A ladder leaning against a wall makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder.
Let AB be the ladder and CA be the wall.
Given: BC = 2.5 m, ∠ABC = 60°
AB = 5 cm and ∠BAC = 30°
From Pythagoras Theorem, we have
AB2 = BC2 + CA2
52 = (2.5)2 + (CA)2
(CA)2 = 25 – 6.25 = 18.75 m
Hence, length of the ladder is
For what value of k will k+9, 2k-1 and 2k+7 are the consecutive terms of an A.P?
If k+9, 2k-1 and 2k+7 are the consecutive terms of A.P, then the common difference will be the same.
∴ (2k – 1) – (k + 9) = (2k + 7) – (2k – 1)
∴ k – 10 = 8
∴ k = 18
A card is drawn at random from a well -shuffled fled pack of 52 playing cards. Find the probability of getting neither a red card nor a queen.
If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x)k = 0 has equal roots, find the value of k.
We Given -5 is a root of the quadratic equation 2x2+px-15 =0
, -5 satisfies the given equation.
∴ 2 (-5)2 +p(-5)-15=0
5p=35 ⇒ p=7
Substituting p=7 in (x2+x)+k =0, we get
The roots of the equation are equal
∴ Discriminant =b2-4ac =0
Here, a= 7, b=7,c=k
The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term.
We have given that
4th term of an A.P.= a4 = 0
∴ a + (4 – 1)d = 0
∴ a + 3d = 0
∴ a = –3d ….(1)
25th term of an A.P. = a25
= a + (25 – 1)d
= –3d + 24d ….[From the equation (1)]
3 times 11th term of an A.P. = 3a11
= 3[a + (11 – 1)d]
= 3[a + 10d]
= 3[–3d + 10d]
= 3 × 7d
∴ a25 = 3a11
i.e., the 25th term of the A.P. is three times its 11th term.
Prove that the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.
Given, the points (3, 0), (6, 4) and (-1, 3) are the vertices of a right-angled isosceles triangle.
Let A (3,0), B (6,4) and C (-1,3) be the given points.
Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.
P and Q are the points of trisection of AB,
Thus, P divides AB internally in the ratio 1:2 and Q divide AB internally in the ratio 2:1.
In Fig.2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.
In the Given figure, since tangents are drawn from an exterior point to a circle are equal in length,
AP = AS ….(1)
BP = BQ ….(2)
CR = CQ ….(3)
DR = DS ….(4)
Adding equations (1), (2), (3) and (4), we get
AP + BP + CR + DS = AS + BQ + CQ + DS
(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
AB + CD = AD + BC
AB + CD = BC + DA
In Fig. , PQ is tangent at point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.
In the given figure,
In Δ ACO,
OA=OC (Radii of the same circle)
ΔACO is an isosceles triangle.
∠CAB = 30° (Given)
∠CAO = ∠ACO = 30° (angles opposite to equal sides of an isosceles triangle are equal)
∠PCO = 90° …(radius is drawn at the point of
contact is perpendicular to the tangent)
Now ∠PCA = ∠PCO – ∠CAO
∠PCA = 90° – 30° = 60°
In Fig, from an external point P, two tangents PT and PS are drawn to a circle with centre O and radius r. If OP=2r, show that ∠ OTS = ∠ OST = 30°.
In the given figure,
∴∠OTS = ∠OST … (Angles opposite to equal sides of an isosceles triangle are equal)
In ΔOTQ and ΔOSQ
OS = OT … (Radii of the same circle)
OQ = OQ ...(side common to both triangles)
∠OTQ = ∠OSQ … (angles opposite to equal sides of an isosceles triangle are
∴ ΔOTQ ≅ ΔOSQ … (By S.A.S)
∴ ∠TOQ = ∠SOQ = 60° … (C.A.C.T)
∴ ∠TOS = 120° … (∠TOS = ∠TOQ + ∠SOQ = 60° + 60° = 120°)
∴ ∠OTS + ∠OST = 180° – 120° = 60°
∴ ∠OTS = ∠OST = 60° ÷ 2 = 30°