Question

यदि त्रिभुज एबीसी में कौन बी बराबर 90 डिग्री व 15 कोर्ट ए = 8 तो सीने क्या होगा​

Answer 1

Answer:

As /_B = 90°

$15cotA = 8 \\ = > cotA = \frac{8}{15} \\ \\ using \:identity \\ {cosec}^{2} A - {cot}^{2} A = 1 \\ {cosec}^{2} A - {( \frac{8}{15}) }^{2} = 1 \\ {cosec}^{2} A = 1 + \frac{64}{225} \\ {cosec}^{2} A = \frac{289}{225} \\ {cosec}^{2} A = {( \frac{17}{15} )}^{2} \\ cosecA \: = \frac{17}{15} \\ sinA = \frac{15}{17}$

Step-by-step explanation:

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Answer 2

Answer:

this is correct...........................