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8. Introduction to Trigonometry
medium
In $\Delta ABC , AC =5, BC =13, m \angle A =90,$ then $\tan B =\ldots \ldots \ldots \ldots$
A
$\frac{5}{13}$
B
$\frac{5}{12}$
C
$\frac{12}{13}$
D
$\frac{12}{5}$
Solution
In $\Delta ABC $, $AC =5,$ hypotenuse $BC =13$, $m \angle A =90$
$\therefore AB ^{2}= BC ^{2}- AC ^{2}$
$=13^{2}-5^{2}=169-25=144=(12)^{2}$
$\therefore A B=12$
Now, $\tan B =\frac{\text { side opposite to } \angle B }{\text { side adjacent to } \angle B }=\frac{ AC }{ AB }=\frac{5}{12}$
Standard 10
Mathematics
