Question

Given ΔACB right angled at C in which AB = 29 units, BC = 21 units and ∠ABC = θ. Determine the value of
$co {s}^{2} θ−si {n}^{2} θ$


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Answer 1

$\underline {\purple {\mathrm{answer}}}$

$AB^2 = AC^2 + CB^2$

$= >  AC= \sqrt {AB^2 + CB^2} = \sqrt{ {29}^{2} - {21}^{2} }$

$\sqrt{(29 + 21)(29 - 21)} = \sqrt{400} = 20units$

$therefore \: \sin \theta = \frac{AC}{AB} = \frac{20}{29} and \: \cos \theta = \frac{BC}{AB}$

$(ii) We \: have,</p><p></p><p>$

${ \cos }^{2} \theta - { \sin}^{2} \theta = ( \frac{21}{29} {)}^{2} - ( \frac{20}{29} {)}^{2} = \frac{ {21}^{2} - {20}^{2} }{ {29}^{2} } = \frac{(21 + 20)(21 - 20)}{841} = \frac{41}{841}$

fig. in ATTACHMENT ✌️