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Law Incorporation [45]

3 years ago

8

In the right triangle ?ABC, leg AC=6 cm and leg BC=8 cm. Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5. Find area of ?MN

C.

1 answer:

Licemer1 [7]3 years ago

4 0

Given : In Right triangle ABC, AC=6 cm, BC=8 cm.Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5.

To find : Area (ΔMNC)

Solution: In Δ ABC, right angled at C,

AC= 6 cm, BC= 8 cm

Using pythagoras theorem

AB² =AC²+ BC²

=6²+8²

= 36 + 64

→AB² =100

→AB² =10²

→AB =10

Also, AM:MN:NB=1:2.5:1.5

Then AM, MN, NB are k, 2.5 k, 1.5 k.

→2.5 k + k+1.5 k= 10

→ 5 k =10

Dividing both sides by 2, we get

→ k =2

MN=2.5×2=5 cm, NB=1.5×2=3 cm, AM=2 cm

As Δ ACB and ΔMNC are similar by SAS.

So when triangles are similar , their sides are proportional and ratio of their areas is equal to square of their corresponding sides.

$\frac{Ar(ACB)}{Ar(MNC)}=[\frac{10}{5}]^{2}$

$\frac{Ar(ACB)}{Ar(MNC)}=4$

But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]

$\frac{24}{Ar(MNC)}=4$

→ Area(ΔMNC)=24÷4

→Area(ΔMNC)=6 cm²

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6 0

3 years ago

Suppose θ is an angle in the standard position whose terminal side is in Quadrant III and sec θ=61/60. Find the exact values of

Nataly_w [17]

Answer:

Part 1) $cos(\theta)=-\frac{60}{61}$

Part 2) $tan(\theta)=\frac{11}{60}$

Part 3) $cot(\theta)=\frac{60}{11}$

Part 4) $csc(\theta)=-\frac{61}{11}$

Part 5) $sin(\theta)=-\frac{11}{61}$

Step-by-step explanation:

we know that

If angle theta lie on Quadrant III

then

The function sine is negative

The function cosine is negative

The function tangent is positive

The function cotangent is positive

The function cosecant is negative

The function secant is negative

step 1

Find $cos(\theta)$

we know that

$cos(\theta)=\frac{1}{sec(\theta)}$

we have

$sec(\theta)=-\frac{61}{60}$ ----> the value must be negative

therefore

$cos(\theta)=-\frac{60}{61}$

step 2

Find $tan(\theta)$

we know that

$tan^{2} (\theta)+1=sec^{2} (\theta)$

we have

$sec(\theta)=-\frac{61}{60}$

substitute

$tan^{2} (\theta)+1=(-\frac{61}{60})^{2}$

$tan^{2} (\theta)+1=\frac{3,721}{3,600}$

$tan^{2} (\theta)=\frac{3,721}{3,600}-1$

$tan^{2} (\theta)=\frac{121}{3,600}$

$tan(\theta)=\frac{11}{60}$

step 3

Find $cot(\theta)$

we know that

$cot(\theta)=\frac{1}{tan(\theta)}$

we have

$tan(\theta)=\frac{11}{60}$

therefore

$cot(\theta)=\frac{60}{11}$

step 4

Find $csc(\theta)$

we know that

$cot^{2} (\theta)+1=csc^{2} (\theta)$

we have

$cot(\theta)=\frac{60}{11}$

substitute

$(\frac{60}{11})^{2}+1=csc^{2} (\theta)$

$\frac{3,600}{121}+1=csc^{2} (\theta)$

$\frac{3,721}{121}=csc^{2} (\theta)$

square root both sides

$csc(\theta)=-\frac{61}{11}$

step 5

Find $sin(\theta)$

we know that

$sin(\theta)=\frac{1}{csc(\theta)}$

we have

$csc(\theta)=-\frac{61}{11}$

therefore

$sin(\theta)=-\frac{11}{61}$

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dangina [55]

Answer:

x = .25

Step-by-step explanation:

6x = 1.5

Divide each side by 6

6x/6 = 1.5/6

Multiply the top and bottom by 10

x = 1.5*10/ (6*10)

x = 15/60

x = 1/4

x = .25

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weqwewe [10]

Answer:

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Step-by-step explanation:

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x pen = 228

48 = 576

x = 228

x = (228 × 48)/576

x = 10,944/576

x = 19

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Answer:

Step-by-step explanation:

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