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3 years ago

5

Exterior Angles and Isosceles Triangles

2 answers:

Naddika [18.5K]3 years ago

7 0

Exterior angles =180-75=105
Interior angles =180_69=111

Ad libitum [116K]3 years ago

3 0

Step-by-step explanation:

this is not an isoceles triangle.

anyway, we need to remember 2 things :

1. the sum of all angles in a triangle is always 180°.

2. the sum of all angles around a single point on a line is 360° (as this point wound be the center of an invisible circle). and as the line cuts this circle in half, the angles on one side of the line around this point is therefore 180°.

so, the internal angle Q is then

180 = 50 + 55 + Q

Q = 180 - 50 - 55 = 180 - 105 = 75°

the exterior angle x at Q is then

180 = Q + x = 75 + x

x = 180 - 75 = 105°

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3 years ago

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tatyana61 [14]

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4 years ago

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In the diagram, ATPR~AXPZ. Find the length of side PZ.

xenn [34]

Answer:

<em>PZ = 15</em>

Step-by-step explanation:

<u>Similar Triangles</u>

Two triangles are similar if all of their corresponding side lengths are proportional to the same scale factor.

It's given triangles TPR and XPZ are similar, thus we can find the scale factor by comparing two corresponding sides with known lengths.

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3 years ago

Two ships set sail in different directions from the same place. Tyler's ship sail 35 miles while Noah's ship sails for 42 miles.

earnstyle [38]

Answer:

<em>The angle between their paths when they started is 93°.</em>

Step-by-step explanation:

<u>The Law of Cosines</u>

It relates the length of the sides of a triangle with one of its internal angles.

Let a,b, and c be the length of the sides of a given triangle, and x the included angle between sides a and b, then the following relation applies:

$c^2=a^2+b^2-2ab\cos x$

When the two ships travel in different directions from the same point in the plane, they form an angle we called x in the image below.

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Since we know the values of all three side lengths, we solve the equation for x:

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Substituting values:

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$\displaystyle \cos x=-\frac{147}{2940}=-\frac{1}{20}$

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6 0

3 years ago

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Alex787 [66]

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