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Q) Prove that: = 2 cosec Ans: Let’s start from LHS LHS = Since sec A = LHS = = = = = = = = = = = = 2 cosec = RHS …………… Hence ProvedÂ
Prove that root[(sec A – 1)/(sec A + 1)] + root[(sec A + 1)/(sec A – 1)] = 2 cosec A Read More »
Q) If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then prove that a2 + b2 = m2 + n2 Ans: Since a cos θ + b sin θ = m By squaring on both sides, we get: (a cos θ + b sin θ)2
Q) Using prime factorisation, find HCF and LCM of 96 and 120. Ans: By prime factorisation, we get: 96 = 25 x 3 and 120 = 23 x 3 x 5 ∴ LCM = 25 x 3 x 5 = 480 And HCF = 23 x3 = 24 Therefore, LCM of given numbers is 480
Using prime factorisation, find HCF and LCM of 96 and 120. Read More »
Q) Find the ratio in which y-axis divides the line segment joining the points (5, – 6) and (- 1, – 4). Ans: Let’s draw the diagram to solve: Let’s consider the coordinates of point P is (0,y) Also consider that the line AB is divided in ratio of m : n. By section formula,
Q) Point P(x, y) is equidistant from points A(5, 1) and B(1,5). Prove that x = y. Ans: Let’s draw the diagram to solve: Given that PAÂ = PB Hence, PA2Â = PB2 Â Â (x – 5)2Â + (y – 1)2 = (x – 1)2 + (y – 5)2 – 10 x – 2y = –
Point P(x, y) is equidistant from points A(5, 1) and B(1, 5). Prove that x = y. Read More »
Q) The line segment joining the points A(4,-5) and B(4,5) is divided by the point P such that AP : AB = 2 : 5. Find the coordinates of P. Ans: Let’s draw the diagram to solve: Given that  (AB = AP + PB) by cross multiplication, we get: 5 AP = 2 (AP
Q) A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article. Ans:Â The curved surface area of the
Q) In the given figure, Δ ABC and ΔDBC are on the same base BC. If AD intersects BC at O, prove that Ans: Let’s draw perpendicular from points A and D on line BC: In Δ AON and Δ DOM, ∠AON = ∠DOM (interior angles) ∠ANO = ∠DMO (given that AN and
Q) In a Δ PQR, N is a point on PR, such that QN is to PR. If PN x NR = QN2, prove that ∠PQR = 900. Ans: Ans: Given that, PN x NR = QN2 Therefore, In Δ PNQ and Δ QNR, ∠QNP = ∠RNQ  (given that QN is to
Q) From an external point, two tangents are drawn to a circle. Prove that the line joining the external point to the centre of the circle bisects the angle between the two tangents. Ans: Step 1: Let’s draw a diagram with a circle of radius r and O as centre. Let the two tangents RP