If αsin A = 5 and 7cosecA = 6 sec A, then the value of α is
(a) 46/45
(b) 46/7
(c)49/46
(d) None of these
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→ ɑ * sin A = 5
→ sin A = 5 / ɑ = P/H
So,
→ B = √(H² - P²) { By pythagoras theorem }
→ B = √(ɑ² - 25)
then,
→ tan A = P/B
→ tan A = 5/√(ɑ² - 25) ----------- Eqn.(1)
now,
→ 7•cosec A = 6•sec A
→ 7/6 = sec A / cosec A
→ 7/6 = (1/cos A)/(1/sin A)
→ 7/6 = (sin A/cos A)
→ 7/6 = tan A -------------- Eqn.(2)
from Eqn.(1) and Eqn.(2) we get,
→ 5/√(ɑ² - 25) = 7/6
→ 7•√(ɑ² - 25) = 5 • 6
→ √(ɑ² - 25) = 30/7
squaring both sides ,
→ (ɑ² - 25) = 900/49
→ ɑ² = (900/49) + 25
→ ɑ² = (900 + 1225)/49
→ ɑ² = (2125/49)
→ ɑ² = (46.09/7)²
→ ɑ = (46.09/7)
→ ɑ ≈ (46/7) (b) (Ans.)
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