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Sonja [21]
2 years ago
12

Question 3 of 10

Mathematics
1 answer:
mestny [16]2 years ago
7 0

Answer:

㋡

Check Answer

♣ Qᴜᴇꜱᴛɪᴏɴ :

If tan θ = \sf{\dfrac{1}{\sqrt{7}}}

7

1

, Show that \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

We know :

\large\boxed{\sf{tan\theta=\dfrac{Height}{Base}}}

tanθ=

Base

Height

So comparing this formula and value of tan θ from question, we get :

Height = 1

Base = √7

Now we need to Prove the value of : \sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

Also :

\large\boxed{\sf{cosec\theta=\dfrac{Hypotenuse}{Height}}}

cosecθ=

Height

Hypotenuse

\large\boxed{\sf{sec\theta=\dfrac{Hypotenuse}{Base}}}

secθ=

Base

Hypotenuse

From this we get :

\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}

cosec

2

θ=(

Height

Hypotenuse

)

2

\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}

sec

2

θ=(

Base

Hypotenuse

)

2

But we have Height and Base, we dont have Hypotenuse.

Hypotenuse can be found by using Pythagoras Theorem

Pythagoras Theorem states that :

Hypotenuse² = Side² + Side²

For our question :

Hypotenuse² = Height² + Base²

Hypotenuse² = 1² + √7²

Hypotenuse² = 1 + 7

Hypotenuse² = 8

√Hypotenuse² = √8

Hypotenuse = √8

➢ Let's find value's of cosec²θ and sec²θ

________________________________________

First cosec²θ :

\large\boxed{\sf{cosec^2\theta=\left(\dfrac{Hypotenuse}{Height}\right)^2}}

cosec

2

θ=(

Height

Hypotenuse

)

2

\sf{cosec^2\theta=\left(\dfrac{\sqrt{8}}{1}\right)^2}cosec

2

θ=(

1

8

)

2

\sf{cosec^2\theta=\dfrac{8}{1}}cosec

2

θ=

1

8

cosec²θ = 8

________________________________________

Now sec²θ :

\large\boxed{\sf{sec^2\theta=\left(\dfrac{Hypotenuse}{Base}\right)^2}}

sec

2

θ=(

Base

Hypotenuse

)

2

\sf{sec^2\theta=\left(\dfrac{\sqrt{8}}{\sqrt{7}}\right)^2}sec

2

θ=(

7

8

)

2

\sf{sec^2\theta=\dfrac{8}{7}}sec

2

θ=

7

8

sec²θ = 8/7

________________________________________

Now Proving :

\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }=\dfrac{3}{4}}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=

4

3

Taking L.H.S :

\sf{\dfrac{cosec ^2 \theta - sec ^2\theta}{cosec^2\theta + sec^2\theta }}

cosec

2

θ+sec

2

θ

cosec

2

θ−sec

2

θ

=\sf{\dfrac{8 - sec ^2\theta}{8 + sec^2\theta }}=

8+sec

2

θ

8−sec

2

θ

=\sf{\dfrac{8 - \dfrac{8}{7}}{8 + \dfrac{8}{7} }}=

8+

7

8

8−

7

8

=\sf{\dfrac{\dfrac{48}{7}}{\dfrac{64}{7} }}=

7

64

7

48

\sf{=\dfrac{48\times \:7}{7\times \:64}}=

7×64

48×7

\sf{=\dfrac{48}{64}}=

64

48

\bf{=\dfrac{3}{4}}=

4

3

= R.H.S

Hence Proved !!!

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In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

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One case can hold 3 boxes.Each box can hold 3 binders.How man cases are needed to hold 126 binders?
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14 because 126 ÷3=42 and 42÷3 =14

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3 years ago
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If C is between A and B, then answer the following if AB=40, AC= 2x+10, CB= 3x-5
Marat540 [252]

If C is between A and B, then, we can write the equation

AB = AC + CB


Substitute the value of AB and the expressions for AC and CB in terms of x.


40 = (2x+10)+(3x-5)


We can now solve for x.



40-10+5= 2x+3x



35=5x


Dividing by 5.


7=x


AB=40



AC=2(7)+10



AC=14+10



AC=24




BC=3(7)-5



BC=21-5


BC=16



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Math Problem:
kirza4 [7]

Answer:

Her answer is not reasonable. Her average is 83.

Step-by-step explanation:

Add all of her test grades together (96+82+78+76=332)

To find the average, you need to add all test grades she got and then divide the sum by the total number of numbers added (in this case, we added 4 tests together, so we need to divide by 4)

\frac{332}{4} =83

So she added wrong and forgot to divide by the total number of tests.

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2 years ago
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