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Sia ? 6 years, 4 months ago
Let the present ages of B and A be x years and y years respectively.
Then,
B's age 5 years ago = {tex}(x - 5) years{/tex}
and A's age 5 years ago = {tex}(y - 5) years.{/tex}
{tex}\therefore{/tex} {tex}(y - 5) = 3(x - 5){/tex} {tex}\Rightarrow{/tex}{tex}3x -y = 10{/tex} .... (i)
B's age 10 years hence = {tex}(x +10) years{/tex}.
A's age 10 years hence = {tex}(y +10) years{/tex}.
{tex}\therefore{/tex} {tex}(y + 10) = 2(x + 10){/tex} {tex}\Rightarrow{/tex} {tex}2x - y = - 10{/tex} ..... (ii)
On subtracting (ii) from (i), we get x = 20.
Putting x = 20 in (i), we get
(3{tex}\times{/tex}20) - y = 10 {tex}\Rightarrow{/tex}y = 60 -10 = 50.
{tex}\therefore{/tex} x = 20 and y = 50.
{tex}\text{Hence, B's present age = 20 years}{/tex}
{tex}\text{and A's present age = 50 years.}{/tex}
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Ram Kushwah 6 years, 9 months ago
cosec2A=Cot2A+1
sin2A=1/(Cot2A+1)----------(1)
sec2A=1+tan2A=1+1/cot2A
tanA=1/cotA
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Sia ? 6 years, 4 months ago
Given equations,
{tex}2 x - y - 2 = 0{/tex}
{tex}4x + 3y - 24 = 0{/tex}
{tex}y + 4 = 0{/tex}
We have, {tex}2x - y - 2 = 0{/tex} or {tex}x = {{y+2}\over2} {/tex}
When y = 0, we have {tex}x = {{0+2}\over2} = 1{/tex}
When x = 0, we have y = -2.
Thus, we obtain the following table giving coordinates of two point on the line represented by the equation {tex}2x - y - 2 = 0{/tex} and its graph is shown below.
| x | 1 | 0 |
| y | 0 | -2 |
Now we have, {tex}\begin{array}{l}4x+3y-24=0\Rightarrow y=\frac{24-4x}3\\\end{array}{/tex}
When y = 0, we have x = 6
When x = 0, we have y = 8
Thus, we obtain the following table giving coordinates of two points on the line represented by the equation {tex}4x + 3y - 24 = 0{/tex} and its graph is shown below.
| x | 6 | 0 |
| y | 0 | 8 |
Also we have {tex}y + 4 = 0{/tex}
Clearly, y = - 4 for every value of x.
So, let E(2, -4) and F(0, -4) be two points on the line represented by y + 4 = 0. Plotting these points on the same graph and drawing a line passing through them, we obtain the graph of the line represented by the equation y + 4 = 0 as shown in Figure.

From Fig. we have {tex}\begin{array}{l}\bigtriangleup PQR\\\end{array}{/tex} having vertices P(3,4), Q(-1,-4) and R(9, -4).
Also, PM = 8 and QR = 10.
{tex}\therefore \quad \text { Area of } \triangle P Q R = \frac { 1 } { 2 } ( \text { Base } \times \text { Height } ){/tex}
{tex}\Rightarrow \quad \text { Area of } \triangle P Q R = \frac { 1 } { 2 } ( Q R \times P M ) = \frac { 1 } { 2 } ( 10 \times 8 )sq.\ units{/tex}
{tex}\Rightarrow \quad \text { Area of } \triangle P Q R = 40 \mathrm { sq } .units.{/tex}
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Aditya Choudhary 6 years, 9 months ago
1Thank You