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Chapter 3: Problem 33

In an acute angled triangle \(A B C\) if \(\cos A, 1-\cos B\), \(\cos C\) are in A.P. and \(\sin A+\sin C=1\), then \(\angle B\) is equal to (a) \(\frac{\pi}{4}\) (b) \(\frac{\pi}{3}\) (c) \(\frac{\pi}{6}\) (d) None of these

Short Answer

Expert verified
The value of \(\angle B\) is (b) \(\frac{\pi}{3}\).

Step by step solution

01

Write the arithmetic progression relation for cosines

Since \(\cos A\), \(1 - \cos B\), and \(\cos C\) are in an arithmetic progression, we can write \(1 - \cos B = \frac{1}{2}(\cos A + \cos C)\). Now let's express \(\sin A\) and \(\sin C\) in terms of \(\cos A\) and \(\cos C\) using the given relation \(\sin A + \sin C = 1\).
02

Express sin A and sin C in terms of cosines

The sine identity sin^2(x) + cos^2(x) = 1 can be used to replace sin A and sin C with their respective cosines. Thus, we have: \(\sin A = \sqrt{1 - \cos^2 A}\) \(\sin C = \sqrt{1 - \cos^2 C}\) From the given relation, \(\sin A + \sin C = 1\), substitute the expressions above: \[\sqrt{1 - \cos^2 A} + \sqrt{1 - \cos^2 C} = 1\]
03

Square the equation and simplify

Squaring both sides gives: \[(1 - \cos^2 A) + 2\sqrt{(1 - \cos^2 A)(1 - \cos^2 C)} + (1 - \cos^2 C) = 1\] This simplifies to: \[2\sqrt{(1 - \cos^2 A)(1 - \cos^2 C)} = 2(\cos^2 A + \cos^2 C - \cos^2 A \cos^2 C)\]
04

Use the AP relation and solve for angle B

Divide both sides of the equation by 2 and then square root: \[\sqrt{(1 - \cos^2 A)(1 - \cos^2 C)} = \cos^2 A + \cos^2 C - \cos^2 A \cos^2 C\] Now use the relation from Step 1: \(1 - \cos B = \frac{1}{2}(\cos A + \cos C)\), to replace \(\cos A + \cos C\): \[(1 - \cos B)^2 = \frac{1}{4}(1 - \cos B)^2 \implies 1 - \cos B = \frac{1}{2}\] From this, we get \(\cos B = \frac{1}{2}\) Since \(\angle B\) is acute, we can determine that \(\angle B = \frac{\pi}{3}\). So, the answer is (b) \(\frac{\pi}{3}\).

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Most popular questions from this chapter

If \(\mathrm{P}\) is a point on the altitude \(\mathrm{AL}\) of \(\Delta A B C\) such that \(\angle \mathrm{PBC}=B / 3\), then the relation between \(\mathrm{Ap}(=\mathrm{x})\) and \(\mathrm{BP}(=\mathrm{y})\) is (a) \(x^{2}+y^{2}=2 c^{2}\) (b) \(x^{2}+c y=c^{2}\) (c) \(x^{2}+c y^{2}=c^{2}\) (d) None of these

In a \(\Delta A B C, a, c, A\) are given and \(b_{1}, b_{2}\) are two values of the third side \(b\) such that \(b_{2}=2 b_{1}\). Then \(\sin A=\) (a) \(\sqrt{\frac{9 a^{2}-c^{2}}{8 a^{2}}}\) (b) \(\sqrt{\frac{9 a^{2}-c^{2}}{8 c^{2}}}\) (c) \(\sqrt{\frac{9 a^{2}+c^{2}}{8 a^{2}}}\) (d) None of these

In a \(\Delta A B C, a, b, A\) are given and \(c_{1}, c_{2}\) are two values of the third side \(c\). The sum of the areas of two triangles with sides \(a, b, c_{1}\) and \(a, b, c_{2}\) is (a) \((1 / 2) b^{2} \sin 2 A\) (b) (1/2) \(a^{2} \sin 2 A\) (c) \(b^{2} \sin 2 A\) (d) None of these

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