Question

find the value of PQ.. ​

Answer 1

Answer:

Solution by TRIGONOMETRY

Step-by-step explanation:

In ΔABQ, ∠B = 90

tan60 = $\frac{AB}{BQ}$  =  $\frac{h}{d}$

$\sqrt{3} = \frac{h}{d}$

$h = \sqrt{3} d$ ........... (1)

Similarly,

In ΔABP, ∠B = 90

tan30 = $\frac{AB}{PB} = \frac{h}{PQ + BQ} = \frac{h}{d + PQ}$  =  $\frac{h}{d+x}$

$\frac{1}{\sqrt{3} } = \frac{h}{d + x}$  

but, $h = \sqrt{3} d$ ........... (1)

so,

$\frac{1}{\sqrt{3} } = \frac{\sqrt{3} d}{d + x}$

3d = d + x

x = 2d

Hence PQ = x = 2d

Answer 2

Answer:

Step-by-step explanation:

In ΔABQ, ∠B = 90

tan60 = \frac{AB}{BQ}

BQ

AB

= \frac{h}{d}

d

h

\sqrt{3} = \frac{h}{d}

3

=

d

h

h = \sqrt{3} dh=

3

d ........... (1)

Similarly,

In ΔABP, ∠B = 90

tan30 = \frac{AB}{PB} = \frac{h}{PQ + BQ} = \frac{h}{d + PQ}

PB

AB

=

PQ+BQ

h

=

d+PQ

h

= \frac{h}{d+x}

d+x

h

\frac{1}{\sqrt{3} } = \frac{h}{d + x}

3

1

=

d+x

h

but, h = \sqrt{3} dh=

3

d ........... (1)

so,

\frac{1}{\sqrt{3} } = \frac{\sqrt{3} d}{d + x}

3

1

=

d+x

3

d

3d = d + x

x = 2d