Question
Download Solution PDFIn Δ ABC ∠C = 90°, AC = 2√3 cm and BC = 1cm. What is the value of (13sin2 A + 6sec2 A + cosec2 A)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
ΔABC, ∠C = 90°
AC = 2√3 cm
BC = 1 cm
Formula used:
Pythagoras theorem: AB2 = AC2 + BC2
sin A = Opposite side / Hypotenuse = BC / AB
sec A = Hypotenuse / Adjacent side = AB / AC
cosec A = Hypotenuse / Opposite side = AB / BC
Calculation:
AB using Pythagoras theorem:
AB2 = (2√3)2 + 12
AB2 = 12 + 1 = 13
AB = √13 cm
Now,
sin A = BC / AB = 1 / √13
sec A = AB / AC = √13 / (2√3)
cosec A = AB / BC = √13 / 1 = √13
Substitute the values in the given expression:
13sin2 A + 6sec2 A + cosec2 A
⇒ 13(1/√13)2 + 6(√13 / (2√3))2 + (√13)2
⇒ 13(1/13) + 6(13/12) + 13
⇒ 1 + 13/2 + 13
⇒ 14 + 6.5 = 20.5 = 41/2 = \(20 \frac{1}{2}\)
∴ The value of (13sin2 A + 6sec2 A + cosec2 A) is \(20 \frac{1}{2}\).
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