In the given figure, PA and PB are tangents to the circle drawn from an external point P.
Question: In the given figure,PAandPBare tangents to the circle drawn from an external point P.CDis a third tangent touching the circle at Q. IfPB= 10 cm andCQ= 2 cm, what is the lengthPC? Solution: Given data is as follows: PB = 10 cm CQ = 2 cm We have to find the length of PC We know that the length of two tangents drawn from the same external point will equal. Therefore, PB = PA It is given that PB = 10 cm Therefore, PA = 10 cm Also, from the same principle we have, CQ = CA It is given that C...
Read More →Evaluate
Question: Evaluate $\sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)$. Solution: We know that $\cos ^{-1} x=2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ $\tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}$ $\therefore \sin \left(\frac{1}{2} \cos ^{-1} \frac{4}{5}\right)=\sin \left(\frac{1}{2} 2 \tan ^{-1} \sqrt{\frac{1-\frac{4}{5}}{1+\frac{4}{5}}}\right)$ $=\sin \left(\tan ^{-1} \sqrt{\frac{\frac{1}{5}}{\frac{9}{5}}}\right)$ $=\sin \left(\tan ^{-1} \frac{1}{3}\right)$ $=\sin \left\{\sin ^{-1}\left(\fr...
Read More →In the given figure, OQ : PQ = 3.4 and perimeter of Δ POQ = 60 cm.
Question: In the given figure, OQ : PQ = 3.4 and perimeter of Δ POQ = 60 cm. Determine PQ, QR and OP. Solution: In the figure, $\angle P Q O=90^{\circ}$. Therefore we can use Pythagoras theorem to find the side $P O$. $P O^{2}=P Q^{2}+O Q^{2} \ldots \ldots(1)$ In the problem it is given that, $\frac{O Q}{P Q}=\frac{3}{4}$ $O Q=\frac{3}{4} P Q$......(2) Substituting this in equation (1), we have, $P O^{2}=\frac{9 P Q^{2}}{16}+P Q^{2}$ $P O^{2}=\frac{25 P Q^{2}}{16}$ $P O=\sqrt{\frac{25 P Q^{2}}{1...
Read More →The angles of a quadrilateral are in the ratio 2: 4 : 5 : 7.
Question: The angles of a quadrilateral are in the ratio 2: 4 : 5 : 7. Find the angles. Solution: LetA = 2x∘.ThenB = (4x)∘;C = (5x)∘andD =(7x)∘Since the sum of the angles of a quadrilateral is360o, we have:2x+ 4x+ 5x+ 7x= 360∘⇒18x= 360∘⇒x= 20∘A= 40∘;B= 80∘;C= 100∘;D= 140∘...
Read More →The sum of three numbers in A.P. is 12 and the sum of their cubes is 288.
Question: The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers. Solution: Let the numbers be $(a-d), a,(a+d)$. Sum $=a-d+a+a+d=12$ $\Rightarrow 3 a=12$ $\Rightarrow a=4$ Also, $(a-d)^{3}+a^{3}+(a+d)^{3}=288$ $\Rightarrow a^{3}-d^{3}-3 a^{2} d+3 a d^{2}+a^{3}+a^{3}+d^{3}+3 a^{2} d+3 a d^{2}=288$ $\Rightarrow 3 a^{3}+6 a d^{2}=288$ $\Rightarrow 3(4)^{3}+6 \times 4 \times d^{2}=288$ $\Rightarrow 192+24 d^{2}=288$ $\Rightarrow 24 d^{2}=96$ $\Rightarrow d^{2}=4$ ...
Read More →Three angles of a quadrilateral are 75°, 90° and 75°.
Question: Three angles of a quadrilateral are 75, 90 and 75. Find the measure of the fourth angle. Solution: Given:Threeanglesofaquadrilateralare75,90and75.Let the fourth angle bex.Using angle sum property of quadrilateral, $75^{\circ}+90^{\circ}+75^{\circ}+x=360^{\circ}$ $\Rightarrow 240^{\circ}+x=360^{\circ}$ $\Rightarrow x=360^{\circ}-240^{\circ}$ $\Rightarrow x=120^{\circ}$ So, the measure of the fourth angle is $120^{\circ}$....
Read More →Write the value
Question: Write the value of $\sin ^{-1}\left(\sin 1550^{\circ}\right)$. Solution: We know that $\sin ^{-1}(\sin x)=x$. Now, $\sin ^{-1}\left(\sin 1550^{\circ}\right)=\sin ^{-1}\left\{\sin \left(1620^{\circ}-1550^{\circ}\right)\right\} \quad\left[\because \sin x=\sin \left(1620^{\circ}-x\right)\right]$ $=\sin ^{-1}\left(\sin 70^{\circ}\right)$ $=70^{\circ}$ $\therefore \sin ^{-1}\left(\sin 1550^{\circ}\right)=70^{\circ}$...
Read More →Equal circles with centres O and O' touch each other at X. OO' produced to meet
Question: Equal circles with centres $\mathrm{O}$ and $\mathrm{O}^{\prime}$ touch each other at $\mathrm{X}$. OO' produced to meet a circle with centre $\mathrm{O}^{\prime}$, at $\mathrm{A}$. AC is a tangent to the circle whose centre is $\mathrm{O}$. O'D is perpendicular to AC. Find the value of $\frac{D O^{\prime}}{C O}$. Solution: Consider the two trianglesand. We have, is a common angle for both the triangles. (Given in the problem) (SinceOCis the radius andACis the tangent to that circle at...
Read More →Write the value
Question: Write the value of $\cos \left(2 \sin ^{-1} \frac{1}{3}\right)$. Solution: Let $y=\sin ^{-1} \frac{1}{3}$ Then, $\sin y=\frac{1}{3}$ Now, $\cos y=\sqrt{1-\sin ^{2} y}$ $\Rightarrow \cos y=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}$ $\cos \left(2 \sin ^{-1} \frac{1}{3}\right)=\cos (2 y)$ $=\cos ^{2} y-\sin ^{2} y \quad\left[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x\right]$ $=\left(\frac{2 \sqrt{2}}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2}$ $=\frac{8}{9}-\frac{1}{9}$ ...
Read More →Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Question: Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least. Solution: Let the four numbers be $a-3 d, a-d, a+d, a+3 d$. Sum $=50$ $\Rightarrow a-3 d+a-d+a+d+a+3 d=50$ $\Rightarrow 4 a=50$ $\Rightarrow a=\frac{25}{2} \ldots(i)$ Also, $a+3 d=4(a-3 d)$ $\Rightarrow a+3 d=4 a-12 d$ $\Rightarrow 3 a=15 d$ $\Rightarrow a=5 d$ $\Rightarrow \frac{25}{2 \times 5}=d \quad(\operatorname{Using}(\mathrm{i}))$ $\Rightarrow \frac{5}{2}=d$ So, the terms are as...
Read More →In the given figure, common tangents PQ and RS to
Question: In the given figure, common tangentsPQandRSto two circles intersect atA. Prove thatPQ=RS. Solution: The figure given in the question is We know from the property of tangents that the length of two tangents drawn from a common external point will be equal. Therefore, PA = RA (1) AQ = AS (2) Let us add equation (1) and (2) PA + AQ = RA + AS PQ = RS Thus we have proved thatPQ = RS....
Read More →Write the value
Question: Write the value of $\sin ^{-1}\left(\sin \left(-600^{\circ}\right)\right)$. Solution: We know that $\sin ^{-1}(\sin x)=x$. Now, $\sin ^{-1}\left\{\sin \left(-600^{\circ}\right)\right\}=\sin ^{-1}\left\{\sin \left(720^{\circ}-600^{\circ}\right)\right\}$ $=\sin ^{-1}\left\{\sin \left(120^{\circ}\right)\right\}$ $=\sin ^{-1}\left\{\sin \left(180^{\circ}-120^{\circ}\right)\right\} \quad[\because \sin x=\sin (\pi-x)]$ $=\sin ^{-1}\left(\sin 60^{\circ}\right)$ $=60^{\circ}$ $\therefore \sin ...
Read More →The length of three concesutive sides of a quadrilateral circumscribing a circle
Question: The length of three concesutive sides of a quadrilateral circumscribing a circle are 4 cm, 5 cm, and 7 cm respectively. Determine the length of the fourth side. Solution: Let us first put the given data in the form of a diagram. From the property of tangents we know that the length of two tangents drawn from the same external point will be equal. Therefore we have, AR = SA Let us representARandSAby a. Similarly, QB = RB Let us represent SD and DP byb PC = CQ Let us representPCandPQby c...
Read More →In the given figure, BC is a tangent to the circle with centre O.
Question: In the given figure,BCis a tangent to the circle with centreO.OEbisectsAP.Prove that ΔAEO~~ΔABC. Solution: The figure given in the question is below Let us first take up. We have, OA = OP(Since they are the radii of the same circle) Therefore,is an isosceles triangle. From the property of isosceles triangle, we know that, when a median drawn to the unequal side of the triangle will be perpendicular to the unequal side. Therefore, Now let us take upand. We know that the radius of the ci...
Read More →Three numbers are in A.P. If the sum of these numbers be 27 and the product 648,
Question: Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers. Solution: Let the three numbers be $a-d, a, a+d$. Their sum $=27$ $\Rightarrow a-d+a+a+d=27$ $\Rightarrow 3 a=27$ $\Rightarrow a=9 \quad \ldots(i)$ Product $=(a-d) a(a+d)=648$ $\Rightarrow a\left(a^{2}-d^{2}\right)=648$ $\Rightarrow 9\left(81-d^{2}\right)=648$ $\Rightarrow\left(81-d^{2}\right)=72$ $\Rightarrow d^{2}=9$ $\Rightarrow d=\pm 3$ When $\mathrm{a}=9, \mathrm{~d}=3$, we have : $6...
Read More →Write the value
Question: Write the value of $\cos ^{-1}\left(\cos 1540^{\circ}\right)$ Solution: We know that $\cos ^{-1}(\cos x)=x$ Now, $\cos ^{-1}\left(\cos 1540^{\circ}\right)=\cos ^{-1}\left\{\cos \left(1440+100^{\circ}\right)\right\}$ $=\cos ^{-1}\left\{\cos \left(100^{\circ}\right)\right\} \quad\left[\because \cos \left(4 \pi+100^{\circ}\right)=\cos 100^{\circ}\right]$ $=100^{\circ}$...
Read More →In the given figure, two tangents AB and AC are drawn to
Question: In the given figure, two tangentsABandACare drawn to a circle with centreOsuch that BAC= 120. Prove thatOA= 2AB. Solution: Considerand. We have, OB = OC(Since they are radii of the same circle) AB = AC(Since length of two tangents drawn from an external point will be equal) OAis the common side. Therefore by SSS congruency, we can say thatandare congruent triangles. Therefore, It is given that, $\angle O A B+\angle O A C=120^{\circ}$ $2 \angle O A B=120^{\circ}$ $\angle O A B=60^{\circ...
Read More →Write the range
Question: Write the range of $\tan ^{-1} x$ Solution: The range of $\tan ^{-1} x$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$....
Read More →Fill in the blanks.
Question: Fill in the blanks.(a) Each angle of an equilateral triangle measures ...... .(b) Medians of an equilateral triangle are ...... .(c) In a right triangle the hypotenuse is the ...... side.(d) Drawing a∆ABCwithAB= 3 cm,BC= 4 cm andCA= 7 cm is ...... . Solution: a) Each angle of an equilateral triangle measures $60^{\circ}$. b) Medians of an equilateral triangle are equal. c) In a right triangle, the hypotenuse is the longest side. d) Drawing a $\triangle A B C$ with $A B=3 \mathrm{~cm}, ...
Read More →Write the value
Question: Write the value of $\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$. Solution: We have $\cos ^{-1} \frac{1}{2}+2 \sin ^{-1} \frac{1}{2}$ $=\cos ^{-1}\left(\cos \frac{\pi}{3}\right)+2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right)$ $\left[\because\right.$ The range of sine is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] ; \frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and the range of cosine is $[0, \pi] ; \frac{\pi}{3} \in[0, \pi]$ $=\frac{\pi}{3}+2\lef...
Read More →In the given figure, PO ⊥ QO. The tangents to the circle at P and Q intersect at a point T
Question: In the given figure,POQO. The tangents to the circle atPandQintersect at a pointT. Prove thatPQandOTare right bisector of each other. Solution: In the given figure, PO = OQ (Since they are the radii of the same circle) PT = TQ(Length of the tangents from an external point to the circle will be equal) Now considering the angles of the quadrilateral PTQO, we have, (Given in the problem) (The radius of the circle will be perpendicular to the tangent at the point of contact) (The radius of...
Read More →Fill in the blanks with < or >.
Question: Fill in the blanks with or .(a) (Sum of any two sides of a triangle) ...... (the third side)(b) (Difference of any two sides of a triangle) ...... (the third side)(c) (Sum of three altitudes of a triangle) ...... (sum of its three side)(d) (Sum of any two sides of a triangle) ...... (twice the median to the 3rd side)(e) (Perimenter of a triangle) ...... (sum of its three medians) Solution: a) Sum of any two sides of a trianglethe third sideb) Difference of any two sides of a triangleth...
Read More →The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6,
Question: The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms. Solution: Let the three terms of the A.P. be $a-d, a, a+d$. Then, we have: $a-d+a+a+d=21$ $\Rightarrow 3 a=21$ $\Rightarrow a=7 \ldots(i)$ Also, $(a-d)(a+d)-a=6$ $\Rightarrow a^{2}-d^{2}-a=6$ $\Rightarrow 49-d^{2}-7=6$ $\Rightarrow 36=d^{2}$ $\Rightarrow \pm 6=d$ When $\mathrm{d}=6, \mathrm{a}=7$, we get: $1,7,13$ When $\mathrm{d}=-6$, a $=7$, we get:...
Read More →Write the value
Question: Write the value of $\sin \left(\cot ^{-1} x\right)$. Solution: We know $\cot ^{-1} x=\tan ^{-1} \frac{1}{x}$ Now, we have $\sin \left(\cot ^{-1} x\right)=\sin \left(\tan ^{-1} \frac{1}{x}\right)$ $=\sin \left[\sin ^{-1}\left(\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x^{2}}}}\right)\right] \quad\left[\because \tan ^{-1} x=\sin ^{-1}\left(\frac{x}{\sqrt{1+x^{2}}}\right)\right]$ $=\sin \left[\sin ^{-1}\left(\frac{\frac{1}{x}}{\frac{\sqrt{x^{2}+1}}{x}}\right)\right]$ $=\sin \left(\sin ^{-1} \fra...
Read More →Which is true?
Question: Which is true?(a) A triangle can have two right angles.(b) A triangle can have two obtuse angles.(c) A triangle can have two acute angles.(d) An exterior angle of a triangle is less than either of the interior opposite angles. Solution: (c) A triangle can have two acute angles.The sum of two acute angles is always less than 180o, thus satisfying the angle sum property of a triangle.Therefore, a triangle can have two acute angles....
Read More →