Question

application of trignometry sol​

Answer 1

Answer:

$\huge{\red{\boxed{\boxed{\green{\sf{HEIGHT\:OF\:MULTI-STOREYED\:BUILDING=4(3+\sqrt{3})m}}}}}}$

$\huge{\red{\boxed{\boxed{\green{\sf{DISTANCE\:BETWEEN\:TWO\:BUILDINGS=4(3+\sqrt{3})m}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}}$

GIVEN:

>>See the above figure(1)and observe that PB is a transversal to the parallel lines PQ and BD.

□∠QPB and ∠PBD are alternate and are equal ∠QPB =∠PBD.

□∠PBD=30°

□∠PAC=45°

To find:

○The height of multi-storeyed building=?

○Distance between the two buildings=?

$in \triangle \: PBD \\ \tan30 \degree = \frac{PD}{BD} \\ = ) \frac{1}{ \sqrt{3} } = \frac{PD}{BD} \\ = ) BD = PD \sqrt{3} \\ \\ in \triangle \: PAC \: \\ \tan45 \degree = \frac{PA}{AC} \\ = )1 = \frac{PA}{AC} \\ = )PC = AC \\ = )PC = PD +DC \\ = ) \therefore \: PD + DC =AC \\ \\ = ) AC= BD \\ = ) DC= AB = 8m \\ \\ PD= \frac{8}{ \sqrt{3} - 1 } \: \: \: \: \: \: \: \: \: (PD + 8 = BD = PD \sqrt{3} ) \\ = )PD = \frac{8( \sqrt{3} + 1) }{ { \sqrt{3} }^{2} - 1 } = \frac{8 (\sqrt{3} + 1) }{3 - 1} \\ = )PD = 4 (\sqrt{3} - 1)m \\ \\ \therefore height \: of \: multi \: storeyed \: building \\ =) 4( \sqrt{3} + 1) + 8m \\ =) 4(3 + \sqrt{3)} m \\ \\ \therefore distance \: between \: two \: buildings \\ =) 4(3 + \sqrt{3} )m$

$\huge{\red{\boxed{\boxed{\green{\sf{HEIGHT\:OF\:MULTI-STOREYED\:BUILDING=4(3+\sqrt{3})m}}}}}}$

$\huge{\red{\boxed{\boxed{\green{\sf{DISTANCE\:BETWEEN\:TWO\:BUILDINGS=4(3+\sqrt{3})m}}}}}}$

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