The diameter of the sun is 1.4 x 10'm and the diameter of earth is 1.2756 x 10'm.
Find the ratio of diameter of sun to the diameter of earth.
Answers
Answer:
We know that : \sf{\sin\theta = Opposite\:Side/Hypotenuse}sinθ=OppositeSide/Hypotenuse
\longrightarrow\sf{a/b = Opposite\:Side/Hypotenuse}⟶a/b=OppositeSide/Hypotenuse
\bigstar★ From this Information ;
\longrightarrow⟶ Opposite Side = a \&& Hypotenuse = b
We can apply Pythagoras Theorem (Pythagoras Theorem states that Square of Hypotenuse is equal to Square of Other two Sides) in The triangle and find the measure of Adjacent Side
\boxed{\rm{(Hypotenuse)^2 = (Opposite\: Side)^2 + (Adjacent\: Side)^2}}
(Hypotenuse)
2
=(OppositeSide)
2
+(AdjacentSide)
2
\longrightarrow\rm{b^2 = a^2 + (Adjacent\: Side)^2\quad...\:\sf{Since\:Opposite\:Side = a\:\&\: Hypotenuse = b }}⟶b
2
=a
2
+(AdjacentSide)
2
...SinceOppositeSide=a&Hypotenuse=b
\longrightarrow\rm{(Adjacent\: Side)^2 = b^2 - a^2\quad...\:\sf{Subtracting\:a^2\:from\:both\:Sides\:of\:Equation}}⟶(AdjacentSide)
2
=b
2
−a
2
...Subtractinga
2
frombothSidesofEquation
\longrightarrow\rm{Adjacent\: Side = \sqrt{b^2 - a^2}\quad...\:\sf{Taking\:Square\:Root\:on\:both\:Sides\:of\:Equation}}⟶AdjacentSide=
b
2
−a
2
...TakingSquareRootonbothSidesofEquation
\bigstar★ Now, From the Calculations ;
\longrightarrow⟶ Opposite Side = a, Adjacent Side = \sqrt{\sf{b^2-a^2}}
b
2
−a
2
\&& Hypotenuse = b
\begin{gathered}\boxed{\begin{aligned}&\bullet\:\sin\theta = \rm{Opposite\:Side/Hypotenuse = a/b}\\&\bullet\:\cos\theta = \rm{Adjacent\:Side/Hypotenuse = \sqrt{b^2-a^2}/b}\\&\bullet\:\tan\theta = \rm{Opposite\:Side/Adjacent\:Side = a/\sqrt{b^2-a^2}}\\&\bullet\:\csc\theta = \rm{Hypotenuse/Opposite\:Side = b/a}\\&\bullet\:\sec\theta = \rm{Hypotenuse/Adjacent\:Side = b/\sqrt{b^2-a^2}}\\&\bullet\:\cot\theta = \rm{Adjacent\:Side/Opposite\:Side = \sqrt{b^2-a^2}/a}\\\end{aligned}}\end{gathered}
∙sinθ=OppositeSide/Hypotenuse=a/b
∙cosθ=AdjacentSide/Hypotenuse=
b
2
−a
2
/b
∙tanθ=OppositeSide/AdjacentSide=a/
b
2
−a
2
∙cscθ=Hypotenuse/OppositeSide=b/a
∙secθ=Hypotenuse/AdjacentSide=b/
b
2
−a
2
∙cotθ=AdjacentSide/OppositeSide=
b
2
−a
2
/a
\large{----------\textbf{ Alternate Method }----------}−−−−−−−−−− Alternate Method −−−−−−−−−−
We know that : \sf{\sin^2\theta+ \cos^2\theta=1}sin
2
θ+cos
2
θ=1
\longrightarrow\rm{(a/b)^2 + \cos^2\theta = 1\longrightarrow\rm{a^2/b^2 + \cos^2\theta = 1}\quad...\:\sf{Since\:\sin\theta\:is\:equal\:to\:a/b}}⟶(a/b)
2
+cos
2
θ=1⟶a
2
/b
2
+cos
2
θ=1...Sincesinθisequaltoa/b
\longrightarrow\rm{\cos^2\theta = 1 - a^2/b^2\quad...\:\sf{Subtracting\:a^2/b^2\:from\:both\:Sides\:of\:the\:Equation}}⟶cos
2
θ=1−a
2
/b
2
...Subtractinga
2
/b
2
frombothSidesoftheEquation
\longrightarrow\rm{\cos^2\theta = b^2/b^2 - a^2/b^2 = (b^2-a^2)/b^2\quad...\:\sf{Taking\:LCM\:of\:1\:and\:b^2}}⟶cos
2
θ=b
2
/b
2
−a
2
/b
2
=(b
2
−a
2
)/b
2
...TakingLCMof1andb
2
\longrightarrow\rm{\cos\theta = \sqrt{(b^2-a^2)/b^2} = \sqrt{b^2-a^2}/b\quad...\:\sf{Taking\:Square\:Root\:on\:both\:sides}}⟶cosθ=
(b
2
−a
2
)/b
2
=
b
2
−a
2
/b...TakingSquareRootonbothsides
We know that : \sf{\tan\theta = \sin\theta/\cos\theta}tanθ=sinθ/cosθ
\longrightarrow\rm{\tan\theta=a/b\div \sqrt{b^2-a^2}/b\quad...\:\sf{Since\:\sin\theta = a/b\:and\:\cos\theta = \sqrt{b^2-a^2}/b}}⟶tanθ=a/b÷
b
2
−a
2
/b...Sincesinθ=a/bandcosθ=
b
2
−a
2
/b
\longrightarrow\rm{\tan\theta=a/b\times b/\sqrt{b^2-a^2} = ab/b\sqrt{b^2-a^2} = a/\sqrt{b^2-a^2}}⟶tanθ=a/b×b/
b
2
−a
2
=ab/b
b
2
−a
2
=a/
b
2
−a
2
We know that : \sf{\csc\theta = 1/\sin\theta}cscθ=1/sinθ
\longrightarrow\rm{\csc\theta = 1\div a/b = 1\times b/a = b/a\quad...\:\sf{Since\:\sin\theta\:is\:equal\:to\:a/b}}⟶cscθ=1÷a/b=1×b/a=b/a...Sincesinθisequaltoa/b
We know that : \sf{\sec\theta = 1/\cos\theta}secθ=1/cosθ
\longrightarrow\rm{\sec\theta = 1\div \sqrt{b^2-a^2}/b} = b/\sqrt{b^2-a^2}\quad...\:\sf{Since\:\cos\theta\:is\:equal\:to\:\sqrt{b^2-a^2}/b}⟶secθ=1÷
b
2
−a
2
/b=b/
b
2
−a
2
...Sincecosθisequalto
b
2
−a
2
/b
We know that : \sf{\cot\theta = 1/\tan\theta}cotθ=1/tanθ
\longrightarrow\rm{\cot\theta = 1\div a/\sqrt{b^2-a^2} = \sqrt{b^2-a^2}/a\quad...\:\sf{Since\:\tan\theta\:is\:equal\:to\:a/\sqrt{b^2-a^2}}}⟶cotθ=1÷a/
b
2
−a
2
=
b
2
−a
2
/a...Sincetanθisequaltoa/
b
2
−a
2
\bigstar★ From the Calculations ;
\begin{gathered}\boxed{\begin{aligned}&\bullet\:\sin\theta = \rm{Opposite\:Side/Hypotenuse = a/b}\\&\bullet\:\cos\theta = \rm{Adjacent\:Side/Hypotenuse = \sqrt{b^2-a^2}/b}\\&\bullet\:\tan\theta = \rm{Opposite\:Side/Adjacent\:Side = a/\sqrt{b^2-a^2}}\\&\bullet\:\csc\theta = \rm{Hypotenuse/Opposite\:Side = b/a}\\&\bullet\:\sec\theta = \rm{Hypotenuse/Adjacent\:Side = b/\sqrt{b^2-a^2}}\\&\bullet\:\cot\theta = \rm{Adjacent\:Side/Opposite\:Side = \sqrt{b^2-a^2}/a}\\\end{aligned}}\end{gathered}
∙sinθ=OppositeSide/Hypotenuse=a/b
∙cosθ=AdjacentSide/Hypotenuse=
b
2
−a
2
/b
∙tanθ=OppositeSide/AdjacentSide=a/
b
2
−a
2
∙cscθ=Hypotenuse/OppositeSide=b/a
∙secθ=Hypotenuse/AdjacentSide=b/
b
2
−a
2
∙cotθ=AdjacentSide/OppositeSide=
b
2
−a
2
/a