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Masteriza [31]
2 years ago
6

Suppose we wish to construct the perpendicular bisector of line segment EF. We can correctly begin by placing the compass point

on E and marking off the distance to point
Mathematics
2 answers:
Bess [88]2 years ago
7 0

The answer is marking of the distence to point A

julia-pushkina [17]2 years ago
3 0
- to a point which is more than halfway to the point F. Then draw an arc  around the point. Next place the compass point at  F and,  keeping the same distance between the points of the compass,  draw another arc so as to intersect the first arc at 2 points above and below the line EF. Finally draw a line passing through these 2 points of intersection.  This is the perpendicular bisector of EF.
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Solve the triangle. Round your answers to the nearest tenth.
wel

Answer:

m∠A ≈ 43°

m∠B ≈ 55°

mBC ≈ 20

Step-by-step explanation:

Law of Sines: \frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC}

Step 1: Find m∠B

\frac{29}{sin82} =\frac{24}{sinB}

Step 2: Solve for ∠B

29sinB = 24sin82°

sinB = 24sin82°/29

B = sin⁻¹(24sin82°/29)

B = 55.038°

Step 3: Find m∠A

180 - (55.038 + 82)

180 - 137.038

m∠A = 42.962°

Step 4: Find BC

\frac{29}{sin82} =\frac{BC}{sin42.962}

Step 5: Solve for BC

29sin42.962° = BCsin82°

BC = 29sin42.962°/sin82°

BC = 19.9581

4 0
2 years ago
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