In the given figure, PR =
Question: In the given figure,PR= (a) 20 cm(b) 26 cm(c) 24 cm(d) 28 cm Solution: We know that the radius will always be perpendicular to the tangent at the point of contact. Therefore, $O Q \perp O P$ Hence we have, $O P^{2}=O Q^{2}+Q P^{2}$ $O P^{2}=4^{2}+3^{2}$ $O P^{2}=16+9$ $O P^{2}=25$ $O P=\sqrt{25}$ $O P=5$ Also, OO= sum of the radii of the two circles OO= 3 + 5 OO= 8 Since radius is perpendicular to the tangent,. Therefore, $O^{\prime} R^{2}=O^{\prime} S^{2}+S R^{2}$ $O^{\prime} R^{2}=12...
Read More →Solve the following
Question: IfSn=n2pandSm=m2p,mn, in an A.P., prove thatSp=p3. Solution: $S_{n}=n^{2} p$ $\Rightarrow \frac{n}{2}[2 a+(n-1) d]=n^{2} p$ $\Rightarrow 2 n p=2 a+(n-1) d \quad \ldots(i)$ $S_{m}=m^{2} p$ $\Rightarrow \frac{m}{2}[2 a+(m-1) d]=m^{2} p$ $\Rightarrow 2 m p=2 a+(m-1) d \quad \ldots(i i)$ Subtracting (ii) from (i), we get: $2 p(n-m)=(n-m) d$ $\Rightarrow 2 p=d \quad \ldots(i i i)$ Substituing the value in (i), we get: $n d=2 a+(n-1) d$ $\Rightarrow n d-n d+d=2 a$ $\Rightarrow a=\frac{d}{2}=...
Read More →In the given figure, if tangents PA and PB are drawn to a circle such that
Question: In the given figure, if tangentsPAandPBare drawn to a circle such that APB= 30 and chordACis drawn parallel to the tangentPB, then ABC= (a) 60(b) 90(c) 30(d) None of these Solution: We know that tangents from an external point will be equal in length. Therefore, PA = PB Now consider. We have, PA = PB We know that angles opposite to equal sides will be equal. Therefore, Also, sum of all angles of a triangle will be equal to. $\therefore \angle A B P+\angle B A P+\angle P=180^{\circ}$ $2...
Read More →Write the value
Question: Write the value of $\cos ^{-1}\left(\cos \frac{14 \pi}{3}\right)$ Solution: $\cos ^{-1}\left(\cos \frac{14 \pi}{3}\right)=\cos ^{-1}\left[\cos \left(4 \pi+\frac{2 \pi}{3}\right)\right]$ $=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)$ $=\frac{2 \pi}{3}$...
Read More →Write the value of
Question: Write the value of $\sec ^{-1}\left(\frac{1}{2}\right)$. Solution: The value of $\sec ^{-1}\left(\frac{1}{2}\right)$ is undefined as it is outside the range i.e., $R-(-1,1)$....
Read More →Write the value
Question: Write the value of $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)$ Solution: $\sin ^{-1}\left(\sin \frac{3 \pi}{5}\right)=\sin ^{-1}\left[\sin \left(\pi-\frac{2 \pi}{5}\right)\right]$ $=\sin ^{-1}\left(\sin \frac{2 \pi}{5}\right)$ $=\frac{2 \pi}{5}$...
Read More →In the given figure, APB is a tangent to a circle with centre O at point P.
Question: In the given figure,APBis a tangent to a circle with centre O at point P. If QPB= 50, then the measure of POQis (a) 100(b) 120(c) 140(d) 150 Solution: We know that the radius of a circle will always be perpendicular to the tangent at the point of contact. ' Therefore, $\angle O P B=90^{\circ}$ That is, $\angle O P Q+\angle Q P B=90^{\circ}$ It is given that, $\angle Q P B=50^{\circ}$ Therefore, we have, $\angle O P Q+50^{\circ}=90^{\circ}$ $\angle O P Q=40^{\circ}$ Now, consider. We ha...
Read More →Write the principal value
Question: Write the principal value of $\cos ^{-1}\left(\cos 680^{\circ}\right)$ Solution: $\cos ^{-1}\left(\cos 680^{\circ}\right)=\cos ^{-1}\left[\cos \left(720^{\circ}-680^{\circ}\right)\right]$ $=\cos ^{-1}\left(\cos 40^{\circ}\right)$ $=40^{\circ}$...
Read More →Write the principal value
Question: Write the principal value of $\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}$ Solution: We know $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$ $\therefore \tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}=\frac{\pi}{2}$...
Read More →Write the value
Question: Write the value of $\tan ^{-1}\left\{2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$ Solution: $\tan ^{-1}\left\{2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$ $=\tan ^{-1}\left\{2 \sin \left[\cos ^{-1} 2\left(\frac{\sqrt{3}}{2}\right)^{2}-1\right]\right\}$ $=\tan ^{-1}\left[\begin{array}{lll}2 \sin \left(\cos ^{-1}\right. \left.\frac{1}{2}\right)\end{array}\right)$...
Read More →Write the value
Question: Write the value of $\tan ^{-1}\left\{2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$ Solution: $\tan ^{-1}\left\{2 \sin \left(2 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$ $=\tan ^{-1}\left\{2 \sin \left[\cos ^{-1} 2\left(\frac{\sqrt{3}}{2}\right)^{2}-1\right]\right\}$ $=\tan ^{-1}\left[\begin{array}{lll}2 \sin \left(\cos ^{-1}\right. \left.\frac{1}{2}\right)\end{array}\right)$...
Read More →The number of terms of an A.P. is even; the sum of odd terms is 24,
Question: The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by $101 / 2$, find the number of terms and the series. Solution: Let total number of terms be2n. According to question, we have: $a_{1}+a_{3}+\ldots+a_{2 n-1}=24 \ldots(1)$ $a_{2}+a_{4}+\ldots+a_{2 n}=30 \quad \ldots(2)$ Subtracting (1) from (2), we get: $\Rightarrow n d=6$ ....(3) Given: $a_{2 n}=a_{1}+\frac{21}{2}$ $\Rightarrow a_{2 n}-a_{1}=\frac{21}{2}$ $...
Read More →If the tangent at a point P to a circle with centre O cuts a line through
Question: If the tangent at a pointPto a circle with centreOcuts a line throughOatQsuch thatPQ= 24 cm andOQ= 25 cm. Find the radius of the circle. Solution: Let us first draw whatever is given so that we can understand the problem better. Since the tangent to a circle is always perpendicular to the radius of the circle at the point of contact, we have drawnOPperpendicular toPQ. Thus we have a right triangle with one of its sides as the radius. To find the radius we have to use Pythagoras theorem...
Read More →In the adjoining figure, ABCD is a square. A line segment CX cuts AB at X and the diagonal BD
Question: In the adjoining figure,ABCDis a square.Aline segmentCXcutsABatXand the diagonalBDatOsuch thatCOD= 80 and OXA=x. Find the value ofx. Solution: The angles of a square are bisected by the diagonals.OBX= 45o [∵ABC= 90oandBDbisectsABC]AndBOX= COD= 80o [Vertically opposite angles] In∆BOX, we have:AXO = OBX+BOX [Exterior angle of∆BOX]⇒ AXO= 45o+ 80o=125ox=125o...
Read More →Write the principal value
Question: Write.the principal value of $\tan ^{-1} 1+\cos ^{-1}\left(-\frac{1}{2}\right)$ Solution: $\tan ^{-1} 1+\cos ^{-1}\left(-\frac{1}{2}\right)=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)+\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)$ $=\frac{\pi}{4}+\frac{2 \pi}{3}$ $=\frac{11 \pi}{12}$...
Read More →O is the centre of a circle of a radius 8 cm.
Question: Ois the centre of a circle of a radius 8 cm. the tangent at a pointAon the circle cuts a line throughOatBsuch thatAB= 15 cm. FindOB. Solution: First let us draw whatever is given in the question. This will help us understand the problem better. Since the tangent will always be perpendicular to the radius we have drawnOAperpendicular toAB. To find the length ofOBwe have to use Pythagoras theorem. $O B^{2}=O A^{2}+A B^{2}$ $O B^{2}=8^{2}+15^{2}$ $O B^{2}=64+225$ $O B^{2}=289$ $O B=\sqrt{...
Read More →In a rhombus ABCD, the altitude from D to the side AB bisects AB.
Question: In a rhombusABCD, the altitude fromDto the sideABbisectsAB. Find the angles of the rhombus. Solution: Given: $A B C D$ is a rhombus, $D E$ is altitude which bisects $A B$ i.e. $A E=E B$ In $\Delta A E D$ and $\Delta B E D$, $D E=D E$ (Common side) $\angle D E A=\angle D E B=90^{\circ}$ (Given) $A E=E B$ (Given) $\therefore \Delta A E D \cong \Delta B E D$ (By SAS congruence Criteria) $\Rightarrow A D=B D$ (CPCT) Also, $A D=A B$ (Sides of rhombus are equal) $\Rightarrow A D=A B=B D$ Thu...
Read More →Write the value
Question: Write the value of $\tan \left(2 \tan ^{-1} \frac{1}{5}\right)$ Solution: $\tan \left(2 \tan ^{-1} \frac{1}{5}\right)=\tan \left[\tan ^{-1} \frac{2 \times \frac{1}{5}}{1-\left(\frac{1}{5}\right)^{2}}\right]$ $=\tan \left(\tan ^{-1} \frac{\frac{2}{5}}{\frac{24}{24}}\right)$ $=\tan \left(\tan ^{-1} \frac{5}{12}\right)$ $=\frac{5}{12}$...
Read More →How many tangents can a circle have?
Question: How many tangents can a circle have? Solution: We know that circle is made of infinite points which are located at a fixed distance from a particular point. Since at each of these infinite points a tangent can be drawn, we can have infinite number of tangent for a given circle....
Read More →Write the principal value
Question: Write the principal value of $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ Solution: We have, $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ $=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{3}\right)\right\}$ $\left[\because\left(\pi-\frac{2 \pi}{3}\right) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]$ $=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^...
Read More →Write the principal value
Question: Write the principal value of $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ Solution: We have, $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ $=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{3}\right)\right\}$ $\left[\because\left(\pi-\frac{2 \pi}{3}\right) \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\right]$ $=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^...
Read More →The first term of an A.P. is 2 and the last term is 50.
Question: The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference. Solution: Given: $a=2, l=50, S_{n}=442$ Now, $S_{n}=442$ $\Rightarrow \frac{n}{2}[a+l]=442$ $\Rightarrow \frac{n}{2}[2+50]=442$ $\Rightarrow n=17$ $\therefore a_{17}=50$ $\Rightarrow a+(17-1) \times d=50$ $\Rightarrow 2+16 d=50$ $\Rightarrow d=3$...
Read More →Fill in the blanks:
Question: Fill in the blanks: (i) The common point of a tangent and the circle is called ..........(ii) A circle may have .............. parallel tangents.(iii) A tangent to a circle intersects in it ............ point(s).(iv) A line intersecting a circle in two points is called a ...........(v) The angle between tangent at a point on a circle and the radius through the point is ........ Solution: (i) We know that the tangent to a circle is that line which touches the circle at exactly one point...
Read More →In the given figure, two equal circles touch each other at T, if OP = 4.5 cm, then QR =
Question: In the given figure, two equal circles touch each other atT, ifOP= 4.5 cm, thenQR= (a) 9 cm(b) 18 cm(c) 15 cm(d) 13.5 cm Solution: We know that tangents drawn from the same external point will be equal in length. Therefore, we have, QP = PT PT = PR From the above two equations, we get, QP = PR It is given that, QP= 4.5 cm Therefore, PR= 4.5 cm From the figure, we have, QR = QP + PR QR= 4.5 + 4.5 QR= 9 The correct answer is option (a)....
Read More →Write the principal value
Question: Write the principal value of $\sin ^{-1}\left(-\frac{1}{2}\right)$ Solution: Let $y=\sin ^{-1}\left(-\frac{1}{2}\right)$ Then, $\sin y=-\frac{1}{2}=\sin \left(-\frac{\pi}{6}\right)$ $y=-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ Here, $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is the range of the principal value branch of the inverse sine function. $\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$...
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