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4 years ago

Copy the problems onto your paper, mark the given and prove the statements asked. Given: marked Prove: ΔABC ≅ ΔDBC, ΔEHF ≅ ΔGHF

Mathematics

1 answer:

hjlf4 years ago

7 0

The proof is given below. Please go through it.

Step-by-step explanation:

To solve Δ ABC ≅ Δ DBC

From Δ ABC and Δ DBC

AB = BD (given)

AC = CD (given)

BC is common side

By SSS condition Δ ABC ≅ Δ DBC ( proved)

To solve Δ EHF ≅ Δ GHF

Δ EHF and Δ GHF

EH = HG ( given)

∠ EFH = ∠ GFH ( each angle is 90°)

HF is common side

By RHS condition

Δ EHF ≅ Δ GHF

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Answer:

b must equal 7 and a second solution to the system must be located at (2, 5).

Step-by-step explanation:

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3 years ago

Towns P,Q,R and S are shown. Q is 35 km due East of P S is 15km due West of P R is 15km due South of P Work out the bearing of R

Anni [7]

Answer:

Part A

The bearing of the point 'R' from 'S' is 225°

Part B

The bearing from R to Q is approximately 293.2°

Step-by-step explanation:

The location of the point 'Q' = 35 km due East of P

The location of the point 'S' = 15 km due West of P

The location of the 'R' = 15 km due south of 'P'

Part A

To work out the distance from 'R' to 'S', we note that the points 'R', 'S', and 'P' form a right triangle, therefore, given that the legs RP and SP are at right angles (point 'S' is due west and point 'R' is due south), we have that the side RS is the hypotenuse side and ∠RPS = 90° and given that $\overline{RP}$ = $\overline{SP}$ , the right triangle ΔRPS is an isosceles right triangle

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The bearing of the point 'R' from 'S' measured from the north of 'R' = 180° + 45° = 225°

Part B

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Therefore the bearing from R to Q = 270 + 90 - 66.8 ≈ 293.2°

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3 years ago

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