In the given figure, BDC is a tangent to the given circle at point D such that BD = 30 cm and CD = 7 cm.
Question: In the given figure,BDCis a tangent to the given circle at pointDsuch thatBD= 30 cm andCD= 7 cm. The other tangentsBEandCFare drawn respectively fromBandCto the circle and meet when produced atAmaking BAC a right angle triangle. Calculate (i)AF(ii) radius of the circle. Solution: The given figure is below (i) The given triangleABCis a right triangle where side BC is the hypotenuse. Let us now apply Pythagoras theorem. We have, $A B^{2}+A C^{2}=B C^{2}$ Looking at the figure we can rewr...
Read More →In ∆ABC and ∆PQR, it is given that AB = AC,
Question: In∆ABCand ∆PQR, it is given thatAB=AC, C= Pand B= Q. Then, the two triangles are(a) isosceles but not congruent(b) isosceles and congruent(c) congruent but not isosceles(d) neither congruent nor isosceles Solution: (a) isosceles but not congruent $A B=A C$ $\Rightarrow \angle C=\angle B$ $\Rightarrow \angle P=\angle Q \quad[\because \angle C=\angle P$ and $\angle B=\angle Q]$ Thus, both the triangles are isosceles but not congruent....
Read More →if
Question: If $-1x0$, then write the value of $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ Solution: Let $x=-\tan y$ where $0y\frac{\pi}{2}$ Then, $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\sin ^{-1}\left(\frac{-2 \tan y}{1+\tan ^{2} y}\right)+\cos ^{-1}\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right)$ $=\sin ^{-1}\{-\sin (2 y)\}+\cos ^{-1}\{\cos (2 y)\}$ $=-\sin ^{-1}\{\sin (2 y)\}+\cos ^{-1}\{\cos (2...
Read More →In ∆ABC and ∆DEF, it is given that ∠B = ∠E and ∠C = ∠F.
Question: In∆ABCand ∆DEF, it is given that B= Eand C= F. In order that ∆ABC ∆DEF, we must have(a)AB=DF(b)AC=DE(c)BC=EF(d) A= D Solution: In order that $\triangle A B C \cong \triangle D E F$, we must have $\mathrm{BC}=\mathrm{EF}$. Hence, the correct answer is option (c)....
Read More →What is the value
Question: What is the value of $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) ?$ Solution: $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ $=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left\{\sin \left(\pi-\frac{\pi}{3}\right)\right\}$ $=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left\{\sin \left(\frac{\pi}{3}\right)\right\}$ $\left[\because\right.$ Range of sine is $\left[-\frac{\pi}{2}, \...
Read More →Solve the following
Question: If $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in $\mathrm{AP}$, whose common difference is $d$, then show that $\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots+\sec \theta_{n-1} \sec \theta_{n}=\frac{\tan \theta_{n}-\tan \theta_{1}}{\sin d}$ Solution: As, $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in AP So, $d=\theta_{2}-\theta_{1}=\theta_{3}-\theta_{2}=\ldots=\theta_{n}-\theta_{n-1} \quad \ldots$ (i) Now, $\mathrm{LHS}=\sec \t...
Read More →In ∆ABC and ∆DEF, it is given that AB = DE and BC = EF.
Question: In ∆ABCand ∆DEF, it is given thatAB=DEandBC=EF. In order that ∆ABC ∆DEF, we must have(a)A= D(b) B= E(c) C= F(d) none of these Solution: (b) $\angle B=\angle E$ In $\triangle A B C$ and $\triangle D E F$, we have : AB = DE (Given)BC = EF (Given) In order that $\triangle A B C \cong D E F$, we must have $\angle B=\angle E$....
Read More →In the given figure, PA and PB are tangents from an external point P
Question: In the given figure,PAandPBare tangents from an external pointPto a circle with centreO.LNtouches the circle at M. Prove thatPL+LM=PN+MN. Solution: The figure given in the question From the property of tangents we know that the length of two tangents drawn from an external point will we be equal. Hence we have, PA = PB PL + LA = PN + NB (1) Again from the same property of tangents we have, LA = LM(where L is the common external point for tangents LA and LM) NB = MN(where N is the commo...
Read More →Write the value
Question: Write the value of $\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right)$ for $x0$. Solution: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ When $x0, \frac{1}{x}0$, then both are negative. Let $x=-y, y0$ Then, $\tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\tan ^{-1}(-y)+\tan ^{-1}\left(-\frac{1}{y}\right)$ $=-\left(\tan ^{-1} y+\tan ^{-1} \frac{1}{y}\right)$ $=-\tan ^{-1}\left(\frac{y+\frac{1}{y}}{1-y \frac{1}{y}}\right), y0$ $=-\tan ^{-1}\left(\frac{y^{2}+1}{0}\right)$ $=-\ta...
Read More →Write the value
Question: Write the value of $\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right)$ for $x0$. Solution: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ When $x0, \frac{1}{x}0$, then both are negative. Let $x=-y, y0$ Then, $\tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\tan ^{-1}(-y)+\tan ^{-1}\left(-\frac{1}{y}\right)$ $=-\left(\tan ^{-1} y+\tan ^{-1} \frac{1}{y}\right)$ $=-\tan ^{-1}\left(\frac{y+\frac{1}{y}}{1-y \frac{1}{y}}\right), y0$ $=-\tan ^{-1}\left(\frac{y^{2}+1}{0}\right)$ $=-\ta...
Read More →In the given figure, AB is a chord of length 16 cm of a circle of radius 10 cm.
Question: In the given figure,ABis a chord of length 16 cm of a circle of radius 10 cm. The tangents atAandBintersect at a pointP. Find the length ofPA. Solution: Considerand. From the property of tangents we know that the length of two tangents drawn form an external point will be equal. Therefore we have, PA = PB OB = OA(They are the radii of the same circle) POis the common side Therefore, from SSS postulate of congruency, we have, Hence, (1) Now considerand. We have, (From (1)) PAis the comm...
Read More →If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is
Question: If the altitudes from two vertices of a triangle to the opposite sides are equal, then the triangle is(a) equilateral(b) isosceles(c) scalene(d) right-angled Solution: (b) isosceles In $\triangle A B C$, BL $\|$ AC. CM $\| \mathrm{AB}$ such that $\mathrm{BL}=\mathrm{CM}$. To prove: AB =AC In $\triangle \mathrm{ABL}$ and $\triangle \mathrm{ACM}$ $\mathrm{BL}=\mathrm{CM} \quad$ (Given) $\angle \mathrm{BAL}=\angle \mathrm{CAM} \quad$ (Common angle) $\angle \mathrm{ALB}=\angle \mathrm{AMC}...
Read More →Write the value
Question: Write the value of $\tan ^{-1} x+\tan ^{-1}\left(\frac{1}{x}\right)$ for $x0$ Solution: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right), x y1$ $\therefore \tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\tan ^{-1}\left(\frac{x+\frac{1}{x}}{1-x \frac{1}{x}}\right), x0=$ $=\tan ^{-1}\left(\frac{x^{2}+1}{0}\right)$ $=\tan ^{-1}(\infty)$ $=\tan ^{-1}\left(\tan \frac{\pi}{2}\right)$ $=\frac{\pi}{2}$ $\therefore \tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\frac{\pi}{2}$...
Read More →In the given figure, AB = AC and OB = OC. Then,
Question: In the given figure,AB=ACandOB=OC. Then,ABO: ACO= ?(a) 1 : 1(b) 2 : 1(c) 1 : 2(d) None of these Solution: (a) 1:1 In $\Delta O A B$ and $\Delta O A C$, we have : $\mathrm{AB}=\mathrm{AC} \quad$ (Given) $\mathrm{OB}=\mathrm{OC} \quad$ (Given) $\mathrm{OA}=\mathrm{OA} \quad$ (Common side) Thus, $\triangle \mathrm{OAB} \cong \mathrm{OAC} \quad$ (SSS criterion) i.e., $\angle A B O=\angle A C O$ $\therefore \angle A B O: \angle A C O=1: 1$...
Read More →If x < 0, then write the value
Question: If $x0$, then write the value of $\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$ in terms of $\tan ^{-1} x$ Solution: Let $x=\tan y$ Then, $\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\cos ^{-1}\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right)$ $=\cos ^{-1}(\cos 2 y) \quad\left[\because \frac{1-\tan ^{2} x}{1+\tan ^{2} x}=\cos 2 x\right]$ $=2 y \quad \ldots(1)$ The value ofxis negative.So, letx=a-awhere a 0. $-a=\tan y$ $\Rightarrow y=\tan ^{-1}(-a)$ Now, $\cos ^{-1}\left(\frac{1-x^...
Read More →Solve the following
Question: If $a_{n}$ is an A.P. such that $\frac{a_{4}}{a_{7}}=\frac{2}{3}$, find $\frac{a_{6}}{a_{8}}$. Solution: Given: an is an A.P. $\frac{a_{4}}{a_{7}}=\frac{2}{3}$ $\Rightarrow \frac{a+(4-1) d}{a+(7-1) d}=\frac{2}{3}$ $\Rightarrow \frac{a+3 d}{a+6 d}=\frac{2}{3}$ $\Rightarrow 3(a+3 d)=2(a+6 d)$ $\Rightarrow 3 a+9 d=2 a+12 d$ $\Rightarrow a=3 d \ldots(i)$ $\therefore \frac{a_{6}}{a_{8}}=\frac{a+(6-1) d}{a+(8-1) d}$ $\Rightarrow \frac{a_{6}}{a_{8}}=\frac{a+5 d}{a+7 d}$ $\Rightarrow \frac{a_{...
Read More →If x > 1, then write the value
Question: If $x1$, then write the value of $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ in terms of $\tan ^{-1} x$. Solution: $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\pi-2 \tan ^{-1} x$ $\left[\because 2 \tan ^{-1} x=\pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right.$ for $\left.x1\right]$...
Read More →If x > 1, then write the value
Question: If $x1$, then write the value of $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ in terms of $\tan ^{-1} x$. Solution: $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\pi-2 \tan ^{-1} x$ $\left[\because 2 \tan ^{-1} x=\pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)\right.$ for $\left.x1\right]$...
Read More →The first and the last terms of an A.P. are a and l respectively.
Question: The first and the last terms of an A.P. areaandlrespectively. Show that the sum ofnthterm from the beginning andnth term from the end isa+l. Solution: Given: First term =a Last term =l nth term from the beginning $=a+(n-1) d$, where $d$ is the common difference. nth term from the end $=l+(n-1)(-d)=l-d n+d$ Their sum $=a+(n-1) d+l-d n+d$ $=a+n d-d+l-n d+d$ $=a+n d-d+l-n d+d$ $=a+l$ Hence, proved....
Read More →In the given figure, there are two concentric circles with centre O
Question: In the given figure, there are two concentric circles with centreOof radii 5 cm and 3 cm. From an external pointP, tangentPAandPBare drawn to these circles. IfAP= 12 cm, find the length ofBP. Solution: The figure given the question is From the property of tangents we know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore,OAis perpendicular toAPand triangleOAPis a right triangle. Therefore, $O P^{2}=A P^{2}+O A^{2}$ $=12^{2}+5^{...
Read More →In the given figure, AB > AC. If BO and CO are the bisectors of ∠B and
Question: In the given figure,ABAC. IfBOandCOare the bisectors ofBand Crespectively, then(a)OB=OC(b)OBOC(c) OBOC Solution: (b) OB OCAB AC (Given) $\Rightarrow \angle C\angle B$ $\Rightarrow \frac{1}{2} \angle C\frac{1}{2} \angle B$ $\Rightarrow \angle O C B\angle O B C$ (Given) $\Rightarrow O BO C$...
Read More →if
Question: If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$, then write the value of $x+y+z$ Solution: $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$ $\Rightarrow \sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}$ $\left[\right.$ As the maximum value in the range of $\sin ^{-1} x$ is $\frac{\pi}{2}$ And here sum of three inverse of sine is 3 times $\frac{\pi}{2}$. i. e., every sin inverse function is equal to $\frac{\pi}{2}$ here. $] \Rightarrow...
Read More →How many numbers are there between 1 and 1000 which when divided by 7 leave remainder 4?
Question: How many numbers are there between 1 and 1000 which when divided by 7 leave remainder 4? Solution: A numberN is divided by 7 leaves a remainder 4. N= 7q + 4 N can take values 4, 11, 18, ..... 998 Now, 4, 11, 18, ..... 998 are in arithmetic progression. First terma= 4 common differenced= 7 last terml = 998 We know that, l=a+ (n 1)d ⇒ 998 = 4 + (n 1)7 ⇒ 998 = 4 + 7n 7 ⇒ 998 = 7n 3 ⇒ 1001 = 7n $\Rightarrow n=\frac{1001}{7}$ $\Rightarrow n=143$ Hence, 143 numbers are there between 1 and 10...
Read More →In the given figure, a circle is inscribed in a quadrilateral ABCD in which
Question: In the given figure, a circle is inscribed in a quadrilateralABCDin which B= 90. It AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radiusrof the circle. Solution: Let us first consider the quadrilateralOPBQ. It is given that. Also from the property of tangents we know that the radius of the circle will always be perpendicular to the tangent at the point of contact. Therefore we have, We know that sum of all angles of a quadrilateral will always be equal to. Therefore, $\angle B+\angle ...
Read More →The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34.
Question: The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P. Solution: Letabe the first term anddbe the common difference. Then, $a_{4}+a_{8}=24$ $\Rightarrow a+(4-1) d+a+(8-1) d=24$ $\Rightarrow a+3 d+a+7 d=24$ $\Rightarrow 2 a+10 d=24$ $\Rightarrow a+5 d=12 \quad \ldots(\mathrm{i})$ Also, $a_{6}+a_{10}=34$ $\Rightarrow a+(6-1) d+a+(10-1) d=34$ $\Rightarrow a+5 d+a+9 d=34$ $\Rightarrow 2 a+14 d=34$...
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