Trigonometric Applications - Pre-Calculus

Card 0 of 292

Question

Find the area of this triangle:

Answer

To solve, use Heron's Formula,

$area = \sqrt{ s (s - a ) (s - b )( s- c )}$

where , , and are the side lengths and

$s = \frac{ 1}{2} ( a + b + c )$ .

In this particular case,

thus,

$s = \frac{ 1}{2} ( 13 + 11 + 7 ) = \frac{ 1}{2} \cdot 31 = 15.5$ .

Plugging these values into the Heron's Formula we arrive at our answer.

$\ area = \sqrt{ 15.5 ( 15.5 - 11 ) (15.5 - 13 ) ( 15.5 - 7 )} \ \= \sqrt{ 15.5 \cdot 4.5 \cdot 2.5 \cdot 8. 5 } \approx 38.50$

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Question

In a right triangle, if the hypotenuse is and a leg is , what is the area of the triangle?

Answer

Use the Pythagorean Theorem to find the other leg.

${}a^2+b^2=c^2$

$a=\sqrt7$

The length of the given leg is 3, and the unknown leg is $\sqrt7$ .

Use the area of a triangle formula and solve.

$A=\frac{bh}{2}= \frac{3\sqrt7}{2}$

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Question

An isosceles right triangle has a hypotenuse of 1. What is the area of this triangle?

Answer

Write the formula for the Pythagorean theorem.

In an isosceles right triangle, both legs of the right triangle are equal.

Substitute the either variable and the known hypotheuse and determine the side length.

$b^2=\frac{1}{2}$

$b=\sqrt\frac{1}{2}$

This length represents both the base and the height of the triangle. Write the area of a triangle and substitute to solve for the area.

$A=\frac{bh}{2}=\frac{\sqrt\frac{1}{2}\sqrt\frac{1}{2}}{2}=\frac{\frac{1}{2}}{2}= \frac{1}{4}$

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Question

In isosceles triangle , $m\angle A = m\angle B = 29^\circ$ . If side , what is the approximate length of the two legs and ?

Answer

In the diagram, AB is cut in half by the altitude.

From here it easy to use right triangle trigonometry to solve for AC.

$\ \cos 29^\circ =\frac{opp}{hyp}= \frac {5}{AC} \* \* AC = \frac {5}{cos 29^\circ} \approx 5.7$

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Question

Given , and the lower angles of the isosceles triangle are $68^\circ$ , what is the length of ? Round to the nearest tenth.

Answer

Since the angle of the isosceles is $68^\circ$ , the larger angle of the right triangle formed by is also $68^\circ$ .

Using , we can find :

$tan(68)=\frac{h}{3}$ .

Then solve for :

.

Simplify: .

Lastly, round and add appropriate units: .

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Question

Find the area of the following Isosceles triangle (units are in cm):

Answer

The formula for the area of a triangle is:

$A=\frac{1}{2}\cdot base\cdot height$

We already know what the base is and we can find the height by dividing the isosceles triangle into 2 right triangles:

From there, we can use the Pathegorean Theorem to calculate height:

$9^{2}=h^{2}+4^{2}$

$h^{2}=81-16=65$

$h=\sqrt{65}$

To find the area, now we just plug these values into the formula:

$A= \frac{1}{2}\cdot8 \cdot \sqrt{65}$

$A=32.25 cm^{2}$

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Question

Find the area of the given isosceles triangle and round all values to the nearest tenth:

Answer

The first step to solve for area is to divide the isosceles into two right triangles:

From there, we can determine the height and base needed for our area equation $A=\frac{1}{2}\cdot b \cdot h$

$sin(67.5^{\circ})=\frac{1/2Base}{12; cm}$

$0.924=\frac{1/2Base}{12; cm}$

From there, height can be easily determined using the Pathegorean Theorem:

$(12; cm)^{2}=(11.1; cm)^{2}+h^{2}$

$h^{2}=144; cm^{2}-123.2; cm^{2}=20.8cm^{2}$

Now both values can be plugged into the Area formula:

$A=\frac{1}{2}\cdot22.2; cm\cdot4.6; cm=51.1cm^{2}$

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Question

Find the area of the given isosceles triangle:

Answer

The first step is to divide this isosceles triangle into 2 right triangles, making it easier to solve:

The equation for area is $A=\frac{1}{2}base\cdot height$

We already know the base, so we need to solve for height to get the area.

$tan20^{\circ}=\frac{height}{7; cm}$

$0.364=\frac{height}{7; cm}$

Then we plug in all values for the equation:

$Area=\frac{1}{2}\cdot14; cm\cdot2.55; cm=17.85; cm^{2}$

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Question

Find the area of the given isosceles triangle:

Answer

The first step toward finding the area is to divide this isosceles triangle into two right triangles:

Trigonometric ratios can be used to find both the height and the base, which are needed to calculate area:

$sin(75^{\circ})=\frac{height}{5: cm}$

$0.9659=\frac{height}{5: cm}$

$sin(15^{\circ})=\frac{1/2base}{5: cm}$

$0.2588=\frac{1/2base}{5: cm}$

With both of those values calculated, we can now calculate the area of the triangle:

$Area=\frac{1}{2}\cdot2.59: cm\cdot4.83: cm=6.25: cm^2$

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Question

Solve the right triangle given that a=5, b=12, and A=22.620°

Answer

C is given as 90°.

A is given as 22.620°

a is given as 5

b is given as 12

$a^{2}+b^{2}=c^{2}$

Therefore...

$5^{2}+12^{2}=c^{2}$

$169=c^{2}$

$\sqrt{169}=\sqrt{c^{2}}$

All angles of a triangle add up to equal 180°.

$90^{\circ}+22.620^{\circ}+B=180^{\circ}$

$112.620^{\circ}+B=180^{\circ}$

$B=180^{\circ}-112.620^{\circ}$

$B=67.380^{\circ}$

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Question

Solve the right triangle.

C=90°

B=45°

a=5

c= $\sqrt{50}$

Answer

Given that:

C=90°

B=45°

a=5

c= $\sqrt{50}$

$a^{2}+b^{2}=c^{2}$

Therefore...

$5^{2}+b^{2}=(\sqrt{50})^{2}$

$25+b^{2}=50$

$b^{2}=50-25$

$b^{2}=25$

$\sqrt{b^{2}} =\sqrt{25}$

All angles of a triangle add up to 180°.

$A+45^{\circ}+90^{\circ}=180^{\circ}$

$A+135^{\circ}=180^{\circ}$

$A=180^{\circ}-135^{\circ}$

$A=45^{\circ}$

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Question

A right triangle has a base of 10 and a hypotenuse of 20. What is the length of the other leg?

Answer

Write the Pythagorean Theorem.

Substitute the values of the leg and hypotenuse. The hypotenuse is the longest side of the right triangle. Solve for the unknown variable.

$a=\sqrt{300} = \sqrt{100\times3} = \sqrt{100}\times \sqrt3 = 10\sqrt3$

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Question

In the right triangle ABC, side AB is cm long, side AC is cm long, and side BC is the hypotenuse. How long is side BC?

Answer

Given that ABC is a right triangle, the length of hypotenuse BC is the root of the sum of the squares of the two other sides (in other words, . Since AB is cm long and AC is cm long, we get that , and so $BC = \sqrt{106}$ .

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Question

The side lengths of right triangle ABC are such that AC > BC > AB. AC = 25 and AB = 9. What is the length of BC?

Answer

When you are using Pythagorean Theorem to calculate the missing side of a right triangle, it is crucial that you identify which side is the hypotenuse, in the Pythagorean equation . Here you're told that side AC is the longest of the three sides, so 25 will serve as the length of the hypotenuse and the value of . This allows you to set up the equation:

$a^{2}+9^{2}=25^{2}$

And then you can perform the calculations on the known values:

Meaning that:

$a^{2}=544$

From there you can simplify, arriving at a = 4 times the square root of 34.

$a^{2}$

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Question

Find the exact area of a triangle with side lengths of , , and .

Answer

Use the Heron's Formula:

$s=\frac{a+b+c}{2}$

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Solve for .

$s=\frac{a+b+c}{2}=\frac{2+3+4}{2}=\frac{9}{2}$

Solve for the area.

$A=\sqrt{s(s-a)(s-b)(s-c)}$

$A=\sqrt{\frac{9}{2}\left(\frac{9}{2}-2\right)\left(\frac{9}{2}-3\right)\left(\frac{9}{2}-4\right)}$

$A=\sqrt{\frac{9}{2}\left(\frac{5}{2}\right)\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)}=\sqrt{\frac{135}{16}}=\frac{\sqrt{135}}{4}$

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Question

What is the area of a triangle with side lengths , , and ?

Answer

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(4+5+7)$

$s=\frac{1}{2}(16)$

So the area is

$area=\sqrt{8(8-4)(8-5)(8-7)}$

$area=\sqrt{96}$

$area=4\sqrt{6}$

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Question

What is the area of a triangle if the sides of a triangle are , , and ?

Answer

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(4+2+4)$

$s=\frac{1}{2}(10)$

So the area is

$area=\sqrt{5(5-4)(5-2)(5-4)}$

$area=\sqrt{15}$

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Question

What is the area of a triangle with sides , , and ?

Answer

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(6+5+9)$

$s=\frac{1}{2}(20)$

So the area is

$area=\sqrt{10(10-6)(10-5)(10-9)}$

$area=\sqrt{200}$

$area=10\sqrt{2}$

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Question

What is the area of a triangle with side lengths of , , and ?

Answer

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(3+4+5)$

$s=\frac{1}{2}(12)$

So the area is

$area=\sqrt{6(6-3)(6-4)(6-5)}$

$area=\sqrt{36}$

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Question

What is the area of a triangle with side lengths , , and ?

Answer

We can solve this question using Heron's Formula. Heron's Formula states that:

The semiperimeter is

$s=\frac{1}{2}(a+b+c)$

where , , are the sides of a triangle.

Then the area is

$area=\sqrt{s(s-a)(s-b)(s-c)}$

So if we plug in

$s=\frac{1}{2}(5+12+13)$

$s=\frac{1}{2}(30)$

So the area is

$area=\sqrt{15(15-5)(15-12)(15-13)}$

$area=\sqrt{900}$

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