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AveGali [126]
2 years ago
7

Please help with this pleaseee

Mathematics
2 answers:
lorasvet [3.4K]2 years ago
6 0

Answer:

∠E= 126°

Step-by-step explanation:

the parallel lines create a transversal (the diagonal line that connects them)

this transversal rule shows us that

∠DCE=∠CBA

so ∠DCE= 38°

let's look at the triangle

∠C+∠D+∠E=180°

38+16+∠E= 180°

∠E= 126°

saw5 [17]2 years ago
5 0

Answer:

CED=126

Step-by-step explanation:

The total of all 3 angles should be 180

so:

180-38-16 (ABC is equal to ECD)

180-38-16=126

126=CED

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Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

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<P + <Q + <S = 180^{o}

048° + 036^{o} + <S = 180^{o}

84^{o} + <S = 180^{o}

<S  = 180^{o} -  84^{o}

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<S = 96^{o}

Applying the Sine rule,

\frac{y}{Sin P} = \frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{y}{Sin P} = \frac{z}{Sin S}

\frac{y}{Sin 48^{o} } = \frac{17}{Sin 96^{o} }

\frac{y}{0.74314} = \frac{17}{0.99452}

⇒ y = \frac{12.63338}{0.99452}

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\frac{x}{Sin 36^{o} } = \frac{17}{Sin 96^{o} }

\frac{x}{0.58779} = \frac{17}{0.99452}

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the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

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3 years ago
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