Math, asked by anithashalu1989, 29 days ago

prove that cos(hx) and e'hx is isomorphism​

Answers

Answered by manojrana3756
0

Step-by-step explanation:

As the angles are complementary for their sum must be 90°.

According to the Attached Figure:

\large \bf In \: \triangle \: SRQIn△SRQ

\begin{gathered} \sf \tan(90 - x) = \dfrac{10 \sqrt{3} }{y} \\ \\ \bf \longrightarrow cot \: x = \frac{10 \sqrt{3} }{y} \: \: \: \:. \: . \: . \: (i) \end{gathered}

tan(90−x)=

y

10

3

⟶cotx=

y

10

3

...(i)

Now ,

\bf \large In \: \triangle \: SRPIn△SRP

\bf tan \: x = \dfrac{10 \sqrt{3} }{y + 20} \: \: \: \: . \: .\: . \: (ii)tanx=

y+20

10

3

...(ii)

Mulyiplying (i) and (ii) We get,

\begin{gathered} \sf \tan x. \cot x = \dfrac{10 \sqrt{3}}{y} \times \frac{10 \sqrt{3} }{20 + y} \\ \\ \longrightarrow \sf 1 = \frac{300}{y(y + 20)} \\ \\ \longrightarrow \sf1 = \frac{300}{y {}^{2} + 20y} \\ \\ \bf \longrightarrow y {}^{2} + 20y - 300 = 0 \\ \\ \longrightarrow \sf y {}^{2} - 10y + 30y - 300 = 0 \\ \\ \longrightarrow \sf y(y - 10) + 30(y - 10) = 0 \\ \\ \longrightarrow\sf (y - 10)(y + 30) = 0 \\ \\ \large \sf \purple{ \longrightarrow \underline {\boxed{{\bf y = 10 \: \: or \: \: y = - 30 } }}}\end{gathered}

tanx.cotx=

y

10

3

×

20+y

10

3

⟶1=

y(y+20)

300

⟶1=

y

2

+20y

300

⟶y

2

+20y−300=0

⟶y

2

−10y+30y−300=0

⟶y(y−10)+30(y−10)=0

⟶(y−10)(y+30)=0

y=10ory=−30

As y is distance So it can't be Negative

Hence,

\Large\purple{ \longrightarrow \underline {\boxed{{\bf y = 10m} }}}⟶

y=10m

So,

Distance of point P from building = 10 + 20

i.e.

\pink{ \huge \mathfrak{Distance=30m}}Distance=30m

\begin{gathered} \Large \red{\mathfrak{ \text{W}hich \: \: is \: \: the \: \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}\end{gathered}

Which is the required

Answer.

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