Geometry - How to find the length of the side of an acute / obtuse triangle

In ΔABC: a = 10, c = 15, and B = 50°.

Find b (to the nearest tenth).

11.5

9.7

12.1

11.9

14.3

Explanation

This problem requires us to use either the Law of Sines or the Law of Cosines. To figure out which one we should use, let's write down all the information we have in this format:

A = ? a = 10

B = 50° b = ?

C = ? c = 15

Now we can easily see that we do not have any complete pairs (such as A and a, or B and b) but we do have one term from each pair (a, B, and c). This tells us that we should use the Law of Cosines. (We use the Law of Sines when we have one complete pair, such as B and b).

Law of Cosines:
$a^2 = b^2 + c^2 - 2bc\cos(A)$
$b^2 = a^2 + c^2 - 2ac\cos(B)$
$c^2 = a^2 + b^2 - 2ab\cos(C)$

Since we are solving for b, let's use the second version of the Law of Cosines.
This gives us:

$b^2 = 10^2 + 15^2 - 2(10)(15)\cos(50)$

A triangle has sides of lengths 12 meters, 1,200 centimeters, and 12 millimeters. Is the triangle scalene, isosceles but not equilateral, or equilateral?

Isosceles, but not equilateral

Equilateral

Scalene

Explanation

Convert each of the three measures to the same unit; we will choose the smallest unit, millimeters.

One meter is equivalent to 1,000 millimeters, so 12 meters can be converted to millimeters by multiplying by 1,000:

$12\textrm{ m }\times 1,000 \textrm{ mm / m}= 12,000 \textrm{ mm}$

One centimeter is equivalent to ten millimeters, so 1,200 cenitmeters can be converted to millimeters by multiplying by 10:

$1,200\textrm{ cm }\times 10 \textrm{ mm / cm}= 12,000 \textrm{ mm}$

These two sides have the same length. However, the third side, which has length 12 millimeters, is of different length. Since the triangle has exactly two congruent sides, it is by definition isosceles, but not equilateral.

A triangle has sides of lengths 18.4, 18.4, and 23.7. Is the triangle scalene or isosceles?

Isosceles

Scalene

Explanation

The triangle has two sides of the same length, 18.4, so, by definition, it is isosceles.

A triangle has sides of lengths one and one half feet, twenty-four inches, and one yard. Is the triangle scalene, isosceles but not equilateral, or equilateral?

Scalene

Equilateral

Isosceles, but not equilateral

Explanation

Convert each of the three measures to the same unit; we will choose the smallest unit, inches.

One foot is equal to twelve inches, so $1\frac{1}{2}$ feet can be converted to inches by multiplying by 12:

$1\frac{1}{2} \textrm{ ft} \times 12 \textrm{ in / ft } = 18 \textrm{ in}$

One yard is equal to 36 inches.

The lengths of the sides in inches are 18, 24, and 36. Since no two sides have the same measure, the triangle is by definition scalene.

True or false: It is possible for a triangle with sides of length $\frac{1}{2}$ , $\frac{1}{3}$ , and $\frac{1}{4}$ to exist.

True

False

Explanation

By the Triangle Inequality Theorem, the sum of the measures of the shortest two sides of a triangle must exceed the length of the longest side.

Write each length in terms of a common denominator; this is . The fractions convert as follows:

$\frac{1}{2} = \frac{1 \times 6}{2 \times 6} = \frac{6}{12}$

$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$

$\frac{1}{4} = \frac{1 \times 3 }{4 \times 3} = \frac{3}{12}$

$\frac{1}{2}$ is the greatest of the three, so for this triangle to be possible it must hold that

$\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$

or, equivalently,

$\frac{4}{12}+ \frac{3}{12} > \frac{6}{12}$

$\frac{7}{12} > \frac{6}{12}$

This is indeed the case, so a triangle with these sidelengths can exist.

True or false: It is possible for a triangle with sides of length five feet, fifty inches, and one and one half yards to exist.

True

False

Explanation

First, convert all of the given sidelengths to the same unit; here, we choose the smallest unit, inches.

One foot is equal to 12 inches, so to convert feet to inches, multiply by 12:

$5 \textrm{ ft} \times 12 \textrm{ in / ft} = 60\textrm{ in}$

One yard is equal to 36 inches, so to convert yards to inches, multiply by 36:

$1\frac{1}{2}\textrm{ yd} \times 36 \textup{in / yd}= 54 \textrm{ in}$

The measures of the sides of the triangle, in inches, are 50, 54, and 60.

By the Triangle Inequality Theorem, the sum of the measures of the shortest two sides of a triangle must exceed the length of the longest side, so for this triangle to be possible it must hold that

This is indeed the case, so a triangle with these sidelengths can exist.

Hinge

Refer to the above diagram. By what statement does it follow that ?

The Triangle Inequality

The Hinge Theorem

The Converse of the Pythagorean Theorem

The Side-Side-Side Postulate

The Side-Side-Side Similarity Theorem

Explanation

In any triangle, the sum of the lengths of any two sides is greater than the length of the third side; the statement is a specific example. This is a direct result of the Triangle Inequality Theorem.

True or false: It is possible for a triangle with sides of length one meter, 250 centimeters, and 1,200 millimeters to exist.

False

True

Explanation

First, convert all of the given sidelengths to the same unit; here, we choose the smallest unit, millimeters.

One meter is equal to 1,000 millimeters.

One centimeter is equal to 10 millimeters, so convert 250 centimeters to millimeters by multiplying by 10:

$250 \textrm{ cm } \times 10 \textrm{mm / cm} = 2,500 \textrm{ mm}$

The measures of the sides of the triangle, in millimeters, are 1,000, 1,200, and 2,500.

By the Triangle Inequality, the sum of the measures of the shortest two sides of a triangle must exceed the length of the longest side, so for this triangle to be possible it must hold that

This is false, so a triangle with these sidelengths cannot exist.

True or false: It is possible for a triangle with sides of length $\frac{1}{2}$ , $\frac{1}{4}$ , and $\frac{1}{8}$ to exist.

False

True

Explanation

By the Triangle Inequality Theorem, the sum of the measures of the shortest two sides of a triangle must exceed the length of the longest side.

Write each length in terms of a common denominator; this is . The fractions convert as follows:

$\frac{1}{2} = \frac{1 \times 4 }{2 \times 4} = \frac{4}{8}$

$\frac{1}{4} = \frac{1 \times 2 }{4 \times 2} = \frac{2}{8}$

$\frac{1}{2}$ is the greatest of the three, so for this triangle to be possible it must hold that

$\frac{1}{8} + \frac{1}{4} > \frac{1}{2}$

or, equivalently,

$\frac{1}{8} +\frac{2}{8}> \frac{4}{8}$

$\frac{3}{8}> \frac{4}{8}$

This is false, so a triangle with these sidelengths cannot exist.

Given: $\bigtriangleup ABC$ ; $m \angle A = 76 ^{\circ }$ ; $m \angle B = 36 ^{\circ }$ .

True or false: $\overline{AB} \cong \overline{BC}$

False

True

Explanation

The sum of the measures of the interior angles of a triangle is $180^{\circ }$ , so

$m \angle A + m \angle B + m \angle C = 180^{\circ }$

Substitute the given two angle measures and solve for $m \angle C$ :

$76 ^{\circ }+36 ^{\circ } + m \angle C = 180^{\circ }$

$112 ^{\circ } + m \angle C = 180^{\circ }$

Subtract $112^{\circ }$ from both sides:

$112 ^{\circ } + m \angle C - 112 ^{\circ } = 180^{\circ } - 112 ^{\circ }$

$m \angle C = 68^{\circ }$

Therefore, $\angle C cong \angle A$

By the Isosceles Triangle Theorem, if $\overline{AB} \cong \overline{BC}$ , their opposite sides are also congruent - that is, $\angle C \cong \angle A$ . Since this is not the case, $\overline{AB} cong \overline{BC}$ .

How to find the length of the side of an acute / obtuse triangle

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