In the given figure, AB > AC. Then which of the following is true?

Question: In the given figure,ABAC. Then which of the following is true?(a)ABAD(b)AB=AD(c)ABAD(d) Cannot be determined Solution: (c)ABAD $A BA C$ is given. $\therefore \angle A C B\angle A B C$ Now, $\angle A D B\angle A C D \quad$ (exterior angle) $\Rightarrow \angle A D B\angle A C B\angle A B C$ $\Rightarrow \angle A D B\angle A B D \Rightarrow A BA D$...

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Two circles touch externally at a point P. From a point T on the tangent at P,

Question: Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR. Solution: We know that the lengths of tangents drawn from an external point to a circle are equal.In the given figure, TQ and TP are tangents drawn to the same circle from an external point T. TQ = TP .....(1)Also, TP and TR are tangents drawn to the same circle from an external point T. TP = TR ........

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In the given figure, O is the centre of the circle and BCD is tangent to it at C.

Question: In the given figure, O is the centre of the circle andBCDis tangent to it atC. Prove that BAC+ ACD= 90. Solution: In the given figure, let us joinDanA. Consider. We have, OC = OA(Radii of the same circle) We know that angles opposite to equal sides of a triangle will be equal. Therefore, (1) It is clear from the figure that Now from (1) Now as BD is tangent therefore Therefore From the figure we can see that Thus we have proved....

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An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively,

Question: An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term. Solution: Given; $a=7, n=60, l=125$ $l=a+(n-1) d$ $\Rightarrow 125=7+(60-1) d$ $\Rightarrow 125=7+59 d$ $\Rightarrow 118=59 d$ $\Rightarrow 2=d$ $a_{32}=a+(32-1) d$ $=a+31 d$ $=7+31 \times 2$ $=7+62$ $=69$...

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Prove that the perpendicular at the point of contact to

Question: Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle. Solution: Let us first put the given data in the form of a diagram. Let us also draw another parallel tangentPQparallel to the given tangentAB. Draw a radiusORto the point of contact of the tangentPQ. Let us also draw a radiusCOparallel to the two tangents. It is given that, We know that sum of angles on the same side of the transversal will be equal to 180. Therefor...

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In ∆ABC, ∠A = 40° and ∠B = 60°.

Question: In∆ABC, A= 40 and B= 60. Then the longest side of ∆ABCis (a)BC(b)AC(c)AB(d) cannot be determined Solution: (c) AB In triangle $\mathrm{ABC}$, we have: $\angle A=40^{\circ}, \angle B=60^{\circ} \quad \ldots$ (Given) Here, $\angle A+\angle B+\angle C=180^{\circ}$ $\Rightarrow 60^{\circ}+40^{\circ}+\angle C=180^{\circ}$ $\Rightarrow \angle C=80^{\circ}$ $\therefore$ The side opposite to $\angle C$, i.e., $\mathrm{AB}$, is the longest side of triangle $\mathrm{ABC}$....

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If $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$, then $x=$

[question] Question. If $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$, then $x=$ (a) 5 (b) 1/5 (c) 5/14 (d) 14/5 [/question] [solution] Solution: (b) $\frac{1}{5}$ We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$. Now, $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$ $\Rightarrow \tan ^{-1}\left(\frac{3+x}{1-3 x}\right)=\tan ^{-1} 8$ $\Rightarrow \frac{3+x}{1-3 x}=8$ $\Rightarrow 3+x=8-24 x$ $\Rightarrow 3-8=-24 x-x$ $\Rightarrow-5=-25 x$ $\Rightarrow x=\frac{5}{25}=\frac{1}{5}$ [/soluti...

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How many numbers of two digit are divisible by 3?

Question: How many numbers of two digit are divisible by 3? Solution: The two digit numbers that are divisible by 3 are: 12, 15, 18...96, 99 This is an A.P. whose first term is 12 and the common difference is 3. We have : $a_{n}=99$ $\Rightarrow 12+(n-1) 3=99$ $\Rightarrow(n-1) 3=87$ $\Rightarrow(n-1)=29$ $\Rightarrow n=30$ Thus, there are 30 such terms....

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Write the difference between maximum and minimum values

Question: Write the difference between maximum and minimum values of $\sin ^{-1} x$ for $x \in[-1,1]$. Solution: The maximum value of $\sin ^{-1} x$ in $x \in[-1,1]$ is at 1 . So, the maximum value is $\sin ^{-1}(1)$ $=\sin ^{-1}\left(\sin \frac{\pi}{2}\right)$ $=\frac{\pi}{2}$ Again, the minimum value is at-1.Thus, the minimum value is $\sin ^{-1}(-1)=-\sin ^{-1}(1)$ $=-\sin ^{-1}\left(\frac{\pi}{2}\right)$ $=-\frac{\pi}{2}$ So, the difference between the maximum and the minimum value is $\frac...

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In the given figure, a circle touches all the four sides of a quadrilateral

Question: In the given figure, a circle touches all the four sides of a quadrilateralABCDwithAB= 6 cm,BC= 7 cm andCD= 4 cm. FindAD. Solution: The figure given in the question is below. 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 the following, SA = AP For our convenience, let us representSAandAPbya PB = BQ Let us representPBandBQbyb QC = CR Let us representQCandCRbyc DR = DS Let us representDRandDSbyd ...

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It is given that ∆ABC ≅ ∆ FDE in which AB = 5 cm, ∠B = 40°, ∠A = 80°

Question: It is given that∆ABC ∆FDEin whichAB= 5 cm, B= 40,A= 80 andFD= 5 cm. Then, which of the following is true?(a) D= 60(b) E= 60(c) F= 60(d)D= 80 Solution: (b) $\angle E=60^{\circ}$ $\Delta A B C \cong \Delta F D E$ $\mathrm{AB}=5 \mathrm{~cm}, \angle B=40^{\circ}, \angle A=80^{\circ}$ and $\mathrm{FD}=5 \mathrm{~cm}$ Then $\angle A+\angle B+\angle C=180^{\circ}$ $\Rightarrow 80^{\circ}+40^{\circ}+\angle C=180^{\circ}$ $\Rightarrow \angle C=60^{\circ}$ Also, $\angle C=\angle E$ $\therefore ...

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Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22.

Question: Find the second term andnthterm of an A.P. whose 6th term is 12 and the 8th term is 22. Solution: Given; $a_{6}=12$ $\Rightarrow a+(6-1) d=12$ $\Rightarrow a+5 d=12$ ...(i) $a_{8}=22$ $\Rightarrow a+(8-1) d=22$ $\Rightarrow a+7 d=22$ ...(ii) Solving (i) and (ii), we get: $2 d=10$ $\Rightarrow d=5$ Putting the value of $d$ in (i), we get: $a+5 \times 5=12$ $\Rightarrow a=12-25=-13$ $\therefore a_{2}=a+(2-1) d=a+d=-13+5=-8$ Also, $a_{n}=a+(n-1) d$ $=-13+(n-1) 5$ $=-13+5 n-5$ $=5 n-18$...

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If Δ ABC is isosceles with AB = AC and C (O, r) is

Question: If ΔABCis isosceles withAB=ACandC(O,r) is the incircle of the ΔABCtouchingBCatL,prove thatLbisectsBC. Solution: Let us first put the given data in the form of a diagram. It is given that triangleABCis isosceles with AB = AC (1) By looking at the figure we can rewrite the above equation as, AM + MB = AN + NC From the property of tangents we know that the length of two tangents drawn to a circle from the same external point will be equal. Therefore, AM = AN Let us substituteANwithAMin th...

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Two tangent segments PA and PB are drawn to a circle

Question: Two tangent segmentsPAand PB are drawn to a circle with centreOsuch that APB= 120. Prove that OP = 2AP. Solution: Let us first put the given data in the form of a diagram. We have, Considerand. We have, Here, PO is the common side. PA = PB (Length of two tangents drawn from the same external point will be equal) OA = OB(Radii of the same circle) By SSS congruency, we haveis congruent to. Therefore, It is given that, That is, $\angle A P O+\angle B P O=120^{\circ}$ $2 \angle A P O=120^{...

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Write the value

Question: Write the value of $\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\cos ^{-1}\left(\frac{-1}{2}\right)$ Solution: $\sin ^{-1}(-x)=-\sin ^{-1} x, x \in[-1,1]$ $\cos ^{-1}(-x)=\pi-\cos ^{-1} x, x \in[-1,1]$ $\therefore \sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)+\cos ^{-1}\left(-\frac{1}{2}\right)$ $=-\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\pi-\cos ^{-1}\left(\frac{1}{2}\right)$ $=-\sin ^{-1}\left(\sin \frac{\pi}{3}\right)+\pi-\cos ^{-1}\left(\cos \frac{\pi}{3}\right)$ $=-\frac{\pi}{3}+\pi...

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In ∆ABC, if ∠C > ∠B, then

Question: In∆ABC, if C B, then(a)BCAC(b)ABAC(c)ABAC(d)BCAC Solution: (b) $A BA C$ In $\Delta A B C$, we have: $\angle C\angle B$ The side opposite to the greater angle is larger. $\therefore A BA C$...

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The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1.

Question: The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference. Solution: Given: Let the first term of the A.P. beaand the common difference bed. $a_{4}=3 a$ $\Rightarrow a+(4-1) d=3 a$ $\Rightarrow a+3 d=3 a$ $\Rightarrow 3 d=2 a$ $\Rightarrow a=\frac{3 d}{2} \quad \ldots(\mathrm{i})$ And, $a_{7}-2 a_{3}=1$$\Rightarrow a+(7-1) d-2[a+(3-1) d]=1$ $\Rightarrow \mathrm{a}+6 d-2(a+2 d)=1$ $\Rightarrow a+6 d...

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Two sides of a triangle are of length 4 cm and 2.5 cm.

Question: Two sides of a triangle are of length 4 cm and 2.5 cm. The length of the third side of the triangle cannot be(a) 6 cm(b) 6.5 cm(c) 5.5 cm(d) 6.3 cn Solution: Since, 4 + 2.5 = 6.5So, 6.5 cm cannot be the third side of the triangle, as the sum of two sides of a triangle is always greater than the third side.Hence, the correct option is (b)....

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From a point P, two tangents PA and PB are drawn to a circle with centre O.

Question: From a pointP, two tangentsPAandPBare drawn to a circle with centreO. IfOP= diameter of the circle, show that ΔAPBis equilateral. Solution: Let us first put the given data in the form of a diagram. Considerand. We have, POis the common side for both the triangles. PA = PB(Tangents drawn from an external point will be equal in length) OB = OA(Radii of the same circle) Therefore, by SSS postulate of congruency, we have $\triangle \mathrm{POA} \cong \triangle \mathrm{POB}$ Hence, $\angle ...

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The value of

Question: The value of $\cot ^{-1}(-x)$ for all $x \in R$ in terms of $\cot ^{-1} x$ is_________________. Solution: We know $\cot ^{-1}(-x)=\pi-\cot ^{-1} x$, for all $x \in \mathrm{R}$ The value of $\cot ^{-1}(-x)$ for all $x \in \mathrm{R}$ in terms of $\cot ^{-1} x$ is $\pi-\cot ^{-1} x$...

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In ∆ABC, ∠C = ∠A, BC = 4 cm and AC = 5 cm. Then, AB = ?

Question: In∆ABC, C= A, BC =4 cm andAC= 5 cm. Then,AB= ?(a) 4 cm(b) 5 cm(c) 8 cm(d) 2.5 cm Solution: Given: In∆ABC, C= A, BC =4 cm andAC= 5 cm.In∆ABC,As, C= A (Given) Therefore, $B C=A B$ (Sides opposite to equal angles.) $\Rightarrow B C=A B=4 \mathrm{~cm}$ Hence, the correct option is (a)....

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Find the 12th term from the end of the following arithmetic progressions:

Question: Find the 12th term from the end of the following arithmetic progressions: (i) 3, 5, 7, 9, ... 201 (ii) 3, 8, 13, ..., 253 (iii) 1, 4, 7, 10, ..., 88 Solution: (i) 3, 5, 7, 9...201 Consider the given progression with 201 as the first term and 2 as the common difference. 12th term from the end $=201+(12-1)(-2)=179$ (ii) 3, 8, 13...253 Consider the given progression with 253 as the first term and 5 as the common difference. 12th term from the end $=253+(12-1)(-5)=198$ (iii) 1, 4, 7, 10......

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If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... find n.

Question: If the nth term of the A.P. 9, 7, 5, ... is same as the nth term of the A.P. 15, 12, 9, ... findn. Solution: Given: nth term of the A.P. 9, 7, 5... is the same as the nth term of the A.P. 15, 12, 9... Considering $9,7,5 \ldots$ $a=9, d=(7-9)=-2$ $\mathrm{n}^{\text {th }}$ term $=9+(n-1)(-2) \quad\left[a_{n}=a+(n-1) d\right]$ $=9-2 n+2$ $=11-2 n \quad \ldots(\mathrm{i})$ Considering $15,12,9, \ldots$ $a=15, d=(12-15)=-3$ $n^{\text {th }}$ term $=15+(n-1)(-3) \quad\left[a_{n}=a+(n-1) d\r...

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The result

Question: The result $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ is true when value of $x y$ is__________________. Solution: We know $\tan ^{-1} x-\tan ^{-1} y= \begin{cases}\tan ^{-1}\left(\frac{x-y}{1+x y}\right), \text { if } x y-1 \\ \pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), \text { if } x0, y0, x y-1 \\ -\pi+\tan ^{-1}\left(\frac{x-y}{1+x y}\right), \text { if } x0, y0, x y-1\end{cases}$ Thus, $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$ ...

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In ∆ABC, BC = AB and ∠B = 80°. Then, ∠A = ?

Question: In∆ABC,BC=ABand B= 80. Then, A= ?(a) 50(b) 40(c) 100(d) 80 Solution: Given:In∆ABC,BC=ABand B= 80.In∆ABC, As, $A B=B C$ $\Rightarrow \angle A=\angle C$ Let $\angle A=\angle C=x$ Using angle sum property of a triangle, $\angle A+\angle B+\angle C=180^{\circ}$ $\Rightarrow x+80^{\circ}+x=180^{\circ}$ $\Rightarrow 2 x=180^{\circ}-80^{\circ}$ $\Rightarrow 2 x=100^{\circ}$ $\Rightarrow x=\frac{100^{\circ}}{2}$ $\Rightarrow x=50^{\circ}$ $\Rightarrow \angle A=50^{\circ}$ Hence, the correct op...

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