Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
3. Section 8.3
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Exercise 94 Page 507

Practice makes perfect
a From the diagram we see a right triangle. Since one of the non-right angles is 60^(∘), this is a 30-60-90 triangle. In such a triangle, the legs and hypotenuse have the following relationship.

The hypotenuse of our triangle is 28 units. With this information we can write and solve the following equation. 2a=28 ⇔ a= 14 Now we can determine the value of x. x=asqrt(3) ⇒ x= 14sqrt(3)

b In this triangle we know two angles and the opposite side of one of them. To find the opposite side of the second angle, we can use the Law of Sines.
Let's solve for x in this equation.
x/sin 50^(∘)=7/sin 68^(∘)
Solve for x
x=7/sin 68^(∘)* sin 50^(∘)
x=7sin 50^(∘)/sin 68^(∘)
x=5.78343...
x≈ 5.78
c In a right triangle the hypotenuse is always greater than either of its legs. From the diagram we see that this is not the case, which means this triangle cannot exist.
d The area of a triangle is calculated by multiplying the base and height of the triangle and dividing the product by 2. From the diagram, we see that the base is 8 units and the height is x units. With this information we can write the following equation.
A=1/2(8)(x) The area of the triangle should equal 96 square units. Therefore, if we set the area equal to 96 we can solve for the triangle's height x.
A=1/2(8)(x)
96=1/2(8)(x)
Solve for x
96=8x/2
96=4x
24=x
x=24