Question

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Answer 1

Solution -

Base = Side adjacent to the angle

Perpendicular = Side opposite to the angle

a) i) cos B =  $\frac{Base}{Hypotenuse}$

 For angle B, Base = 8, Hypotenuse = 17

 cos B =  $\frac{8}{17}$

ii) tan C = $\frac{Perpendicular}{Base}$

  For angle C, Perpendicular = 8

  Base = AC = $\sqrt{(17)^{2} - (8)^{2} } = \sqrt{225} = 15$

  tan C = $\frac{8}{15}$

iii) sinB. cosC + cosB.sinC = 1

    cos B = $\frac{8}{17}$,  sin B = $\frac{15}{17}$

    cos C = $\frac{15}{17}$,  sin C = $\frac{8}{17}$

    Putting the values in above equation we get,

    $(\frac{15}{17}) (\frac{15}{17} ) + (\frac{8}{17})(\frac{8}{17} ) = 1 \\\\ (\frac{15}{17})^{2} + (\frac{8}{17})^{2} = 1 \\ (\frac{225+64}{289} ) = 1 \\ (\frac{289}{289} ) = 1\\1 = 1$

    Hence proved.