The angles of a triangle are arranged in ascending order of magnitude.
Question: The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is 10, find the three angles. Solution: Given that, The difference between two consecutive angles is10 Let x, x+10, x+20be the consecutive angles that differ by10 We know that, Sum of all angles in a triangle is180 x + x + 10 + x + 20=180 3x + 30=180 ⇒ 3x =180 -30 ⇒ 3x =150 ⇒ x =50 Therefore, the required angles are x =50 x + 10=50+10=60 x + 20=50+20=70 As the differe...
Read More →The angles of a triangle are
Question: The angles of a triangle are (x 40),(x 20) and (12x 10). Find the value of x. Solution: Given that, The angles of a triangle are (x 40),(x 20) and (1/2x 10) We know that, Sum of all angles of triangle is180 (x 40) + (x 20) + (1/2x 10) = 180 2x + 1/2x 70 = 180 5/2 x = 180 + 70 5x = 2(250) x = 500/5 x = 100...
Read More →Show that
Question: $\frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$ Solution: Let $5 x+3=A \frac{d}{d x}\left(x^{2}+4 x+10\right)+B$ $\Rightarrow 5 x+3=A(2 x+4)+B$ Equating the coefficients ofxand constant term, we obtain $2 A=5 \Rightarrow A=\frac{5}{2}$ $4 A+B=3 \Rightarrow B=-7$ $\therefore 5 x+3=\frac{5}{2}(2 x+4)-7$ $\Rightarrow \int \frac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \frac{\frac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} d x$ $=\frac{5}{2} \int \frac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \frac{1}{\sqrt{x^{2}...
Read More →If the angles of a triangle are in the ratio 1: 2: 3, determine three angles.
Question: If the angles of a triangle are in the ratio 1: 2: 3, determine three angles. Solution: Given that, Angles of a triangle are in the ratio 1: 2: 3 Let the angles be x, 2x, 3x We know that, Sum of all angles of triangles is180 x + 2x + 3x =180 ⇒ 6x =180 ⇒ x = 180/6 ⇒ x = 30 Since x = 30 2x = 2(30)=60 3x = 3(30)=90 Therefore, angles are 30, 60, 90...
Read More →In a ΔABC, if ∠A = 55°, ∠B = 40°, Find ∠C.
Question: In a ΔABC, if A = 55, B = 40, Find C. Solution: Given Data: B = 55, B = 40, then C = ? We know that In a ΔABCsum of all angles of a triangle is 180 i.e.,A + B + C = 180 ⇒ 55 + 40 + C = 180 ⇒ 95 +C = 180 ⇒ C = 180 95 ⇒ C = 850...
Read More →A relation R is defined from [2, 3, 4, 5] to [3, 6, 7, 10]
Question: A relation R is defined from [2, 3, 4, 5] to [3, 6, 7, 10] by :xRy⇔xis relatively prime toy. Then, domain of R is(a) [2, 3, 5] (b) [3, 5] (c) [2, 3, 4] (d) [2, 3, 4, 5] Solution: (d) {2, 3, 4, 5}Given: From {2, 3, 4, 5} to {3, 6, 7, 10},xRy⇔x is relatively prime to y 2 is relatively prime to 3,7 3 is relatively prime to 7,10 4 is relatively prime to 3,7 5 is relatively prime to 3,6,7So, domain of R is {2,3,4,5}...
Read More →Solve each of the following systems of equations by the method of cross-multiplication :
Question: Solve each of the following systems of equations by the method of cross-multiplication : $\frac{5}{x+y}-\frac{2}{x-y}=-1$ $\frac{15}{x+y}+\frac{7}{x-y} 10$ where $x \neq 0$ and $y \neq 0$ Solution: GIVEN: $\frac{5}{x+y}-\frac{2}{x-y}=-1$ $\frac{15}{x+y}+\frac{7}{x-y}=10$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation Rewriting the equation again $\frac{5}{x+y}-\frac{2}{x-y}+1=0$ $\frac{15}{x+y}+\fra...
Read More →Fill in the blanks in each of the following to make the statement true:
Question: Fill in the blanks in each of the following to make the statement true: (i) If two parallel lines are intersected by a transversal, then each pair of corresponding angles are ____________ (ii) If two parallel lines are intersected by a transversal, then interior angles on the same side of the transversal are _____________ (iii) Two lines perpendicular to the same line are _______ to each other (Iv) Two lines parallel to the same line are __________ to each other. (v) If a transversal i...
Read More →If R =
Question: If R = {(x,y) :x,y Z,x2+y2 4} is a relation on Z, then the domain of R is (a) [0, 1, 2] (b) [0, 1, 2] (c) {2, 1, 0, 1, 2] (d) None of these Solution: (c) {2, 1, 0, 1, 2} R = {(x,y) :x,y Z,x2+y2 4} We know that, $(-2)^{2}+0^{2} \leq 4$ $\Rightarrow(2)^{2}+0^{2} \leq 4$ $\Rightarrow(-1)^{2}+0^{2} \leq 4$ $\Rightarrow(-1)^{2}+0^{2} \leq 4$ $\Rightarrow(1)^{2}+0^{2} \leq 4$ $\Rightarrow 0^{2}+0^{2} \leq 4$ $\Rightarrow(1)^{2}+(1)^{2} \leq 4$ $\Rightarrow(-1)^{2}+(-1)^{2} \leq 4$ Hence, dom...
Read More →With of the following statements are true (T) and which are false (F)? Give reasons.
Question: With of the following statements are true (T) and which are false (F)? Give reasons. (1) If two lines are intersected by a transversal, then corresponding angles are equal. (ii) If two parallel lines are intersected by a transversal, then alternate interior angles are equal. (ii) Two lines perpendicular to the same line are perpendicular to each other. (iv) Two lines parallel to the same line are parallel to each other. (v) If two parallel lines are intersected by a transversal, then t...
Read More →If R =
Question: If R = {(x,y) :x,y Z,x2+y2 4} is a relation on Z, then the domain of R is (a) [0, 1, 2] (b) [0, 1, 2] (c) {2, 1, 0, 1, 2] (d) None of these Solution: (c) {2, 1, 0, 1, 2} R = {(x,y) :x,y Z,x2+y2 4} We know that, $(-2)^{2}+0^{2} \leq 4$ $\Rightarrow(2)^{2}+0^{2} \leq 4$ $\Rightarrow(-1)^{2}+0^{2} \leq 4$ $\Rightarrow(-1)^{2}+0^{2} \leq 4$ $\Rightarrow(1)^{2}+0^{2} \leq 4$ $\Rightarrow 0^{2}+0^{2} \leq 4$ $\Rightarrow(1)^{2}+(1)^{2} \leq 4$ $\Rightarrow(-1)^{2}+(-1)^{2} \leq 4$ Hence, dom...
Read More →Prove
Question: $\frac{x+3}{x^{2}-2 x-5}$ Solution: Let $(x+3)=A \frac{d}{d x}\left(x^{2}-2 x-5\right)+B$ $(x+3)=A(2 x-2)+B$ Equating the coefficients ofxand constant term on both sides, we obtain $2 A=1 \Rightarrow A=\frac{1}{2}$ $-2 A+B=3 \Rightarrow B=4$ $\therefore(x+3)=\frac{1}{2}(2 x-2)+4$ $\Rightarrow \int \frac{x+3}{x^{2}-2 x-5} d x=\int \frac{\frac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x$ $=\frac{1}{2} \int \frac{2 x-2}{x^{2}-2 x-5} d x+4 \int \frac{1}{x^{2}-2 x-5} d x$ Let $I_{1}=\int \frac{2 x-2}{...
Read More →In the below fig, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF Prove that ∠ABC + ∠DEP = 180°
Question: In the below fig, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF Prove that ABC + DEP = 180 Solution: Given: AB ∥ DE, BC ∥ EF To prove:ABC + DEF = 180 Construction: Produce BC to intersect DE at M Proof: Since AB || EM and BL is the transversal ABC = EML [Corresponding angle] ... (i) Also, EF || ML and EM is the transversal By the property of co-interior angles are supplementary DEF + EML = 180 .... (ii) From (i) and (ii) we have Therefore DEF + ABC = 180...
Read More →In the below fig, arms BA and BC of ∠ABC are respectively parallel to arms ED and EF of ∠DEF.
Question: In the below fig, arms BA and BC of ABC are respectively parallel to arms ED and EF of DEF. Prove that ABC = DEF. Solution: Given AB ∥ DE and BC ∥ EF To Prove:ABC = DEF Construction: Produce BC to x such that it intersects DE at M. Proof: Since AB ∥ DE and BX is the transversal ABC = DMX [Corresponding angle] .... (i) Also, BX ∥ EF and DE is the transversal DMX = DEF [Corresponding angles] -----(ii) From (i) and (ii) ABC = DEF...
Read More →Solve each of the following systems of equations by the method of cross-multiplication :
Question: Solve each of the following systems of equations by the method of cross-multiplication : $\frac{x}{a}=\frac{y}{b}$ $a x+b y=a^{2}+b^{2}$ Solution: GIVEN: $\frac{x}{a}=\frac{y}{b}$ $a x+b y=a^{2}+b^{2}$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $\frac{x}{a}-\frac{y}{b}=0$ $a x+b y-\left(a^{2}+b^{2}\right)=0$ By cross multiplication method we get $\frac{-y}{\left(\frac{-\left(a^{2}+b^{2}\right)}{...
Read More →If I, m, n are three lines such that I∥ m and n perpendicular to l, prove that n perpendicular to m.
Question: If I, m, n are three lines such that I∥ m and n perpendicular to l, prove that n perpendicular to m. Solution: Given, l ∥ m, n perpendicular to I To prove: n perpendicular to m Since l ∥ m and n intersects 1 = 2 [Corresponding angles] But, U = 90 ⟹2 = 90 Hence n is perpendicular to m...
Read More →Which pair of lines in the below fig. is parallel? Give reasons.
Question: Which pair of lines in the below fig. is parallel? Give reasons. Solution: A + B = 115 + 65 = 180 Therefore, AB ∥ BC [As sum of co interior angles are supplementary] B + C = 65 + 115 = 180 Therefore, AB ∥ CD (As sum of interior angles are supplementary]...
Read More →If A = [1, 2, 3], B = [1, 4, 6, 9] and R is a relation from A to B defined by 'x' is greater than y.
Question: If A = [1, 2, 3], B = [1, 4, 6, 9] and R is a relation from A to B defined by 'x' is greater thany. The range of R is(a) {1, 4, 6, 9} (b) (4, 6, 9) (c) [1] (d) none of these. Solution: (c) {1} A = {1, 2, 3} and B = {1, 4, 6, 9} R is a relation from A to B defined by:xis greater thany. Then R = {(2,1),(3,1)} Range (R) = {1}...
Read More →In the below fig. transversal t intersects two lines m and n, ∠4 = 110° and ∠7 = 65° is m ∥ n?
Question: In the below fig. transversal t intersects two lines m and n, 4 = 110 and 7 = 65 is m ∥ n? Solution: Given: 4 = 110and7 = 65 To find: is m ∥ n Here7 = 5 = 65 [Vertically opposite angle] Now.4 + 5 = 110 + 65 = 175 Therefore, m is not parallel to n as the pair of co interior angles is not supplementary....
Read More →Prove
Question: $\frac{x+2}{\sqrt{x^{2}+2 x+3}}$ Solution: $\int \frac{(x+2)}{\sqrt{x^{2}+2 x+3}} d x=\frac{1}{2} \int \frac{2(x+2)}{\sqrt{x^{2}+2 x+3}} d x$ $=\frac{1}{2} \int \frac{2 x+4}{\sqrt{x^{2}+2 x+3}} d x$ $=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\frac{1}{2} \int \frac{2}{\sqrt{x^{2}+2 x+3}} d x$ $=\frac{1}{2} \int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x+\int \frac{1}{\sqrt{x^{2}+2 x+3}} d x$ Let $I_{1}=\int \frac{2 x+2}{\sqrt{x^{2}+2 x+3}} d x$ and $I_{2}=\int \frac{1}{\sqrt{x^{2...
Read More →Solve each of the following systems of equations by the method of cross-multiplication :
Question: Solve each of the following systems of equations by the method of cross-multiplication : $\frac{x}{a}+\frac{y}{b}=a+b$ $\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$ Solution: GIVEN: $\frac{x}{a}+\frac{y}{b}=a+b$ $\frac{x}{a^{2}}+\frac{y}{b^{2}}=2$ To find: The solution of the systems of equation by the method of cross-multiplication: Here we have the pair of simultaneous equation $\frac{x}{a}+\frac{y}{b}-(a+b)=0$ $\frac{x}{a^{2}}+\frac{y}{b^{2}}-2=0$ By cross multiplication method we get $\frac{...
Read More →Let A = [1, 2, 3], B = [1, 3, 5].
Question: Let A = [1, 2, 3], B = [1, 3, 5]. If relation R from A to B is given by = {(1, 3), (2, 5), (3, 3)}, Then R1is(a) {(3, 3), (3, 1), (5, 2)} (b) {(1, 3), (2, 5), (3, 3)} (c) {(1, 3), (5, 2)} (d) None of these Solution: (a) {(3, 3), (3, 1), (5, 2)} A = {1, 2, 3}, B ={1, 3, 5} R = {(1, 3), (2, 5), (3, 3)} R1= {(3,1),(5,2),(3,3)}...
Read More →In the below figure, p is a transversal to lines m and
Question: In the below figure, p is a transversal to lines m and n, 2 = 120 and 5 = 60. Prove that m|| n. Solution: Given that 2 = 120and5 = 60 To prove, 2 + 1 = 180 [Linear pair] 120 + 1 = 180 1 = 180 120 1 = 60 Since1 = 5 = 60 Therefore, m ∥ n [As pair of corresponding angles are equal]...
Read More →Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, find the measures of ∠AOC, ∠COB, ∠BOD, ∠DOA
Question: Two lines AB and CD intersect at O. If AOC + COB + BOD = 270, find the measures of AOC, COB, BOD, DOA Solution: Given:AOC + COB + BOD = 270 To find:AOC, COB, BOD, DOA Here,AOC + COB + BOD = 270 [Complete angle] ⟹ 270 + AOD = 360 ⟹ AOD = 360 - 270 ⟹ AOD = 90 Now, AOD + BOD = 180 [Linear pair] 90 + BOD = 180 ⟹ BOD = 180 - 90 ⟹ BOD = 90 AOD = BOC = 90 [Vertically opposite angles] BOD = AOC = 90 [Vertically opposite angles]...
Read More →The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60°. Find the other angles.
Question: The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is 60. Find the other angles. Solution: Given AB ∥ CD AD ∥ BC Since AB ∥ CD and AD is the transversal Therefore, A + D = 180 (Co-interior angles are supplementary) 60 + D = 180 D = 180 - 60 D = 120 Now. AD ∥ BC and AB is the transversal A + B = 180 (Co-interior angles are supplementary) 60 +B = 180 B = 180 - 60 = 120 Hence, A = C = 60andB = D = 120...
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