First slide

Compound angles

Question

$A triangle ABC satisfies the relation 2 \sec 4 C + \sin^{2} 2 A +$ $\sqrt{sinB} = 0 and a point P is taken on the longest side of the$ $triangle such that it divides the side in the ratio 1 : 3 . Let Q$ $and R be the circumcentre and orthocentre of Δ ABC . IfPQ :$ $QR : RP = 1 : α : β, then the value of α^{2} + β^{2} .$

Moderate

A

9
B

8
C

6
D

7

Solution

$\begin{array}{l} Given 2 s e c 4 C + s i n^{2} 2 A + \sqrt{\sin B} = 0 \\ This is possible when the triangle ABC \\ is right angled Isosceles triangle with right angle at B \\ Since R is orthocenter, it is coincide with B \\ Since Q is circum center it is midpoint of AB \\ Suppose that QR = 2 x, A Q = 2 x \\ Since the point P divides the hypotenuse in the ration 1:3 \\ Hence, P Q = x \\ The triangle PQR is right angled triangle \\ Hence, (PR) = \sqrt{{(P Q)}^{2} + {(Q R)}^{2}} = \sqrt{4 x^{2} + x^{2}} = \sqrt{5} x \\ Therefore, PQ:QR:RP = 1 : 2 : \sqrt{5} \\ hence, α^{2} + β^{2} = 4 + 5 = 9 \end{array}$

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