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Chapter 8: Problem 685

In a \(\triangle \mathrm{ABC}\) angles \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) are in increasing A. P. and \(\sin (B+2 C)=[(-1) / 2]\) then \(A=\) (A) \((3 \pi / 4)\) (B) \((\pi / 4)\) (C) \((5 \pi / 6)\) (D) \((\pi / 6)\)

Short Answer

Expert verified
Angle A, denoted by α, is equal to \(\frac{2}{3}\)·\(\pi\) or (C) \(\frac{5}{6}\)·\(\pi\).

Step by step solution

01

Write the geometrical conditions

All angles of a triangle add up to π radians or 180°. So, we have: α + β + γ = π
02

Find the relationship between the angles

Since angles A, B, and C are in increasing arithmetic progression, α < β < γ We can also find the common difference d: β = α + d γ = β + d = α + 2d
03

Express sin(B + 2C) in terms of angles α and d

The given condition is: sin(β + 2γ) = -\(\frac{1}{2}\) Plugging in the expressions for β and γ in the equation above, we get: sin(α + d + 2(α+2d)) = -\(\frac{1}{2}\) sin(3α + 5d) = -\(\frac{1}{2}\)
04

Find the range of sin function for given output

Since sin(3α + 5d) is negative, we know sin function is negative in the third and fourth quadrants i.e., \(\frac{3\pi}{2}\) < 3α + 5d < 2\(\pi\) or \(\frac{3}{2}\)·\(\pi\)< 3α + 5d < 2·\(\pi\)
05

Use the sine arcsin function to find the angle

As sin(3α + 5d) = -\(\frac{1}{2}\), we have: 3α + 5d = arcsin\(\left(-\frac{1}{2}\right)\) arcsin\(\left(-\frac{1}{2}\right)\) = -\(\frac{1}{6}\)·\(\pi\) So, 3α + 5d = -\(\frac{1}{6}\)·\(\pi\)
06

Solve for α in terms of the common difference d

We have the equation, 3α + 5 d = -\(\frac{1}{6}\)·\(\pi\) Using equation α + β + γ = π and substituting values of β and γ we get, α + α + d+ α + 2d = π => α = \(\frac{1}{3}\)·\(\pi\) - d We can substitute this value of α in the equation 3α + 5d = -\(\frac{1}{6}\)·\(\pi\): 3 * (\(\frac{1}{3}\)·\(\pi\) - d) + 5d = -\(\frac{1}{6}\)·\(\pi\) \(\pi\) - 3d + 5d = -\(\frac{1}{3}\)·\(\pi\) => 2d = -\(\frac{2}{3}\)·\(\pi\) => d = -\(\frac{1}{3}\)·\(\pi\)
07

Solve for α

We can substitute the value of d back into the formula for α: α = \(\frac{1}{3}\)·\(\pi\) - (-\(\frac{1}{3}\)·\(\pi\)) α = \(\frac{2}{3}\)·\(\pi\) So, α(Angle A) = \(\frac{2}{3}\)·\(\pi\) or (C) \(\frac{5}{6}\)·\(\pi\)

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If \((1 / a),(1 / H),(1 / b)\) are in A. P. then \([(H+a) /(H-a)]\) \(+[(\mathrm{H}+\mathrm{b}) /(\mathrm{H}-\mathrm{b})]=\) (A) 2 (B) 4 (C) 0 (D) 1
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