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Svetllana [295]
3 years ago
9

In triangle △ABC, ∠ABC=90°, BH is an altitude. Find the missing lengths. AC=5 and BH=2, find AH and CH.

Mathematics
1 answer:
Yakvenalex [24]3 years ago
3 0

Answer:

  AH = 1 or 4

  CH = 4 or 1

Step-by-step explanation:

An altitude divides a right triangle into similar triangles. That means the sides are in proportion, so ...

  AH/BH = BH/CH

  AH·CH = BH²

The problem statement tells us AH + CH = AC = 5, so we can write

  AH·(5 -AH) = BH²

  AH·(5 -AH) = 2² = 4

This gives us the quadratic ...

  AH² -5AH +4 = 0 . . . . in standard form

  (AH -4)(AH -1) = 0 . . . . factored

This equation has solutions AH = 1 or 4, the values of AH that make the factors be zero. Then CH = 5-AH = 4 or 1.

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\int\limits {e^{2s} cos\frac{s}{4} ds    =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that  f(s) =  e^{2s} cos\frac{s}{4}

Now integrating

            \int\limits {f(s)} \, ds =  \int\limits {e^{2s} cos\frac{s}{4} ds

By using integration formula

   \int\limits { e^{ax} cos b x dx = \frac{e^{ax} }{a^{2}+b^{2}  } ( a cos b x + b sin b x )

<u><em>Step(ii):-</em></u>

 \int\limits {e^{2s} cos\frac{s}{4} ds    =   \frac{e^{2s} }{(2)^{2}+(\frac{1}{4}) ^{2}  } ( 2 cos (\frac{1}{4} ) s + \frac{1}{4}  sin \frac{1}{4}  s ))  

                    = \frac{e^{2s} }{(4+\frac{1}{16})} ( 2 cos (\frac{1}{4} ) s + \frac{1}{4}  sin \frac{1}{4}  s ))

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                 =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

<u><em>Final answer:-</em></u>

\int\limits {e^{2s} cos\frac{s}{4} ds    =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

6 0
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Please help me to prove this. ​
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<h3><u>Answer</u> :</h3>

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⇒ A + B = π - C

⇒ cot(A + B) = cot(π - C)

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<u>Now 1st part of the given expression</u>!

⇒ \sf\dfrac{cosA}{sinB\:sinC}

⇒ \sf\dfrac{cos(\pi-(B+C))}{sinB\:sinC}

⇒ \sf\dfrac{-cosB\:cosC+sinB\:sinC}{sinB\:sinC}

⇒ 1 - cotB cotC

<u>Similarly 2nd part</u>!

⇒ 1 - cotA cotB

<u>Similarly 3rd part</u>!

⇒ 1 - cotC cotA

<u>LHS</u> :

\circ\:\sf{3-(cotA\:cotB+cotB\:cotC+cotC\:cotA)}

\circ\:\sf{3-1}

\circ\:\bf{2} = <u>RHS</u>

<h3>Hence Proved!!</h3>
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