Question

paragraph \t\ for q.no. 11 and 12
consider an obtuse - angled triangle abc in which the difference between the largest and the smallest angle is $\frac{pi}{2}$ and whose sides are in arithmetic progression. suppose that the vertices of this triangle lie on a circle of radius 1.
(there are two questions based on paragraph \t\, the question given below is one of them)

1. let a be the area of the triangle abc. then the value of $(64a)^2$ is
2. then the in - radius of the triangle abc is

jee(advanced) 2023

Explanation:

1. For question 11:

• Explanation:
• Step 1: Let the angles and sides of the triangle
• Let the angles of the triangle be \(A,B,C\) such that \(A > B > C\). Given \(A - C=\frac{\pi}{2}\). Also, since the sides \(a,b,c\) are in arithmetic - progression, \(2b=a + c\). By the sine rule, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\), and since \(R = 1\), \(a = 2\sin A\), \(b = 2\sin B\), \(c = 2\sin C\). So, \(2\times2\sin B=2\sin A+2\sin C\), which simplifies to \(2\sin B=\sin A+\sin C\).
• We know that \(A + B + C=\pi\), and \(A - C=\frac{\pi}{2}\), so \(A=\frac{\pi}{2}+C\) and \(B=\pi-(A + C)=\frac{\pi}{2}-2C\).
• Substitute into \(2\sin B=\sin A+\sin C\): \(2\sin(\frac{\pi}{2}-2C)=\sin(\frac{\pi}{2}+C)+\sin C\).
• Using the trigonometric identities \(\sin(\frac{\pi}{2}-\alpha)=\cos\alpha\) and \(\sin(\frac{\pi}{2}+\alpha)=\cos\alpha\), we get \(2\cos2C=\cos C+\sin C\).
• Using the double - angle formula \(\cos2C=\cos^{2}C-\sin^{2}C=(\cos C+\sin C)(\cos C - \sin C)\), the equation becomes \(2(\cos C+\sin C)(\cos C - \sin C)-(\cos C+\sin C)=0\).
• Factor out \(\cos C+\sin C\): \((\cos C+\sin C)(2\cos C - 2\sin C - 1)=0\).
• Since \(A\) is obtuse, \(C\) is acute, so \(\cos C+\sin C

eq0\). Then \(2\cos C-2\sin C - 1 = 0\), or \(\cos C-\sin C=\frac{1}{2}\). Squaring both sides, \(\cos^{2}C - 2\sin C\cos C+\sin^{2}C=\frac{1}{4}\), and \(1-\sin2C=\frac{1}{4}\), so \(\sin2C=\frac{3}{4}\).

• Step 2: Find the area of the triangle
• The area of the triangle \(a=\frac{1}{2}ac\sin B\). Since \(a = 2\sin A\), \(c = 2\sin C\), and \(B=\frac{\pi}{2}-2C\), \(\sin B=\cos2C\).
• We know that \(\cos2C=\frac{\sqrt{7}}{4}\) (because \(C\) is acute). Also, \(a = 2\sin(\frac{\pi}{2}+C)=2\cos C\) and \(c = 2\sin C\).
• The area \(a=\frac{1}{2}(2\cos C)(2\sin C)\cos2C=\sin2C\cos2C\). Since \(\sin2C=\frac{3}{4}\) and \(\cos2C=\frac{\sqrt{7}}{4}\), \(a=\frac{3\sqrt{7}}{16}\).
• Step 3: Calculate \((64a)^{2}\)
• Substitute \(a=\frac{3\sqrt{7}}{16}\) into \((64a)^{2}\). \(64a = 64\times\frac{3\sqrt{7}}{16}=12\sqrt{7}\). Then \((64a)^{2}=(12\sqrt{7})^{2}=1008\).
• Answer: \(1008\)

1. For question 12:

• Explanation:
• Step 1: Recall the formula for in - radius
• The formula for the in - radius \(r\) of a triangle is \(r = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\). Since \(R = 1\), \(r = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\).
• We know that \(A=\frac{\pi}{2}+C\) and \(B=\frac{\pi}{2}-2C\).
• \(\sin\frac{A}{2}=\sin(\frac{\pi}{4}+\frac{C}{2})\), \(\sin\frac{B}{2}=\sin(\frac{\pi}{4}-C)\), \(\sin\frac{C}{2}\).
• First, from \(\cos C-\sin C=\frac{1}{2}\), we can find \(\sin C\) and \(\cos C\). Solving \(\cos C-\sin C=\frac{1}{2}\) gives \(\cos C=\frac{1 + \sqrt{7}}{4}\) and \(\sin C=\frac{\sqrt{7}-1}{4}\).
• \(\sin\frac{A}{2}=\sin(\frac{\pi}{4}+\frac{C}{2})=\frac{\sqrt{2}}{2}(\cos\frac{C}{2}+\sin\frac{C}{2})\), \(\sin\frac{B}{2}=\sin(\frac{\pi}{4}-C)=\frac{\sqrt{2}}{2}(\cos C-\sin C)\).
• Using the half - angle formulas \(\sin\frac{C}{2}=\sqrt{\frac{1 - \cos C}{2}}\) and \(\cos\frac{C}{2}=\sqrt{\frac{1+\cos C}{2}}\).
• After substituting the values of \(\sin C\) and \(\cos C\) and simplifying, we get \(r=\frac{\sqrt{7}-1}{4}\).
• Answer: \(\frac{\sqrt{7}-1}{4}\)

Answer:

1. For question 11:

• Explanation:
• Step 1: Let the angles and sides of the triangle
• Let the angles of the triangle be \(A,B,C\) such that \(A > B > C\). Given \(A - C=\frac{\pi}{2}\). Also, since the sides \(a,b,c\) are in arithmetic - progression, \(2b=a + c\). By the sine rule, \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R\), and since \(R = 1\), \(a = 2\sin A\), \(b = 2\sin B\), \(c = 2\sin C\). So, \(2\times2\sin B=2\sin A+2\sin C\), which simplifies to \(2\sin B=\sin A+\sin C\).
• We know that \(A + B + C=\pi\), and \(A - C=\frac{\pi}{2}\), so \(A=\frac{\pi}{2}+C\) and \(B=\pi-(A + C)=\frac{\pi}{2}-2C\).
• Substitute into \(2\sin B=\sin A+\sin C\): \(2\sin(\frac{\pi}{2}-2C)=\sin(\frac{\pi}{2}+C)+\sin C\).
• Using the trigonometric identities \(\sin(\frac{\pi}{2}-\alpha)=\cos\alpha\) and \(\sin(\frac{\pi}{2}+\alpha)=\cos\alpha\), we get \(2\cos2C=\cos C+\sin C\).
• Using the double - angle formula \(\cos2C=\cos^{2}C-\sin^{2}C=(\cos C+\sin C)(\cos C - \sin C)\), the equation becomes \(2(\cos C+\sin C)(\cos C - \sin C)-(\cos C+\sin C)=0\).
• Factor out \(\cos C+\sin C\): \((\cos C+\sin C)(2\cos C - 2\sin C - 1)=0\).
• Since \(A\) is obtuse, \(C\) is acute, so \(\cos C+\sin C

eq0\). Then \(2\cos C-2\sin C - 1 = 0\), or \(\cos C-\sin C=\frac{1}{2}\). Squaring both sides, \(\cos^{2}C - 2\sin C\cos C+\sin^{2}C=\frac{1}{4}\), and \(1-\sin2C=\frac{1}{4}\), so \(\sin2C=\frac{3}{4}\).

• Step 2: Find the area of the triangle
• The area of the triangle \(a=\frac{1}{2}ac\sin B\). Since \(a = 2\sin A\), \(c = 2\sin C\), and \(B=\frac{\pi}{2}-2C\), \(\sin B=\cos2C\).
• We know that \(\cos2C=\frac{\sqrt{7}}{4}\) (because \(C\) is acute). Also, \(a = 2\sin(\frac{\pi}{2}+C)=2\cos C\) and \(c = 2\sin C\).
• The area \(a=\frac{1}{2}(2\cos C)(2\sin C)\cos2C=\sin2C\cos2C\). Since \(\sin2C=\frac{3}{4}\) and \(\cos2C=\frac{\sqrt{7}}{4}\), \(a=\frac{3\sqrt{7}}{16}\).
• Step 3: Calculate \((64a)^{2}\)
• Substitute \(a=\frac{3\sqrt{7}}{16}\) into \((64a)^{2}\). \(64a = 64\times\frac{3\sqrt{7}}{16}=12\sqrt{7}\). Then \((64a)^{2}=(12\sqrt{7})^{2}=1008\).
• Answer: \(1008\)

1. For question 12:

• Explanation:
• Step 1: Recall the formula for in - radius
• The formula for the in - radius \(r\) of a triangle is \(r = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\). Since \(R = 1\), \(r = 4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\).
• We know that \(A=\frac{\pi}{2}+C\) and \(B=\frac{\pi}{2}-2C\).
• \(\sin\frac{A}{2}=\sin(\frac{\pi}{4}+\frac{C}{2})\), \(\sin\frac{B}{2}=\sin(\frac{\pi}{4}-C)\), \(\sin\frac{C}{2}\).
• First, from \(\cos C-\sin C=\frac{1}{2}\), we can find \(\sin C\) and \(\cos C\). Solving \(\cos C-\sin C=\frac{1}{2}\) gives \(\cos C=\frac{1 + \sqrt{7}}{4}\) and \(\sin C=\frac{\sqrt{7}-1}{4}\).
• \(\sin\frac{A}{2}=\sin(\frac{\pi}{4}+\frac{C}{2})=\frac{\sqrt{2}}{2}(\cos\frac{C}{2}+\sin\frac{C}{2})\), \(\sin\frac{B}{2}=\sin(\frac{\pi}{4}-C)=\frac{\sqrt{2}}{2}(\cos C-\sin C)\).
• Using the half - angle formulas \(\sin\frac{C}{2}=\sqrt{\frac{1 - \cos C}{2}}\) and \(\cos\frac{C}{2}=\sqrt{\frac{1+\cos C}{2}}\).
• After substituting the values of \(\sin C\) and \(\cos C\) and simplifying, we get \(r=\frac{\sqrt{7}-1}{4}\).
• Answer: \(\frac{\sqrt{7}-1}{4}\)