Solution: Given that cos(x) = -(5 13 \frac{5}{13} 13 5 ) = − 5 13 \frac{-5}{13} 13 − 5 and π / 2 ≤ x ≤ π \pi /2 \leq x \leq \pi π /2 ≤ x ≤ π
∴ \therefore ∴ x is in second quadrant.
{ we know that if cos(x) = b h \frac{b}{h} h b then sin(x) = a h \frac{a}{h} h a , where a a a = h 2 − b 2 \sqrt{h^2 - b^2} h 2 − b 2 & tan(x) = s i n ( x ) c o s ( x ) \frac{sin(x)}{cos(x)} cos ( x ) s in ( x ) }
therefore if we compare with the given cos(x) value we get:
b = -5 & h = 13
⟹ \implies ⟹ a a a = 1 3 2 − ( − 5 ) 2 \sqrt{13^2 - (-5)^2} 1 3 2 − ( − 5 ) 2
⟹ \implies ⟹ a a a = 169 − 25 \sqrt{169 - 25} 169 − 25
⟹ \implies ⟹ a a a = 144 \sqrt{144} 144
⟹ \implies ⟹ a a a = ± \pm ± 12
in the second quadrant value of sin(x) is always positive.
therefore, we take a a a = 12
sin(x) = a h \frac{a}{h} h a = 12 13 \frac{12}{13} 13 12
tan(x) = s i n ( x ) c o s ( x ) \frac{sin(x)}{cos(x)} cos ( x ) s in ( x ) = 12 / 13 − 5 / 13 \frac{12/13}{-5/13} − 5/13 12/13 = - 12 5 \frac{12}{5} 5 12
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