If cosec A = 10, then what is the value of 20 sin A + $9\sqrt{11}$ sec A ?
Given that A is an acute angle ?
Answer & explanation
Correct answer: option 3
cosec A = \(\frac{10 }{1}\)
{ cosec A = \(\frac{H}{P}\) }
P² + B² = H²
1² + B² = 10²
B = 3\(\sqrt {11 }\)
Now,
20 sinA + 9\(\sqrt {11 }\) secA
= 20 × \(\frac{P}{H}\) + 9\(\sqrt {11 }\)× \(\frac{H}{B}\)
= 20 × \(\frac{1}{10}\) + 9\(\sqrt {11 }\)× \(\frac{10}{3√11}\)
= 2 + 30
= 32