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Olegator [25]
3 years ago
5

Can 1,2,5 form a right triangle​

Mathematics
1 answer:
malfutka [58]3 years ago
8 0
No 1+2=3 which is less than 5
The two sides must equal more than the third one when added
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Please show the working and answer. you can take a picture for the working.
baherus [9]

Answer:

(a) The area of the triangle is approximately 39.0223 cm²

(b) ∠SQR is approximately 55.582°

Step-by-step explanation:

(a) By sin rule, we have;

SQ/(sin(∠SPQ)) = PQ/(sin(∠PSQ)), which gives;

5.4/(sin(52°)) = 6.8/(sin(∠PSQ))

∴ (sin(∠PSQ)) = (6.8/5.4) × (sin(52°)) ≈ 0.9923

∠PSQ = sin⁻¹(0.9923) ≈ 82.88976°

Similarly, we have;

5.4/(sin(52°)) = SP/(sin(180 - 52 - 82.88976))

Where, 180 - 52 - 82.88976 = ∠PQS = 45.11024

SP = 5.4/(sin(52°))×(sin(180 - 52 - 82.88976)) ≈ 4.8549

Given that RS : SP = 2 : 1, we have;

RS = 2 × SP = 2 × 4.8549 ≈ 9.7098

We have by cosine rule, \overline {RQ}² =  \overline {SQ}² +  \overline {SR}² - 2 × \overline {SQ} × \overline {SR} × cos(∠QSR)

∠QSR and ∠PSQ are supplementary angles, therefore;

∠QSR = 180° - ∠PSQ = 180° - 82.88976° = 97.11024°

∠QSR = 97.11024°

Therefore;

\overline {RQ}² =  5.4² +  9.7098² - 2 ×  5.4×9.7098× cos(97.11024)

\overline {RQ}² ≈ 136.42

\overline {RQ} = √(136.42) ≈ 11.6799

The area of the triangle = 1/2 ×\overline {PQ} × \overline {PR} × sin(∠SPQ)

By substituting the values, we have;

1/2 ×\overline {PQ} × \overline {PR} × sin(∠SPQ)

1/2 × 6.8 × (4.8549 + 9.7098) × sin(52°) ≈ 39.0223 cm²

The area of the triangle ≈ 39.0223 cm²

(b) By sin rule, we have;

\overline {RS}/(sin(∠SQR)) = \overline {RQ}/(sin(∠QSR))

By substituting, we have;

9.7098/(sin(∠SQR)) = 11.6799/(sin(97.11024))

sin(∠SQR) = 9.7098/(11.6799/(sin(97.11024))) ≈ 0.82493

∠SQR = sin⁻¹(0.82493) ≈ 55.582°.

8 0
3 years ago
Idk why I can’t get this ?! number two plz
Arturiano [62]
The only possible outcome to this would be any number greater or less than 361 in the 300’s
7 0
3 years ago
(-33.28 ÷ -5.2) ÷ -1.6 = ?
tigry1 [53]

Answer:

-4

Step-by-step explanation:

Use PEMDAS

5 0
3 years ago
Read 2 more answers
Which sequence of transformations will map figure K onto figure K′? Two congruent kites, figure K and figure K prime, are drawn
elixir [45]

Answer:

The sequence of transformations will map figure K onto figure K′

is the first sequence <u>option (1)</u>

======================================================

Step-by-step explanation:

See the attached figure

as shown in the figure the K' is the image of K by reflection over x-axis

<u>But </u> We need to know which sequence of transformations will give the same result.

So, we will test the options by any point from K and its image from K'

i.e: we will test the options using the points (6,5) , (6,-5)

(6,5) ⇒ (6,-5)

<u>option (1):</u>

Reflection across x = 4, 180° rotation about the origin, and a translation of (x + 8, y)

(6,5) ⇒ (2,5) ⇒(-2,-5) ⇒ (6,-5)

<u>option (2):</u>

Reflection across x = 4, 180° rotation about the origin, and a translation of (x − 8, y)

(6,5) ⇒ (2,5) ⇒(-2,-5) ⇒ (-10,-5)

<u>option (3):</u>

Reflection across y = 4, 180° rotation about the origin, and a translation of (x + 8, y)

(6,5) ⇒ (6,3) ⇒ (-6,-3) ⇒ (2,-3)

<u>option (4):</u>

Reflection across y = 4, 180° rotation about the origin, and a translation of (x − 8, y)

(6,5) ⇒ (6,3) ⇒ (-6,-3) ⇒ (-14,-3)

As shown: The sequence of transformations will map figure K onto figure K′

is the first sequence <u>option (1)</u>

7 0
3 years ago
Read 2 more answers
What fraction is greater than 7/12
lapo4ka [179]
5/9 is grater than 7/12. same with 9/12, 6/7, 6/9
3 0
3 years ago
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