If ( x + A )a = ( x + 2A )b = ( x + 3A )c and A = 60°, x = 15°, then the value of ( a + cb )^2 + ( a - cb )^2 is . . . . . . . .

Option B. 4. Answer & Solution Answer: Option B Solution: & ( x + A )a = ( x + 2A )b = ( x + 3A )c & A = 60^ ,\,\,x = 15^ given, & ( x + A )a = ( x + 2A )b & ( 15^ + 60^ ) ( 15^ + 120^ ) = a/b & 75^ 135^ = a/b & 15^ - 45^ = a/b & a^2b^2 = ^215^ 1/2 & a^2b^2 = 2 ^215^ ........ ( i ) & ( x + 2…

If ( x + A )a = ( x + 2A )b = ( x + 3A )c and A = 60°, x = 15°, then…

Answer: Option B

Solution:

\frac{cos ( x + A )}{a} = \frac{cos ( x + 2 A )}{b} = \frac{cos ( x + 3 A )}{c} A = 6 0^{\circ}, x = 1 5^{\circ} given, \Rightarrow \frac{cos ( x + A )}{a} = \frac{cos ( x + 2 A )}{b} \Rightarrow \frac{cos ( 15 ^{\circ} + 60 ^{\circ} )}{cos ( 15 ^{\circ} + 120 ^{\circ} )} = \frac{a}{b} \Rightarrow \frac{cos 75 ^{\circ}}{cos 135 ^{\circ}} = \frac{a}{b} \Rightarrow \frac{sin 15 ^{\circ}}{- sin 45 ^{\circ}} = \frac{a}{b} \Rightarrow \frac{a ^{2}}{b ^{2}} = \frac{sin ^{2} 15 ^{\circ}}{\frac{1}{2}} \Rightarrow \frac{a ^{2}}{b ^{2}} = 2 sin^{2} 1 5^{\circ} ........ (i) \Rightarrow \frac{cos ( x + 2 A )}{b} = \frac{cos ( x + 3 A )}{c} \Rightarrow \frac{cos ( 15 ^{\circ} + 120 ^{\circ} )}{cos ( 15 ^{\circ} + 180 ^{\circ} )} = \frac{b}{c} \Rightarrow \frac{- sin 45 ^{\circ}}{- cos 15 ^{\circ}} = \frac{b}{c} \Rightarrow \frac{c ^{2}}{b ^{2}} = \frac{cos ^{2} 15 ^{\circ}}{\frac{1}{2}} \Rightarrow \frac{c ^{2}}{b ^{2}} = 2 cos^{2} 1 5^{\circ} ........ (ii) Now, (\frac{a}{b} + \frac{c}{b})^{2} + (\frac{a}{b} - \frac{c}{b})^{2} = 2 [\frac{a ^{2}}{b ^{2}} + \frac{c ^{2}}{b ^{2}}] = 2 [2 sin^{2} 15^{\circ} + 2 cos^{2} 15^{\circ}] = 4