Triangles - Trigonometry

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Question

In a triangle, the side opposite the degree angle is . How long is the side opposite the degree angle?

Answer

Based on the 30-60-90 identity, the measure of the side opposite the 30 degree angle is doubled to get the hypotenuse.

Therefore,

$h=15\cdot2$

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Question

In a 30-60-90 triangle, the length of the side opposite the $30^\circ$ angle is . What is the length of the hypotenuse?

Answer

By definition, the length of the hypotenuse is twice the length of the side opposite the $30^\circ$ angle.

Recall that the hypotenuse is the side opposite the $90^\circ$ angle.

Thus, using the equation below, where ss represents the short side (that opposite the $30^\circ$ angle) we get:

$H=2\cdot SS$

Plugging in our values for the short side we find the hypotenuse as follows:

$5\cdot2 = 10$

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Question

A triangle has three angles , and such that and . The side opposite to measures units in length. How long is the side opposite of ?

Answer

A triangle with a angle relation is a , , degree triangle. The side opposite the smallest angle of a triangle is the shortest side, of length . The side opposite the largest angle is the longest side, measuring twice the length of the shortest side for this triangle, units.

$\frac{1}{2}=\sin{30}$

$=\frac{opp}{hyp}=\frac{3}{h}$

Therefore, to make the above statement true .

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Question

Triangle is equilateral with a side length of .

What is the height of the triangle?

Answer

An equilateral triangle has internal angles of 60°, so the sin of one of those angles is equivalent to the height of the triangle divided by the side length,

$\sin(60)=\frac{height}{side}$

so..

$h=s*\sin(60)=s\frac{\sqrt3}{2}$

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Question

What is the height of an equilateral triangle with side length 8?

Answer

The altitude of an equilateral triangle splits it into two 30-60-90 triangles. The height of the triangle is the longer leg of the 30-60-90 triangle. If the hypotenuse is 8, the longer leg is $4\sqrt{3}$ .

To double check the answer use the Pythagorean Thereom:

$\(4\sqrt{3})^2+4^2= \16(3)+16= \48+16=64 \ \sqrt{64}=8$

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Question

What is the ratio of the side opposite the $30^{\circ}$ angle to the hypotenuse?

Answer

Step 1: Locate the side that is opposite the $30^{\circ}$ side..

The shortest side is opposite the $30^\circ$ angle. Let's say that this side has length .

Step 2: Recall the ratio of the sides of a $30^\circ-60^\circ-90^\circ$ triangle:

From the shortest side, the ratio is $1:\sqrt {3}:2$ .

is the hypotenuse, which is twice as big as the shortest side..

The ratio of the short side to the hypotenuse is

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Question

It is known that the smallest side of a 30-60-90 triangle is 5.

Find $\sin(30\degree)$ .

Answer

We know that in a 30-60=90 triangle, the smallest side corresponds to the side opposite the 30 degree angle.

Additionally, we know that the hypotenuse is 2 times the value of the smallest side, so in this case, that is 10.

The formula for

$\sin=\frac{opposite}{hypotenuse}$ , so $\frac{5}{10}$ or $\frac{1}{2}$ .

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Question

It is known that for a 30-60-90 triangle,

$\sin(30\degree)=\frac{6}{12}$ .

Find the area of the triangle.

Note:

$\textup{Area}=\frac{1}{2}bh$

Answer

First, we know that in a 30-60-90 triangle,

$\sin(30)=\frac{height}{hypotenuse}$ .

Also, the base is the smallest side times $\sqrt3$ , so in our case it is $6\sqrt3$ .

The height is just the smallest side, .

Substituting these values into the formula given for area of a triangle, we obtain the answer $18\sqrt3$ .

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Question

In a triangle, if one leg is . What is the measure of the hypotenuse?

Answer

One option is to use the Pythagorean Theorem.

Since we have an isosceles triangle, both legs must be congruent.

Plug in to get your answer.

$h=\sqrt{6^2+6^2}$

$h=\sqrt{72}=\sqrt{36\cdot2}=6\sqrt2$

Or, you can remember the 45-45-90 identity, which states that the hypotenuse is $\sqrt{2}$ times the leg.

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Question

A triangle has three angles , , such that and together are as much as . What is the ratio of the longest side to the shortest?

Answer

A triangle with the sum of two angles equaling the third is a triangle

$\sin{A}=\frac{s}{h}, \frac{h}{s}=\frac{1}{\sin{A}}, A=45$

$\frac{h}{s}=\frac{1}{\frac{1}{\sqrt2}}=\sqrt2$

$h:s=\sqrt2:1$

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Question

Find the value of in the triangle below.

Answer

The first two things to recognize regarding our tirangle are 1) it is a right triangle and 2) it is an isosceles triangle. The two congruent sides tell us that the two non-right angles are also congruent, and a little quick math tells us that they each equal 45 degrees. This means our right triangle is not just any right triangle but a 45-45-90 triangle.

This is important because the sides of every 45-45-90 triangle follow the same ratio. The two legs are obviously always congruent to each other (being isosceles), but to find the hypotenuse, we simply have to multiply the length of a leg by $\sqrt{2}$ .

Given this fact we would be in good shape if we had the length of a leg and needed the hypotenuse. But we have the hypotenuse and need the leg, which we means we need to work backwards going this way, we need to divide the length of the hypotenuse by $\sqrt{2}$ . Therefore,

$x=\frac{16}{\sqrt{2}}$

However, general practice in mathematics doesn't allow us to leave a square root in the denominator. We solve this problem by rationalizing the denominator, which is accomplished by multiplying the numerator and the denominator by $\sqrt{2}$ .

$x=\frac{16}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}$

This effectively eliminates the square root in the denominator and provides our answer.

$x=\frac{16}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{16\sqrt{2}}{2}=8\sqrt{2}$

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Question

The following figure was made by beginning with a square. The midpoints of the four sides of the square were then joined to form another square. The process was repeated to form a third square and finally once more to form the fourth and smallest square in the middle, which has a side length of . Find the value of .

Answer

We begin by realizing that the midpoints of the sides of our outer square divide each side in half. Furthermore, the sides of our second square connecting these midpoints form four right triangles in each corner of our largest square.

But these right triangles are special right triangles. They are 45-45-90 triangles, which means we can find the hypotenuse (and thus the side of our second square) by multiplying the length of the leg by $\sqrt{2}$ . Therefore the length of a side of our second square is $20\sqrt{2}$ .

We now repeat the process, beginning by forming four new 45-45-90 triangles

To find the hypotenuse of each of these triangles (and thus the side length of our third square), we simply multiply by $\sqrt{2}$ again.

$10\sqrt{2}\cdot\sqrt{2}=20$

We then repeat the process one final time, multiplying by $\sqrt{2}$ again.

Our final hypotenuse and thus the side of our innermost square is $10\sqrt{2}$ .

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Question

Find the value of

Answer

Solving this problem begins with realizing that all three of our triangles are not only right triangles but isosceles and are therefore 45-45-90 triangles. That means in each triangle to get from the length of a leg to the length of the hypotenuse, we simply multiply by $\sqrt{2}$ . Therefore, the hypotenuse of our bottom triangle is

$7\cdot \sqrt{2}=7\sqrt{2}$

However, the hypotenuse of the bottom triangle is also the leg of the middle triangle. To find the hypotenuse of this triangle, we simply repeat the process.

$7\sqrt2 \cdot \sqrt{2}=14$

However, again the hypotenuse of the middle triangle is also the leg of the upper triangle. To find , the hypotenuse of the upper triangle, we simply repeat the process one last time.

$14\cdot \sqrt{2}=14\sqrt{2}$

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Question

In a 45-45-90 triangle, if the hypotenuse is 10, what is the perimeter of the triangle?

Answer

Write the Pythagorean Theorem.

In a 45-45-90 triangle, the length of the legs are equal, which indicates that:

Rewrite the formula and substitute the known sides.

$a=\sqrt{50}=5\sqrt2$

The lengths of the triangle are: $5\sqrt{2},5\sqrt{2},10$

Sum the three lengths for the perimeter.

$10\sqrt{2}+10$

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Question

One side of a -- triangle has a length of 3. Which cannot be the length of one of the other sides?

Answer

If 3 is one of the legs, then the hypotenuse is $3 \sqrt2$ .

If 3 is the hypotenuse, then the legs are $\frac{3}{\sqrt2}$ or equivalently $\frac{3 \sqrt2}{2}$

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Question

The hypotenuse of a -- triangle is 4. What is the length of each of the legs?

Answer

Divide the length of the hypotenuse by $\sqrt{2}$ to get the length of the legs:

$\frac{4 } {\sqrt2} = \frac{4 \sqrt2}{2} = 2 \sqrt2$

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Question

The perimeter of a square is 56 feet. What is the length of the diagonal in feet?

Answer

The perimeter of a square can be found using the formula , where P is the perimeter and s is the length of the side of the square.

The diagonal of a square forms the hypotenuse of a 45-45-90 triangle, where each leg is the side of the square. In a 45-45-90 triangle, the ratio of the hypotenuse to the leg is $\sqrt2$ , so the diagonal of this square is $14\sqrt2$ .

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Question

If $\small B$ = $\small 90^o$ , $\small a=7$ , and $\small b=8$ find $\small \small \angle A$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin B}{b}$ , we get $b\cdot \frac{\sin A}{a}=\sin B$ . In this equation, if $\sin B > 1$ , no angle A that satisfies the triangle can be found. If $\sin B = 1$ , $B=90^\circ$ and there is a right triangle determined. Finally, if $\sin B < 1$ , two measures of angle B can be calculated: an acute angle B and an obtuse angle $B'=180^\circ-B$ . In this case, there may be one or two triangles determined. If $B'+A\geq 180^\circ$ , then the angle B' is not a solution.

In this problem, $\sin B = 1$ , $B=90^\circ$ and there is one right triangle determined. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

$\small \frac{b}{\sin\angle B} = \frac{a}{\sin \angle A}$

Inputting the lengths of the triangle into this equation

$\small \frac{8}{\sin(90^o)} = \frac{7}{\sin\angle A}$

Isolating $\small \angle A$

$\small \angle A = \sin^{-1}(\frac{7\sin(90^o)}{8})$

$\small \angle A = 61^o$

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Question

If $\small c=10.3$ , $\small a=7.4$ , and $\small \angle A$ = $\small 34^o$ find $\small \angle C$ to the nearest degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin C}{c}$ , we get $c\cdot \frac{\sin A}{a}=\sin C$ . In this equation, if $\sin C >1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin C=1, C=90^\circ$ and there is a right triangle determined. Finally, if $\sin C <1$ , two measures of $\angle C$ can be calculated: an acute $\angle C$ and an obtuse angle $C'=180^\circ-C$ . In this case, there may be one or two triangles determined. If $C'+A \geq 180^\circ$ , then the $\angle C'$ is not a solution.

In this problem, $\sin C =.559<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

$\small \small \frac{c}{\sin\angle C} = \frac{a}{\sin\angle A}$

Inputting the values of the problem

$\small \frac{10.3}{\sin\angle C} = \frac{7.4}{\sin\angle 34^o}$

Rearranging the equation to isolate $\small \angle C$

$\small \angle C = \sin^{-1}(\frac{10.3\sin\angle34^o}{7.4})$

$\small \angle C =$ $\small 51^o$

When the original given angle ( $\angle A=34^\circ$ ) is acute, there will be:

One solution if the side opposite the given angle is equal to or greater than the other given side

No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side

In this problem, the side opposite the given angle is , which is less than the other given side . Therefore, we have a second solution. Find it by following the below steps:

$\angle C'=180^\circ-C=180^\circ-51^\circ=129^\circ$ .

$C'+A=129^\circ+34^\circ=163^\circ\leq 180^\circ$ , so $\angle C'$ is a solution.

Therefore there are two values for an angle, $\angle C=51^\circ$ and $\angle C'=129^\circ$

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Question

If $A=5^\circ$ , $\small b=10$ , $\small a=4$ , find $\small \angle B$ to the nearest tenth of a degree.

Answer

Notice that the given information is Angle-Side-Side, which is the ambiguous case. Therefore, we should test to see if there are no triangles that satisfy, one triangle that satisfies, or two triangles that satisfy this. From $\frac{\sin A}{a}=\frac{\sin B}{b}$ , we get $b\cdot \frac{\sin A}{a}=\sin B$ . In this equation, if $\sin B > 1$ , no $\angle A$ that satisfies the triangle can be found. If $\sin B = 1$ , $B=90^\circ$ and there is a right triangle determined. Finally, if $\sin B < 1$ , two measures of $\angle B$ can be calculated: an acute $\angle B$ and an obtuse $\angle$ $B'=180^\circ-B$ . In this case, there may be one or two triangles determined. If $B'+A\geq 180^\circ$ , then the $\angle B'$ is not a solution.

In this problem, $\sin\angle A=.087<1$ , so there may be one or two angles that satisfy this triangle. Since we have the length of two sides of the triangle and the corresponding angle of one of the sides, we can use the Law of Sines to find the angle that we are looking for. This goes as follows:

$\small \frac{b}{\sin\angle B} = \frac{a}{\sin\angle A}$

Inputting the values from the problem

$\small \frac{10}{\sin\angle B} = \frac{4}{\sin\angle 5^o}$

$\small \angle B = \sin^{-1}(\frac{10\sin\angle 5^o}{4})$

$\small \small \angle B = 12.6^o$

When the original given angle ( $\angle A=5^\circ$ ) is acute, there will be:

One solution if the side opposite the given angle is equal to or greater than the other given side

No solution, one solution (right triangle), or two solutions if the side opposite the given angle is less than the other given side

In this problem, the side opposite the given angle is , which is less than the other given side . Therefore, we have a second solution. Find it by following the below steps:

$\angle B'=180^\circ-B=180^\circ-12.6^\circ=167.4^\circ$

$\angle B'+\angle A=167.4^\circ+5^\circ=172.4^\circ$ $\leq180^\circ$ , so $\angle B'$ is a solution.

Therefore there are two values for an angle, $\angle B=12.6^\circ$ and $\angle B'=167.4^\circ$ .

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