Question

1.The point which divides the line segment joining the points A (0, 5) and
B (5, 0) internally in the ratio 2:3 is _____________

2.The pair of lines represented by the equations 2x+y+3 = 0 and 4x+ky+6 =
0 will be parallel if value of k is ______.

3.The value of sin 60theta
cos 30theta + sin 30theta
cos 60theta
is_____


4.Value of cos 0°. Cos 30° .cos 45° . cos 60° . cos 90° is ___________.

5.The sides of two similar triangles are in the ratio 2:3, then the areas of
these triangles are in the ratio ______________

Answer 1

Answer:

1)$(x,y)=(2,3)$

2)$k=2$

3)sin(90theta)

4)$cos0$°$=1$

$cos30$°$=\frac{\sqrt{3} }{2}$

$cos45$°$=\frac{1}{\sqrt{2} }$

$cos60$°$=\frac{1}{2}$

$cos90$°$=0$

5)\frac{4}{9}

Step-by-step explanation:

1)The given points are $A(0,5)$ and $B(5,0)$.

the given ratio is $2:3$.

Use section formula

$x=\frac{mx_{2} +nx_{1} }{m+n}$

$y=\frac{my_{2}+ny_{1}  }{m+n}$

$x_{1} =0,y_{1} =5,x_{2} =5,y_{2} =0$

$x=\frac{2*5+3*0}{2+3} =\frac{10}{5} =2$

$y=\frac{2*0+3*5}{2+3} =\frac{15}{5} =3$

hence the point is $(x,y)=(2,3)$.

2)the given equations are,

$2x+y+3=0$ and $4x+ky+6=0$

The slope of parallel lines are equal.

Write the given equation in slope-intercept form.

slope of first equation is

$2x+y+3=0$

⇒$y+3=-2x$

⇒$y=-2x-3$

$m_{1} =-2$

Slope of second equation is

$4x+ky+6=0$

⇒$ky+6=-4x$

⇒$ky=-4x-6$

⇒$y=\frac{-4}{k} x-\frac{6}{k}$

$m_{2} =\frac{-4}{k}$

now,

$-2=\frac{-4}{k}$

⇒$k=\frac{-4}{-2} =2$

hence the value of $k=2$

3)$sin(60theta)cos(30theta)+sin(30theta)cos(60theta)=sin((60theta)+(30theta))=sin(90theta)$

4)$cos0$°$=1$

$cos30$°$=\frac{\sqrt{3} }{2}$

$cos45$°$=\frac{1}{\sqrt{2} }$

$cos60$°$=\frac{1}{2}$

$cos90$°$=0$

5)$\frac{area of triangle 1}{area of triangle2} =\frac{(side of triangle 1)^{2} }{(side of triangle 2)^{2}} =\frac{(2)^{2} }{(3)^{2} } =\frac{4}{9}$