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Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Question: Insert five numbers between 8 and 26 such that the resulting sequence is an A.P. Solution: Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ be five numbers between 8 and 26 . Letdbe the common difference. Then, we have: 26 =a7 $\Rightarrow 26=8+(7-1) d$ $\Rightarrow 26=8+6 d$ $\Rightarrow d=3$ $A_{1}=8+d=8+3=11$ $A_{2}=8+2 d=8+6=14$ $A_{3}=8+3 d=8+9=17$ $A_{4}=8+4 d=8+12=20$ $A_{5}=8+5 d=8+15=23$ Therefore, the five numbers are 11, 14, 17, 20, 23....

Evaluate:

Question: Evaluate: (i) $\cos \left\{\sin ^{-1}\left(-\frac{7}{25}\right)\right\}$ (ii) $\sec \left\{\cot ^{-1}\left(-\frac{5}{12}\right)\right\}$ (iii) $\cot \left\{\sec ^{-1}\left(-\frac{13}{5}\right)\right\}$ Solution: (i) $\cos \left\{\sin ^{-1}\left(-\frac{7}{25}\right)\right\}=\cos \left\{-\sin ^{-1}\left(\frac{7}{25}\right)\right\}$ $=\cos \left\{\sin ^{-1}\left(\frac{7}{25}\right)\right\}$ $=\cos \left\{\cos ^{-1} \sqrt{1-\left(\frac{7}{25}\right)^{2}}\right\}$ $=\cos \left\{\cos ^{-1} \...

If x, y, z are in A.P.

Question: Ifx,y,zare in A.P. andA1is the A.M. ofxandyandA2is the A.M. ofyandz, then prove that the A.M. ofA1andA2isy. Solution: x, y, zare in A.P. $\therefore y=\frac{x+z}{2}$ Now, $A_{1}$ is the arithmetic mean of $\mathrm{x}$ and $\mathrm{y}$. $A_{1}=\frac{x+y}{2}=\frac{x+\frac{x+z}{2}}{2}=\frac{3 x+z}{4}$ And, $A_{2}$ is the arithmetic mean of $y$ and $z$. $A_{2}=\frac{y+z}{2}=\frac{\frac{x+z}{2}+z}{2}=\frac{3 z+x}{4}$ Let $A_{3}$ be the arithmetic mean of $A_{1}$ and $A_{2}$. $A_{3}=\frac{A_...

From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are 30°

Question: From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are 30 and 45 respectively. If bridge is at the height of 30 m from the banks, find the width of the river. Solution: Let $B D$ be the width of river. And the angle of depression of the bank on opposite side of the river are $30^{\circ}$ and $45^{\circ}$ respectively. It is given that $A C=30 \mathrm{~m}$. Let $B C=x$ and $C D=y$. And $\angle A B C=30^{\circ}$, $\angle A D C=45^{...

If n A.M.s are inserted between two numbers,

Question: IfnA.M.s are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant. Solution: Let $A_{1}, A_{2} \ldots \ldots A_{n}$ be $n$ A.M.s between two numbers $a$ and $b$. Then, $a, A_{1}, A_{2} \ldots \ldots A_{n}, b$ are in A.P. with common difference, $d=\frac{b-a}{n+1}$. $\therefore A_{1}+A_{2}+\ldots \ldots+A_{n}=\frac{n}{2}\left[A_{1}+A_{n}\right]$ $=\frac{n}{2}\left[A_{1}-d+A_{n}+d\right]$ $=\frac{n}{2}[a+b]$ $=n \times\left[...

In the given figure, ABCD is a || gm and E is the mid-point of BC.

Question: In the given figure,ABCDis a || gm andEis the mid-point ofBC. Also,DEandABwhen produced meet atF. Then, (a) $A F=\frac{3}{2} A B$ (b) $A F=2 A B$ (c) $A F=3 A B$ (d) $A F^{2}=2 A B^{2}$ Solution: (b)AF= 2ABExplanation:In parallelogramABCD,we have:AB||DCDCE= EBF (Alternate interior angles)In∆DCEand ∆BFE, we have:DCE=EBF (Proved above) DEC=BEF (Vertically opposite angles)BE=CE ( Given)i.e., ∆DCE ∆BFE (By ASA congruence rule)DC=BF (CPCT)ButDC= AB, asABCDis a parallelogram.DC=AB= BF......

Solve this

Question: Solve: $\cos \left\{2 \sin ^{-1}(-x)\right\}=0$ Solution: Given, $\cos \left\{2 \sin ^{-1}(-x)\right\}=0$ $\Rightarrow \cos \left(-2 \sin ^{-1} x\right)=0$ $\left[\because \sin ^{-1}(-\theta)=-\sin ^{-1} \theta\right]$ $\Rightarrow \cos \left(2 \sin ^{-1} x\right)=0$ $[\because \cos (-\theta)=\cos \theta]$ We know, $-\frac{\pi}{2} \leq \sin ^{-1} \theta \leq \frac{\pi}{2}$ Therefore, $2 \sin ^{-1} x=\pm \frac{\pi}{2}$ $\Rightarrow \sin ^{-1} x=\pm \frac{\pi}{4}$ $\Rightarrow x=\pm \fra...

The angle of elevation of the top of the building from the foot of the tower is 30°

Question: The angle of elevation of the top of the building from the foot of the tower is 30 and the angle of the top of the tower from the foot of the building is 60. If the tower is 50 m high, find the height of the building. Solution: LetADbe the building of heighthm. and an angle of elevation of top of building from the foot of tower is 30 and an angle of the top of tower from the foot of building is 60. LetAD = h,AB = xandBC= 50 and, So we use trigonometric ratios. In a triangle, $\Rightarr...

Insert A.M.s between 7 and 71 in such a way that the 5th A.M. is 27.

Question: Insert A.M.s between 7 and 71 in such a way that the 5thA.M. is 27. Find the number of A.M.s. Solution: Let $A_{1}, A_{2}, A_{3}, A_{4}, 27, A_{6} \ldots A_{n}$ be the $n$ arithmetic means between 7 and 71 . Thus, there are $(n+2)$ terms in all. Thus, there are $(n+2)$ terms in all. Letdbe the common difference of the above A.P. Now,a6= 27 $\Rightarrow a+(6-1) d=27$ $\Rightarrow a+5 d=27$ $\Rightarrow d=4$ Also, $71=a_{n+2}$ $71=7+(n+2-1) 4$ $\Rightarrow 71=7+(n+1) 4$ $\Rightarrow n=15...

If area of a || gm with sides a and b is A and that of a rectangle with sides a and b is B, then

Question: If area of a || gm with sidesaandbisAand that of a rectangle with sidesaandbisB, then(a)AB(b)A=B(c)AB(d)AB Solution: (c)ABExplanation:Lethbe the height of parallelogram.Then clearly,hbA=a⨯ha⨯b= BHence,AB...

In the given figure, ABCD is a parallelogram in which ∠BDC = 45°

Question: In the given figure,ABCDis a parallelogram in whichBDC= 45 and BAD= 75. Then, CBD= ?(a) 45(b) 55(c) 60(d) 75 Solution: (c) 60Explanation:BAD = BCD = 75o [Opposite angles are equal]In∆ BCD, C = 75o CBD = 180o(75o+ 45o) = 60o...

As observed from the top of a 75 m tall light house,

Question: As observed from the top of a 75 m tall light house, the angle of depression of two ships are 30 and 45. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. Solution: Let $O C$ be the height of light house $75 \mathrm{~m}$. and $A$ and $B$ the position of two ships and angle of depression are $A=30^{\circ}$ and $B=45^{\circ}$. Let $O C=75$ and $B C=h, A B=x$ Here we have to find distance between two ships. The correspondi...

If APB and CQD are two parallel lines, then the bisectors of ∠APQ,

Question: IfAPBandCQDare two parallel lines, then the bisectors ofAPQ, BPQ, CQPand PQDenclose a(a) square(b) rhombus(c) rectangle(d) kite Solution: (c) RectangleIfAPBandCQDare two parallel lines, then the bisectors ofAPQ, BPQ, CQPand PQD enclose a rectangle....

Which of the following is not true for a parallelogram?

Question: Which of the following is not true for a parallelogram?(a) Opposite sides are equal.(b) Opposite angles are equal.(c) Opposite angles are bisected by the diagonals.(d) Diagonals bisect each other. Solution: (c)Opposite angles are bisected by the diagonals....

If ∠A, ∠B, ∠C and ∠D of a quadrilateral ABCD taken in order, are in the ratio 3 : 7 : 6 : 4

Question: IfA, B, Cand Dof a quadrilateralABCDtaken in order, are in the ratio 3 : 7 : 6 : 4 thenABCDis a(a) rhombus(b) kite(c) trapezium(d) parallelogram Solution: (c) TrapeziumExplanation:Let the angles be (3x), (7x), (6x) and (4x).Then 3x+ 7x+ 6x+4x= 360ox= 18oThus, the angles are 3 ⨯18o= 54o,7⨯ 18o= 126o, 6⨯ 18o= 108oand 4⨯18o= 72o.But 54o+ 126o= 180oand 72o+ 108o = 180oABCDis a trapezium....

From the top of a 7 m high building,

Question: From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 and the angle of depression of its foot is 45. Determine the height of the tower. Solution: LetOCbe the tower of heightHm andm high building makes an angle of elevation of top of cable wire isand an angle of depression from the its foot is. Let $B C=x, A D=x$ and $C D=7, A B=7$ and $\angle O A D=60^{\circ}, \angle A C B=45^{\circ}$ So we use trigonometric ratios. In a triangle $A B C$, $\Right...

If one angle of a parallelogram is 24° less than twice the smallest angle,

Question: If one angle of a parallelogram is 24 less than twice the smallest angle, then the largest angle of the parallelogram is(a) 68(b) 102(c) 112(d) 136 Solution: (c)112Explanation:LetABCDis a parallelogram. A= CandB= D (Opposite angles)LetAbe the smallest angle whose measure isx.B = (2x 24)oNow, A+ B=180o (Adjacent angles are supplementary) ⇒x+2x 24o=180o ⇒3x= 204o ⇒x=68oB= 2 ⨯ 68o 24o= 112oHence,A=C=68oandB=D=112o...

There are n A.M.s between 3 and 17. The ratio of the last mean to the first mean is 3 : 1.

Question: There arenA.M.s between 3 and 17. The ratio of the last mean to the first mean is 3 : 1. Find the value ofn. Solution: Let $A_{1}, A_{2}, A_{3}, A_{4} \ldots A_{n}$ be the $n$ arithmetic means between 3 and 17 . Let $d$ be the common difference of the A.P. $3, A_{1}, A_{2}, A_{3}, A_{4} \ldots A_{n}$ and17. Then, we have: $d=\frac{17-3}{n+1}=\frac{14}{n+1}$ Now, $A_{1}=3+d=3+\frac{14}{n+1}=\frac{3 n+17}{n+1}$ And, $A_{n}=3+n d=3+n\left(\frac{14}{n+1}\right)=\frac{17 n+3}{n+1}$ $\theref...

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is

Question: If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is(a) 108(b) 54(c) 72(d) 81 Solution: (c)72Explanation:LetABCDbe a parallelogram. A= Cand B= D (Opposite angles) Let $\angle A=x$ and $\angle B=\frac{2}{3} x$ $\therefore \angle A+\angle B=180^{\circ}$ (Adjacent angles are supplementary) $\Rightarrow x+\frac{2}{3} x=180^{\circ}$ $\Rightarrow \frac{5}{3} x=180^{\circ}$ $\Rightarrow x=108^{\circ}$ $\therefore \angle B=\frac{2}{3} \...

Insert six A.M.s between 15 and −13.

Question: Insert six A.M.s between 15 and 13. Solution: Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ be the 6 A.M.s between 15 and $-13$. Then, $15, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ and $-13$ are in A.P. whose common difference is as follows: $d=\frac{-13-15}{6+1}$ =-4 $A_{1}=15+d=15+(-4)=11$ $A_{2}=15+2 d=15+(-8)=7$ $A_{3}=15+3 d=15+(-12)=3$ $A_{4}=15+4 d=15+(-16)=-1$ $A_{5}=15+5 d=15+(-20)=-5$ $A_{6}=15+6 d=15+(-24)=-9$ Hence, the required A.M.s are $11,7,3,-1,-5,-9$....

Solve:

Question: Solve: $\cos \left(\sin ^{-1} x\right)=\frac{1}{6}$ Solution: $\cos \left(\sin ^{-1} x\right)=\frac{1}{6}$ $\Rightarrow \cos \left(\cos ^{-1} \sqrt{1-x^{2}}\right)=\frac{1}{6}$ $\Rightarrow \sqrt{1-x^{2}}=\frac{1}{6}$ $\Rightarrow 1-x^{2}=\frac{1}{36}$ $\Rightarrow 1-\frac{1}{36}=x^{2}$ $\Rightarrow x^{2}=\frac{35}{36}$ $\Rightarrow x=\pm \frac{\sqrt{35}}{6}$...

A T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly

Question: A T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60. From a point 20 m away this point on the same bank, the angle of elevation of the top of the tower is 30. Find the height of the tower and the width of the river. Solution: LetABbe the T.V tower of heighthm on a bank of river andDbe the point on the opposite of the river. An angle of elevation at top of tower is 60 and fr...

Insert 7 A.M.s between 2 and 17.

Question: Insert 7 A.M.s between 2 and 17. Solution: Let $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ be the seven A.M.s between 2 and 17 . Then, $2, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}, A_{7}$ and 17 are in A.P. whose common difference is as follows: $d=\frac{17-2}{7+1}$ $=\frac{15}{8}$ $A_{1}=2+d=2+\frac{15}{8}=\frac{31}{8}$ $A_{2}=2+2 d=2+\frac{15}{4}=\frac{23}{4}$ $A_{3}=2+3 d=2+\frac{45}{8}=\frac{61}{8}$ $A_{4}=2+4 d=2+\frac{15}{2}=\frac{19}{2}$ $A_{5}=2+5 d=2+\frac{75}{8}=\frac{9...

The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square, only if

Question: The figure formed by joining the midpoints of the sides of a quadrilateralABCD, taken in order, is a square, only if(a)ABCDis a rhombus(b) diagonals ofABCDare equal(c)diagonals ofABCDare perpendicular(d) diagonals ofABCDare equal and perpendicular Solution: Since,The quadrilateral formed by joining the mid-points of the sides of a rhombus is rectangle,The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonalsequal is rhombus,The quadrilateral for...

Prove the following results

Question: Prove the following results (i) $\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\frac{17}{6}$ (ii) $\cos \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)=\frac{6}{5 \sqrt{13}}$ (iii) $\tan \left(\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\right)=\frac{63}{16}$ (iv) $\sin \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)=\frac{63}{65}$ Solution: (i) LHS $=\tan \left(\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right)=\tan \left(\tan ^{-1} \frac{\...

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