Consider the bigger triangle as ABC , such that, ∠ ABC = 90^∘ AC = 17 , AB = 8 and a line from A meets BC at D, BD = 6 Now, In AB ...

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Range of Trigonometric Ratios from 0 to 90 Degrees

sin x∘ is 8...

Question

$sin x^{\circ}$ is $\frac{8}{m}$ , m is

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Solution

Consider the bigger triangle as $△ A B C$ , such that, $∠ A B C = 90^{\circ}$
$A C = 17$ , $A B = 8$ and a line from A meets BC at D, $B D = 6$
Now, In $△ A B C$ ,
Using Pythagoras Theorem,
$A B^{2} + B C^{2} = A C^{2}$
$8^{2} + B C^{2} = 17^{2}$
$B C^{2} = 289 - 64$
$B C = 15$
Now, In $△ A B D$ ,
Using Pythagoras Theorem,
$A B^{2} + B D^{2} = A D^{2}$
$8^{2} + 6^{2} = A D^{2}$
$A D^{2} = 100$
$A D = 10$
$s i n x^{\circ} = \frac{P}{H} = \frac{A B}{A C} = \frac{8}{17}$

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