Mathematics Questions and Answers

Let the given points be:

P(-3, -14) = (x\(_1\), y\(_1\))

Q(t, -5) = (x\(_2\), y\(_2\))

PQ = 9 units (given)

Using the distance formula,

d = √ [ \((x_2 - x_1)^2 + (y_2 - y_1)^2\)]

PQ = √ [ \((t - (-3))^2 + (-5 + 14)^2\)]

\(\implies\) √ [ \((t + 3)^2 + 81\)] = 9

Squaring on both sides,

⇒ (t + 3)\(^2\) + 81 = 81

⇒ (t + 3)\(^2\)  = 0

⇒ t + 3 = 0

∴ t = -3

\(\frac {3 - \sqrt 3}{2 + \sqrt 3} = a + b\sqrt 3\)

Rationalize

= \(\frac {3 - \sqrt 3}{2 + \sqrt 3} \times \frac {2 - \sqrt 3}{2 - \sqrt 3}\)

= \(\frac {(3 - \sqrt 3)}{(2 + \sqrt 3)} \frac {(2 - \sqrt 3)}{(2 - \sqrt 3)}\)

= \(\frac {6 - 3 \sqrt 3 - 2 \sqrt 3 + (\sqrt 3)^2}{4 - 2 \sqrt 3 + 2 \sqrt 3 - (\sqrt 3)^2}\)

= \(\frac {6 - 5 \sqrt 3 + 3}{4 - 3}\)

= \(\frac {9 - 5 \sqrt 3}{1} = 9 - 5 \sqrt 3\)

= 9 + (-5) \(\sqrt 3\)

\(\therefore a = 9, b = - 5\)

No explanation has been provided for this answer.

Let T be the time taken and D is the distance

For the case given in the problem, the distance covered is directly proportional to the time taken i.e D α T

⇒ D = ST where S is the constant speed

D = 225 km and T = 4.5 hrs

∴ 225 = S x 4.5

S = \(\frac {225}{4.5}\) = 50km/hr

When D = 150km

D = ST

150 = 50 \(\times\) T

T = \(\frac {150}{50}\) = 3 hrs

So it will take 3 hrs

From ∆PSR

|PS| = |SR| (If two tangents are drawn from an external point of the circle, then they are of equal lengths)

∴ ∆PSR is isosceles

∠PSR + ∠SRP + ∠SPR = 180\(^o\) (sum of angles in a triangle)

Since |PS| = |SR|; ∠SRP = ∠SPR

⇒ ∠PSR + ∠SRP + ∠SRP = 180\(^o\)

∠PSR + 2∠SRP = 180\(^o\)

36\(^o\) + 2∠SRP = 180\(^o\)

2∠SRP = 180\(^o\) - 36\(^o\)

2∠SRP = 144\(^o\)

∠SRP = \(\frac {144^o}{2} = 72^0\)

∠SRP = ∠PQR (angle formed by a tangent and chord is equal to the angle in the alternate segment)

∴ ∠PQR = 72\(^0\)