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alukav5142 [94]
3 years ago
11

2. Use the following picture for this question:

Mathematics
1 answer:
Art [367]3 years ago
4 0

Answer:

a.  24\sqrt{3}  and 41.6

b.  52.1

Step-by-step explanation:

a.

Considering the left side triangle the blue dotted side is the side "opposite" to the angle given and the side 24 is the side that is "adjacent" to the angle given. The trigonometric ratio tan relates opposite to adjacent. Also, let the blue dotted side be y.

<u>Note:</u> the exact value of tan 60 is \sqrt{3}

Thus, we can write Tan(60)=\frac{y}{24}\\y=24*Tan(60)\\y=24*\sqrt{3} \\y=24\sqrt{3}

Approximate value (rounded to nearest tenth):  24\sqrt{3} =41.6

b.

Considering the triangle to the right, the side "opposite" to the angle given (53 degrees) is 41.6 (just found in part (a)) and the side "hypotenuse" (side opposite to 90 degree angle) is x. The trigonometric ratio sine relates opposite and hypotenuse.

Thus we can write and solve:

Sin(53)=\frac{41.6}{x}\\x=\frac{41.6}{Sin(53)}=52.1

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(4x2+x+7)(8x−5)

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4x2(8x)+4x2⋅−5+(x(8x)+x⋅−5)+(7(8x)+7⋅−5)

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32x3−12x2+51x−35

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Given: KLMN is a trapezoid, m∠N=m∠KML, FD=8, LM KN = 3 5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3 5 Find: KM
denis23 [38]

Answer:

The length of side KM is \sqrt{109} units.

Step-by-step explanation:

Given information:  KLMN is a trapezoid, ∠N= ∠KML, FD=8, LM:KN=3:5, F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3\sqrt{5}.

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is 3\sqrt{5}.

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since LM:KN=3:5, therefore LM is 3x and KN is 5x.

\frac{3x+5x}{2}=8

\frac{8x}{2}=8

x=2

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

KN=KA+AE+EN

10=6+2(EN)                (KA=EN, isosceles trapezoid)

EN=2

KE=KN-EN=10-2=8

Therefore the length of KE is 8.

Use pythagoras theorem is triangle EKM.

Hypotenuse^2=base^2+perpendicular^2

(KM)^2=(KE)^2+(ME)^2

(KM)^2=(8)^2+(3\sqrt{5})^2

KM^2=64+9(5)

KM=\sqrt{109}

Therefore the length of side KM is \sqrt{109} units.

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3 years ago
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