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ozzi
3 years ago
8

9. Complete the reasons in the proof below.

Mathematics
1 answer:
Arada [10]3 years ago
4 0
For a), this is clearly a given as it is literally to the right of where it says “Given:”
For b), since ON bisects ∠JOH, this means that it splits it into two separate angles - JON and HON, which are similar due to that bisects mean that it splits it equally into two halves
For c), since NO is the same thing as NO, it is equal to itself
For d), since AAS (angle-angle-side) congruence states that if there are two angles that are congruent (proved in a) and b) ) as well as that a side is congruent (proved in c) ), two triangles are congruent
For e), since two triangles are congruent, every side must have one side that it matches up to in the other triangle. As the opposite side of angle H is JO and the opposite side of angle J is OH, and ∠J=∠H, those two are congruent. As JN and HN are the two sides left, they must be congruent.

Feel free to ask further questions!
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6) Two coastguard stations P and Q are 17km apart, with due East of P. A ship S is observed in distress on a bearing 048° fromP
adelina 88 [10]

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°

<Q = 360^{o} - 324^{o}

     = 036^{o}

Sum of angles in a triangle = 180^{o}

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

048° + 036^{o} + <S = 180^{o}

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

<S  = 180^{o} -  84^{o}

    = 96^{o}

<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}

       = 12.703

y = 12.70 km

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

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

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

⇒ x = \frac{9.992430}{0.99452}

      = 10.0475

x = 10.05 km

Thus,

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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