In Delta ABC , m ∠ C =90 and cos B =frac{1}{2}, then operatorname{cosec} A =ldots ldots ldot

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8. Introduction to Trigonometry

easy

In $\Delta ABC , m \angle C =90$ and $\cos B =\frac{1}{2},$ then $\operatorname{cosec} A =\ldots \ldots \ldots \ldots$

A

$\frac{1}{2}$

B

$\sqrt{3}$

C

$\frac{2}{\sqrt{3}}$

D

$2$

Solution

$\cos B =\frac{ BC }{ AB }=\frac{1}{2}$

$\therefore BC =k$ and $AB =2 k$ (where, $k > 0$)

Now, $\operatorname{cosec} A=\frac{\text { hypotenuse }}{\text { side opposite to } \angle A }$

$=\frac{A B}{B C}=\frac{2 k}{k}=2$

Standard 10

Mathematics

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