prove that cos(hx) and e'hx is isomorphism
Answers
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.