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astraxan [27]
3 years ago
14

If x + sin^2 theta + cos^2 theta = 4, then what is the value of x?

Mathematics
1 answer:
Ray Of Light [21]3 years ago
4 0

Answer:

x = 3

Step-by-step explanation:

x + sin^2  \theta + cos^2  \theta = 4 \\ x + 1 = 4 \\ ( \because \: sin^2  \theta + cos^2  \theta = 1) \\ x = 4 - 1 \\  x = 3

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Answer:

m∠P = 82°

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m∠R = 49°

Step-by-step explanation:

<em>In the isosceles triangle, the base angles are equal in measures</em>

In Δ PQR

∵ PQ = PR

∴ Δ PQR is an isosceles triangle

∵ ∠Q and ∠R are the base angles

→ By using the fact above

∴ m∠Q = m∠R

∵ m∠Q = (3x + 25)°

∵ m∠R = (2x + 33)°

→ Equate them

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∵ 3x - 2x + 25 = 2x - 2x + 33

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→ Subtract 25 from both sides

∵ x + 25 - 25 = 33 - 25

∴ x = 8

→ Substitute the value of x in the measures of angles Q and R

∵ m∠Q = 3(8) + 25 = 24 + 25

∴ m∠Q = 49°

∵ m∠R = 2(8) + 33 = 16 + 33

∴ m∠R = 49°

∵ The sum of the measures of the interior angles of a Δ is 180°

∴ m∠P + m∠Q + m∠R = 180°

→ Substitute the measures of angles Q and R

∵ m∠P + 49 + 49 = 180

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∵ m∠P + 98 - 98 = 180 - 98

∴ m∠P = 82°

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2 years ago
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