Question

$\huge\red\bigstar$ Hello!!
•Solve sum with explanation .
•No irrevalent/unnecessary answers...otherwise reported .​

Answer 1

GIVEN :-

• angle P = 30°

• angle Q = 45°

• let AB be the tree and h is the hieght of tree let X be the distance to the bottom of the tree from points Q

• distance from the tree is 100 - x

TO FIND :-

• hieght of tree

SOLUTION :-

$\implies \rm{in \: \triangle \: abq \: }$

we know that

$\implies \boxed{ \rm{tan \: x= \dfrac{p}{b} }}$

$\implies \rm{tan \: q = \dfrac{ab}{bq} }$

now we know that angle Q = 45°

HENCE ,

$\implies \rm{tan \: 45 = \dfrac{ab}{bq} }$

now we know that according to trignometric values :-

$\implies \boxed{ \rm{tan \: 45 = 1 }}$

HENCE ,

$\implies \rm{1 = \dfrac{ab}{bq} }$

now it's given that

AB = hieght ( h )

BQ = x

$\implies \rm{ 1 = \dfrac{h}{x} }$

$\implies \rm{ \bf \: h = x} \: \: \: \: \: \: \: \: (1)$

$\implies \rm{in \: \triangle \: abp \: }$

we know that

$\implies \boxed{ \rm{tan \: x= \dfrac{p}{b} }}$

$\implies \rm{tan \: p = \dfrac{ab}{bp} }$

NOW we know that angle P = 30°

$\implies \rm{tan \: 30 = \dfrac{ab}{bp} }$

now ,

AB = hieght of tree (h )

and

PB + BQ = PQ

now PQ = 100 ( distance between poles )

PB + BQ = 100

PB + x = 100 ( we had taken BQ = x )

PB = 100 - x

$\implies \rm{tan \: 30 = \dfrac{h}{100 - x} }$

now we know that according to trignometric values

$\implies \boxed{ \rm{tan \: 30= \dfrac{1}{ \sqrt{3} \: } }}$

HENCE,

$\implies \rm{ \dfrac{1}{ \sqrt{3} } = \dfrac{h}{100 - x} }$

$\implies \rm{ \bf \: 100 - x = h\sqrt{3} } \: \: \: \: \: \: \: \: (2)$

now putting eq 1 in eq 2

$\implies \rm{ \: 100 - x = h\sqrt{3} }$

$\implies \rm{ \: 100 - h = h\sqrt{3} }$

( as according to eq 1 , h = x )

$\implies \rm{ \: 100 = h\sqrt{3} + h }$

$\implies \rm{ \: 100 = h(\sqrt{3} + 1)}$

$\implies \rm{ h = \dfrac{100}{( \sqrt{3} + 1)} }$

hence,

$\implies \boxed{ \boxed{ \rm{hieght \: of \: tree \: = \dfrac{100}{ (\sqrt{3} + 1) } \: m }}}$

OTHER INFORMATION :-

TRIGNOMETRIC IDENTITIES

• sin²∅ + cos²∅ = 1

• sec²∅ - tan²∅ = 1

• cosec²∅ - cot²∅ = 1

TRIGNOMETRIC RATIOS

• sin ∅ = 1 / cosec ∅

• cos ∅ = 1 / sec ∅

• tan ∅ = 1 / cot ∅

TRIGNOMETRIC COMPLEMENTRY ANGLES

• sin ∅ = cos ( 90 - ∅ )

• cos ∅ = sin ( 90 - ∅ )

• sec ∅ = cosec ( 90 - ∅ )

• cosec ∅ = sec ( 90 - ∅ )

• tan ∅ = cot ( 90 - ∅ )

• cot ∅ = tan ( 90 - ∅ )