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Whitepunk [10]
3 years ago
7

IM just a little bit slow

Mathematics
2 answers:
viva [34]3 years ago
5 0

Answer:

71°

Step-by-step explanation:

31-3x+19x-5=90

16x=90-26

16x=64

x=4

m<R=19x-5 = 19(4)-5 = 76-5 = 71

FromTheMoon [43]3 years ago
5 0

Answer:

x = 4

∠R = 71°

Step-by-step explanation:

<u>COMPLEMENTARY ANGLES:</u> TWO ANGLES WITH A SUM OF 90°.

31 - 3x + 19x - 5 = 90

To make this problem easier to solve, we can combine like terms.

26 + 16x = 90.

Now, we can proceed.

First, subtract 26 from 26 and 90.

26 - 26 = 0. (cancels itself out).

90 - 26 = 64.\\

Now, we are left with 16x = 64.

Lastly, we divide the number in front of the variable by the answer.

64/16 = 4

Therefore, the value of x is 4.

-------------------------------------------------------------------------------------------------------------

Now that we have the value of x, we can use this value to replace the variable (x).

19x - 5

19 * 4 - 5 = 71.

Therefore, the value of ∠R is 71°.

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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
Licemer1 [7]

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

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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

4 0
3 years ago
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