Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
2. Section 6.2
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Exercise 55 Page 368

Practice makes perfect
a The triangle is isosceles which means it has a pair of congruent legs and a pair of congruent base angles. We know that one of these base angles is 52^(∘). With this we can find the measure of the vertex angle.
θ + 52^(∘) + 52^(∘) = 180^(∘) ⇔ θ = 76^(∘)

Let's add this information to the diagram.

Now we have enough information to solve for x with the Law of Sines.
sin A/a=sin B/b
sin 52^(∘)/37=sin 76^(∘)/x
Solve for x
sin 52^(∘)/37* x=sin 76^(∘)
sin 52^(∘) * x=37 sin 76^(∘)
x=37 sin 76^(∘)/sin 52^(∘)
x=45.55894...
x≈ 45.56
b With the given information, we can use the Law of Cosines to determine x.
c^2=a^2+b^2-2abcos C
x^2=16^2+7^2-2(16)(7)cos 31^(∘)
Solve for x
x^2=256+49-2(16)(7)cos 31^(∘)
x^2=256+49-192.00547
x^2=112.99452
x=± 10.62988...

x > 0

x=10.62988...
x≈ 10.63
c Since we have been given two angles in this triangle, we can find the measure of the last angle by equating the sum of the known angle measures with 180^(∘) and solving for the unknown angle.

θ +5^(∘)+85^(∘) = 180^(∘) ⇔ θ = 90^(∘)

With the extra piece of information, we can use the Law of Sines to determine x.
sin A/a=sin B/b
sin 90^(∘)/3046=sin 5^(∘)/x
Solve for x
sin 90^(∘)/3046* x=sin 5^(∘)
sin 90^(∘) * x=3046 sin 5^(∘)
x=3046 sin 5^(∘)/sin 90^(∘)
x=265.47639...
x≈ 265.48
d Since this is a right triangle, we can use the Pythagorean Theorem to find the value of x.
a^2+b^2=c^2
x^2+(x+7)^2=13^2
Solve for x
x^2+(x+7)^2=169
x^2+x^2+14x+49=169
2x^2+14x+49=169
2x^2+14x-120=0
x^2+7x-60=0
To solve this equation, we will use the Quadratic Formula.
x=- b± sqrt(b^2-4ac)/2a

a= 1, b= 7, c= - 60

x=- 7± sqrt(7^2-4( 1)( - 60))/2( 1)
Simplify right-hand side
x=- 7± sqrt(7^2+240)/2
x=- 7± sqrt(49+240)/2
x=- 7± sqrt(289)/2
x=- 7± 17/2
Finally, to find the solutions, we have to split the fraction into the positive and negative case. Positive:& - 7 + 17/2= 5 [0.5em] Negative:& - 7 - 17/2=- 12 Note that a side in a triangle cannot be negative making x=5 the only viable solution.