Let points be A(0,0) and B(36,15)
The distance between two points is
\(AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)
=\(\sqrt{(36-0)^2+(15-0)^2}\)
=\(\sqrt{1296+225}=\sqrt{1521}=39\)

We have, 26 and 91

\(\therefore \)  Prime factors of 26 = 2 x 13
and prime factors of 91 = 7 x 13
Now, LCM of 26 and 91 = 2 x 7 x 13 = 182
and HCF of 26 and 91 = 13
For verification
LCM x HCF = 182 x 13 = 2366
and product of two numbers = 26 x 91 = 2366
Thus, LCM x HCF = Product of two numbers.

Recall the identity a² - b² = (a - b)(a + b). Using it, we can write:
x² - 3 = (x - \(\sqrt{3}\))(x + \(\sqrt{3}\))
So, the value of x² - 3 is zero when x = \(\sqrt{3}\) or x = -\(\sqrt{3}\)
Therefore, the zeroes of x² - 3 are \(\sqrt{3}\) and -\(\sqrt{3}\)
Now,
sum of zeroes = \(\sqrt{3}\) - \(\sqrt{3}\) = 0 = \(\frac{-(Coefficient \quad of \quad x)}{Coefficient \quad of \quad x^{2}}\)
product of zeroes = (\(\sqrt{3}\))(-\(\sqrt{3}\)) = -3 = \(\frac{-3}{1}=\frac{Constant \quad term}{Coefficient \quad of \quad x^{2}}\)

Given, pair of linear equations is x - y + 1 = 0 and 3x + 2y -12 = O. Table for x r- y+l = 0 or y = x + 1 is

Table for 3x+2y-12=0 or \(y=\frac {12-3x}{2}\) is

Now, plot the points A (0,1), B(4, 5), C (-1,0) and join them to get a line CB. Similarly, plot the points D(0, 6), E(2,3), F (4,0) and join them to get a line DF. 

Clearly, the two lines intersect each other at the point E (2, 3). Hence, x = 2 and y = 3 is the solution of the given pair of equations. The line DE cuts X-axis at the point F (4, 0) and the line AB cuts X-axis at the point C (-1, 0).
Hence, the coordinates of the vertices of the triangle, so formed are E(2, 3), F( 4,0) and C (-1,0).

In \(\triangle OPQ\),  \(AB\parallel PQ\) [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OB }{ BQ } \)  ... (i)
[by basic proportionality theorem]
Also, in \(\triangle OPR\), \(AC\parallel PR\)  [given]
\(\therefore \quad \frac { OA }{ AP } =\frac { OC }{ CR } \)  ... (ii)
From Eqs. (i) and (ii), 
\(\frac { OB }{ BQ } =\frac { OC }{ CR } \Rightarrow \quad BC\parallel QR\)
[by converse of basic proportionality theorem]
Hence proved.

False, let A = 60⁰ and B = 30⁰
Then, sin (A + B) = sin (60⁰ + 30⁰) = sin 90⁰ = 1
and sin A+sin B=sin 60⁰ + sin 30⁰
\(=\frac { \sqrt { 3 } }{ 2 } +\frac { 1 }{ 2 } =\frac { \sqrt { 3 } +1 }{ 2 } \)
So, sin (A + B) \(\ne\)  sin A + sin B

RS.145.20

AB²=(-2-7)²+(5-10)² = (-9)²+(-5)² =81+25 =106
BC²=(3-(-2))²+(-4-5)²=(5)²+(-9)²=25+81=106
AC²=(3-7)²+(-4-10)²=(4)^2₊(14)²=16+196=212
Since AB²+BC²=AC²
∴ ABC is a right triangle.
AB = \(\sqrt { 106 } \) and BC = \(\sqrt { 106 } \)
∵ AB = BC
∴ ABC is an isosceles right triangle. 

P(3,-1)

Given capacities of two tankers are 850 L and 680 L.
Here, 850 > 680
Now, 850 = (680 x 1) + 170
[by Euclid's division lemma]
Here, remainder = 170 \(\neq \) 0. So, new dividend is 680 and divisor is 170.
Now, 680 = (170 x 4) + 0
[by Euclid's division lemma]
Here, remainder is zero and divisor is 170.
So, the HCF of 850 and 680 is 170.
Hence, the maximum capacity of the required container is 170 L.

We have, (3 x 5 x 7) + 7 = 105 + 7 = 112
\(\therefore \)  Prime factors of 112 = 2 x 2 x 2 x 2 x 7 = 2⁴ x 7
So, it is the product of prime factors 2 and 7.
Hence, it is a composite number.

Here, n is a natural number and let 6ⁿ ends with 0.
Hence, 6ⁿ is divisible by 5.
But the prime factors of 6 are 2 and 3, so 5 is not a factor.
\(\Rightarrow \)  6ⁿ = (2 x 3)ⁿ
In the prime factorisation of 6ⁿ , 5 is not a factor.
By using the fundamental theorem of arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order, in which the prime factors occur.
So, our assumption, 6ⁿ ends with 0, is wrong.
Thus, there does not exist any natural number n, for which 6ⁿ ends with zero.

Given, f(x)=ax+b
For zero of f(x), put p(x) = 0⇒ ax+b=0
⇒ x=b/a
So, the zero of f(x) is -b/a.

The given system can be written as
ax+by=a-b   ...(i)
bx-ay=a+b   ...(ii)
From Eq. (i), we get
by=a-b-ax
\(\Rightarrow \quad y=\frac { a-b-ax }{ b } \quad \quad ...(iii)\)
On substituting the value of y in Eq. (ii), we get
\(bx-a\left[ \frac { a-b-ax }{ b } \quad \right] =a+b\)
\(\Rightarrow\) b²x-a(a-b-ax)=b(a+b) [multiplying both sides by b]
\(\Rightarrow\) (b²+a²)x=ab+b²+a²-ab \(\Rightarrow \quad x=\frac { { a }^{ 2 }+{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } } =1\)
On substituting x=1 in Eq. (iii), we get
\(y=\frac { a-b-a }{ b } \quad \Rightarrow \quad y=\frac { -b }{ b } =-1\)
Hence, solution of the given system is x=1 and y=-1

Draw the figure according to the question and get two triangles. Then, show both triangles are similar by AAA similarity criterion and then calculate the required distance.
She is at 9 m from the base of the pole.

Let us join AC to intersect EF at G

AB || DC and EF || AB (Given)
So, EF || DC (Lines parallel to the same line are parallel to each other)
Now, in \(\Delta\) ADC,
EG || DC (As EF || DC)
So, \(\frac{AE}{ED}=\frac{AG}{GC}\)
Similarly, from  \(\Delta\)CAB,
\(\begin{aligned} & \frac{C G}{A G}=\frac{C F}{B F} \\ \end{aligned}\)
\(\begin{aligned} & \frac{A G}{G C}=\frac{B F}{F C} \end{aligned}\)
Therefore, from (1) and (2),
\(\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}\)

Since we will apply the identity involving sec \(\theta\) and tan \(\theta\), let us first convert the LHS (of the identity we need to prove) in terms of sec \(\theta\) and tan \(\theta\) by dividing numerator and denominator by cos \(\theta\).
\(\begin{aligned} \text { LHS } & =\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{\tan \theta-1+\sec \theta}{\tan \theta+1-\sec \theta} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{(\tan \theta+\sec \theta)-1}{(\tan \theta-\sec \theta)+1}=\frac{\{(\tan \theta+\sec \theta)-1\}(\tan \theta-\sec \theta)}{\{(\tan \theta-\sec \theta)+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\left(\tan ^2 \theta-\sec ^2 \theta\right)-(\tan \theta-\sec \theta)}{\{\tan \theta-\sec \theta+1\}(\tan \theta-\sec \theta)} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{-1-\tan \theta+\sec \theta}{(\tan \theta-\sec \theta+1)(\tan \theta-\sec \theta)} \end{aligned}\)
\(=\frac{-1}{\tan \theta-\sec \theta}=\frac{1}{\sec \theta-\tan \theta}\)
which is the RHS of the identity, we are required to prove.

Given, a pair of linear equations is :
3x - y = 3 or y = 3x - 3   ....(i)
and 9x-3y = 9  ....(ii)
On substituting y from eqn. (i) in eqn. (ii),
9x - 3(3x- 3) = 9
i.e., 9 = 9
It is a true statement. Hence, eqn. (i) and (ii) have infinitely many solutions.

We have: 6 = 2 \(\times\) 3, 72 = 2³ \(\times\) 3², 120 = 2³ \(\times\) 3 \(\times\) 5
Here, 2¹ and 3¹ are the smallest powers of the common factors 2 and 3, respectively.
So, HCF (6, 72, 120) = 2¹ \(\times\) 3¹ = 2 \(\times\) 3 = 6
2³, 3² and 5¹ are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers.
So,    LCM (6, 72, 120) = 2³ \(\times\) 3² \(\times\) 5¹ = 360
Remark : Notice, 6 × 72 × 120 ≠ HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numbers is not equal to the product of their HCF and LCM.

 Let a = 7\(\sqrt5\) be a rational number.
\(\Rightarrow \frac{a}{7}=\sqrt{5}\)
Now, \(\frac{a}{7}\) is a rational number since product of two rational number is a rational number.
The above will imply that \(\sqrt5\) is a rational number. But \(\sqrt5\)  is an irrational number.
his contradicts our assumption. Therefore we can conclude that 7\(\sqrt5\) is an irrational number and hence the result.

Let p(t) = t² - 15 = t² - \(\left ( \sqrt{15} \right )^{2}\)
= (t - \(\sqrt{15}\))(t + \(\sqrt{15}\))     [\(\because\) a² - b² = (a - b) (a + b)]
To find zeroes, put p(t) = 0
\(\Rightarrow (t-\sqrt{15})(t+\sqrt{15})=0\)
\(\Rightarrow t-\sqrt{15}=0\) or t + \(\sqrt{15}\) = 0 \(\Rightarrow\) t = \(\sqrt{15}\) or t = -\(\sqrt{15}\)
Hence, zeroes of the given polynomial are -\(\sqrt{15}\) and \(\sqrt{15}\).
Verification
Hence, sum of zeroes = -\(\sqrt{15}\) + \(\sqrt{15}\) = 0 = -(0/1)
\(=-\frac{Coefficient \quad of \quad t}{Coefficient \quad of \quad t^{2}}\)
and product of zeroes = -\(\sqrt{15}\) \(\times\)\(\sqrt{15}\)= -15 = \(\frac{-15}{1}\)
\(=\frac{Constant \quad term}{Coefficient \quad of \quad t^{2}}\)
So, the relationship between the zeroes and its coefficients is verified.

Given, pair of linear equations is 
x - y = 8 \(\Rightarrow\) x - y - 8 = 0            ....(i)
and 3x - 3y = 16 \(\Rightarrow\) 3x - 3y - 16 = 0 ...(ii)
On comparing with standard form of pair of linear equations, we get
a₁= 1, b₁ = -1, c₁ = -8
and a₂ = 3, b₂ = -3, c₂ = -16
Here,  \(\frac{a_{1}}{a_{2}}=\frac{1}{3},\frac{b_{1}}{b_{2}}=\frac{1}{3}\) and \(\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}\)
Thus, \(\frac{1}{3}=\frac{1}{3}\neq \frac{1}{2}\) i.e. \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}\)
So, the pair of linear equations is inconsistent.