In Fig. 1, PA and PB are tangents to the circle with centre O
such that ∠APB = 50°. Write the measure of ∠OAB.
Figure 1 
Answer :
PA and PB are tangents drawn from an external point P to the circle.
∴ PA = PB (Length of tangents drawn from an external point to the circle are equal.)
In ΔPAB,
PA = PB
⇒ $\angle $PBA = $\angle $PAB .....(1) (Angles opposite to equal sides are equal.)
Now,
$\angle $APB + $\angle $PBA + $\angle $PAB = 180°
⇒ 50º + $\angle $PAB + $\angle $PAB = 180° [Using (1)]
⇒ $\angle $PAB = 130°
⇒ $\angle $PAB = $\frac{130\xb0}{2}$ = 65°
We know that radius is perpendicular to the tangent at the point of contact.
∴ $\angle $OAP = 90° (OA ⊥ PA)
⇒ $\angle $PAB + $\angle $OAB = 90°
⇒ 65° + $\angle $OAB = 90°
⇒$\angle $OAB = 90° − 65° = 25°
Hence, the measure of $\angle $OAB is 25°.
If $x=\frac{1}{2},$ is a solution of the quadratic equation $3{x}^{2}+2kx3=0,$ find the value of k.
Answer :
It is given that $x=\frac{1}{2}$ is the solution of the quadratic equation $3{x}^{2}+2kx3=0$.
$\therefore 3{\left(\frac{1}{2}\right)}^{2}+2k\left(\frac{1}{2}\right)3=0$
$\Rightarrow \frac{3}{4}k3=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{3}{4}3=\frac{9}{4}$
Hence, the value of k is $\frac{9}{4}$.
The tops of two towers of height x and y, standing on level ground, subtend angles of 30° and 60° respectively at the centre of the line joining their feet, then find x, y.
Answer :
Let AB and CD be the two towers of heights x and y, respectively.
Suppose E is the centre of the line joining the feet of the two towers i.e. BD.
Now, in ΔABE,
$\frac{\mathrm{AB}}{\mathrm{BE}}=\mathrm{tan}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{\mathrm{BE}}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BE}=\sqrt{3}x.....\left(1\right)$
Also,
In ΔCDE,
$\frac{\mathrm{CD}}{\mathrm{DE}}=\mathrm{tan}60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{y}{\mathrm{DE}}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DE}=\frac{y}{\sqrt{3}}.....\left(2\right)$
Now, BE = DE .....(3) (E is midpoint of BD.)
So, from (1), (2) and (3), we get
$\sqrt{3}x=\frac{y}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{y}=\frac{1}{3}$
Hence, the ratio of x and y is 1 : 3.
A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is consonant.
Answer :
From a point T outside a circle of centre O, tangents TP and TQ are drawn to the circle. Prove that OT is the right bisector of line segment PQ.
Answer :
Find the middle term of the A.P. 6, 13, 20, ... , 216.
Answer :
In Fig. 2, AB is the diameter of a circle with centre O and AT is
a tangent. If ∠AOQ = 58°, find ∠ATQ.

Figure 2 
Answer :
If A(5, 2), B(2, −2) and C(−2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t.
Answer :
Find the ratio in which the point $\mathrm{P}\left(\frac{3}{4},\frac{5}{12}\right)$ divides the line segment joining the points $\mathrm{A}\left(\frac{1}{2},\frac{3}{2}\right)$ and B(2, −5).
Answer :
Solve the following quadratic equation for x :
9x^{2} − 6b^{2}x
− (a^{4} − b^{4}) =
0
Answer :
In Fig. 3, APB and AQO are semicircles, and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. $\left[\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right]$
Figure 3 
Answer :
A solid wooden toy is in the form of a hemisphere surrounded by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is $166\frac{5}{6}{\mathrm{cm}}^{3}$. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of Rs 10 per cm^{2}. $\left[\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right]$
Answer :
In Fig. 4, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. $\left(\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\mathrm{and}\sqrt{5}=2.236\right)$
Figure 4 
Answer :
In Fig. 5, from a cuboidal solid metallic block, of dimensions
15cm âœ• 10cm âœ• 5cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block $\left[\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right]$
Figure 5 
Answer :
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 45°. If the tower is 30 m high, find the height of the building.
Answer :
In Fig. 6, find the area of the shaded region [Use π =
3.14]
Figure 6 
Answer :
Find that nonzero value of k, for which the quadratic equation kx^{2} + 1 − 2(k − 1)x + x^{2} = 0 has equal roots. Hence find the roots of the equation.
Answer :
All red face cards are removed from a pack of playing cards. The
remaining cards were well shuffled and then a card is drawn at
random from them. Find the probability that the drawn card is
(i) a red card
(ii) a face card
(iii) a card of clubs
Answer :
Find the area of the triangle PQR with Q(3,2) and the midpoints of the sides through Q being (2,−1) and (1,2).
Answer :
If S_{n} denotes the sum of first n terms of an A.P., prove that S_{30} = 3[S_{20} − S_{10}]
Answer :
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Answer :
In Fig. 7, tangents PQ and PR are drawn from an external point P
to a circle with centre O, such that ∠RPQ = 30°. A chord
RS is drawn parallel to the tangent PQ. Find ∠RQS.
Figure 7 
Answer :
From a point P on the ground the angle of elevation of the top of a tower is 30° and that of the top of a flag staff fixed on the top of the tower, is 60°. If the length of the flag staff is 5 m, find the height of the tower.
Answer :
Ramkali required Rs 2,500 after 12 weeks to send her daughter to school. She saved Rs 100 in the first week and increased her weekly saving by Rs 20 every week. Find whether she will be able to send her daughter to school after 12 weeks.
What value is generated in the above situation?
Answer :
The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and original fraction is $\frac{29}{20}$. Find the original fraction.
Answer :
Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe.
Answer :
If A(−4, 8), B(−3, −4), C(0, −5) and D(5, 6) are the vertices of a quadrilateral ABCD, find its area.
Answer :
A 21 m deep well with diameter 6 m is dug and the earth from digging is evenly spread to form a platform 27 m âœ• 11 m. Find the height of the platform. $\left[\mathrm{Use}\mathrm{\pi}=\frac{22}{7}\right]$
Answer :
A bag contains 25 cards numbered from 1 to 25. A card is drawn at
random from the bag. Find the probability that the number on the
drawn card is:
(i) divisible by 3 or 5
(ii) a perfect square number
Answer :
Draw a line segment AB of length 7 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.Steps
Answer :
Solve for x :
$\frac{3}{x+1}+\frac{4}{x1}=\frac{29}{4x1};x\ne 1,1,\frac{1}{4}$
Answer :
10th Maths Paper Solutions Set 3 : CBSE Delhi Previous Year 2013 will be available online in PDF book soon. The solutions are absolutely Free. Soon you will be able to download the solutions.