ \sqrt{a {}^{2} + b {}^{2} } } = \frac{p}{h} }$

we know that Pythagoras trangle rule

$\implies \star \pink{h {}^{2} = p {}^{2} + b {}^{2} } \\ \\ \\ \implies \star \red{b = \sqrt{h {}^{2} - p {}^{2} } }$

$\implies \red {b} = \sqrt{ (\sqrt{a {}^{2} + b {}^{2}) {}^{2} - b {}^{2} } } \\ \\ \\ \implies \red{b} = \sqrt{a {}^{2} + b {}^{2} - b {}^{2} } \\ \\ \\ \implies \: \red{b} = \sqrt{a {}^{2} } \\ \\ \\ \implies \red{b} = a$

we know that

$\implies \: \green{ \cot \theta = \frac{ \red{b}}{p} = \frac{a}{b} }$

$\implies \: \cot \alpha = \frac{a}{b}$

cross multiplie

$\implies \orange{b \: \cot \alpha = a } \: \: \: \: \: \: \: \: \: \: {(hence \: proved)}$

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note

$\implies \red{b} = base \\ \\ \\ \implies \: b = parpendicular$

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I hope it helps you

Answer 2

Question:

If $\displaystyle{\sf\:\sin\:\alpha\:=\:\dfrac{b}{\sqrt{a^2\:+\:b^2}}}$ then prove that $\displaystyle{\sf\:b\:\cot\:\alpha\:=\:a}$

Answer:

$\displaystyle{\boxed{\red{\sf\:b\:\cot\:\alpha\:=\:a\:}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\sin\:\alpha\:=\:\dfrac{b}{\sqrt{a^2\:+\:b^2}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\sin^2\:\alpha\:+\:\cos^2\:\alpha\:=\:1\:}}$

$\displaystyle{\implies\sf\:\cos^2\:\alpha\:=\:1\:-\:\sin^2\:\alpha}$

$\displaystyle{\implies\sf\:\cos^2\:\alpha\:=\:1\:-\:\left(\:\dfrac{b}{\sqrt{a^2\:+\:b^2}}\:\right)^2}$

$\displaystyle{\implies\sf\:\cos^2\:\alpha\:=\:1\:-\:\dfrac{b^2}{a^2\:+\:b^2}}$

$\displaystyle{\implies\sf\:\cos^2\:\alpha\:=\:\dfrac{a^2\:+\:b^2\:-\:b^2}{a^2\:+\:b^2}}$

$\displaystyle{\implies\sf\:\cos^2\:\alpha\:=\:\dfrac{a^2}{a^2\:+\:b^2}}$

$\displaystyle{\implies\boxed{\blue{\sf\:\cos\:\alpha\:=\:\dfrac{a}{\sqrt{a^2\:+\:b^2}}}}}$

Now, we know that,

$\displaystyle{\pink{\sf\:\cot\:\alpha\:=\:\dfrac{\cos\:\alpha}{\sin\:\alpha}}}$

$\displaystyle{\implies\sf\:\cot\:\alpha\:=\:\dfrac{\dfrac{a}{\sqrt{a^2\:+\:b^2}}}{\dfrac{b}{\sqrt{a^2\:+\:b^2}}}}$

$\displaystyle{\implies\sf\:\cot\:\alpha\:=\:\dfrac{a}{\cancel{\sqrt{a^2\:+\:b^2}}}\:\times\:\dfrac{\cancel{\sqrt{a^2\:+\:b^2}}}{b}}$

$\displaystyle{\implies\sf\:\cot\:\alpha\:=\:\dfrac{a}{b}}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:b\:\cot\:\alpha\:=\:a\:}}}}$

Hence proved!