find the value of PQ..
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Answer:
Solution by TRIGONOMETRY
Step-by-step explanation:
In ΔABQ, ∠B = 90
tan60 = [tex]\frac{AB}{BQ}[/tex] = [tex]\frac{h}{d}[/tex]
[tex]\sqrt{3} = \frac{h}{d}[/tex]
[tex]h = \sqrt{3} d[/tex] ........... (1)
Similarly,
In ΔABP, ∠B = 90
tan30 = [tex]\frac{AB}{PB} = \frac{h}{PQ + BQ} = \frac{h}{d + PQ}[/tex] = [tex]\frac{h}{d+x}[/tex]
[tex]\frac{1}{\sqrt{3} } = \frac{h}{d + x}[/tex]
but, [tex]h = \sqrt{3} d[/tex] ........... (1)
so,
[tex]\frac{1}{\sqrt{3} } = \frac{\sqrt{3} d}{d + x}[/tex]
3d = d + x
x = 2d
Hence PQ = x = 2d
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