Trigonometric Applications

Help Questions

Pre-Calculus › Trigonometric Applications

Questions 1 - 10

Solve the triangle

Law_of_cosines__sss_

$A = 115^{\circ}, B = 22^{\circ}, C=43^{\circ}$

$A = 100^{\circ}, B = 31^{\circ}, C=49^{\circ}$

$A = 95^{\circ}, B = 23^{\circ}, C=62^{\circ}$

None of the other answers

$A = 99^{\circ}, B = 21^{\circ}, C=54^{\circ}$

Explanation

Since we are given all 3 sides, we can use the Law of Cosines in the angle form:

$cos(A) = \frac{-a^2+b^2+c^2}{2bc}$

$cos(B) = \frac{a^2-b^2+c^2}{2ac}$

$cos(C) = \frac{a^2+b^2-c^2}{2ab}$

Let's start by finding angle A:

$cos(A) = \frac{-(12)^2+5^2+9^2}{2\cdot5\cdot9} = \frac{-38}{90} = -0.4\overline{22}$

$A = cos^{-1}(-0.4\overline{22}) = 115^{\circ}$

Now let's solve for B:

$cos(B) = \frac{12^2-(5)^2+9^2}{2\cdot12\cdot9} = \frac{200}{216} = 0.\overline{925}$

$B = cos^{-1}(0.\overline{925}) = 22^{\circ}$

We can solve for C the same way, but since we now have A and B, we can use our knowledge that all interior angles of a triangle must add up to 180 to find C.

$C = 180^{\circ} - 115^{\circ} - 22^{\circ} = 43^{\circ}$

Solve the right triangle.

Pcq1

C=90°

B=45°

a=5

c= $\sqrt{50}$

A=45°

b=5

A=135°

b=5

A=135°

b=2.07

A=45°

b= $\sqrt{50}$

None of these answers are correct.

Explanation

Pcq1

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°.

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

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

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

$A=45^{\circ}$

Solve the triangle

Law_of_cosines__sss_

$A = 115^{\circ}, B = 22^{\circ}, C=43^{\circ}$

$A = 100^{\circ}, B = 31^{\circ}, C=49^{\circ}$

$A = 95^{\circ}, B = 23^{\circ}, C=62^{\circ}$

None of the other answers

$A = 99^{\circ}, B = 21^{\circ}, C=54^{\circ}$

Explanation

Since we are given all 3 sides, we can use the Law of Cosines in the angle form:

$cos(A) = \frac{-a^2+b^2+c^2}{2bc}$

$cos(B) = \frac{a^2-b^2+c^2}{2ac}$

$cos(C) = \frac{a^2+b^2-c^2}{2ab}$

Let's start by finding angle A:

$cos(A) = \frac{-(12)^2+5^2+9^2}{2\cdot5\cdot9} = \frac{-38}{90} = -0.4\overline{22}$

$A = cos^{-1}(-0.4\overline{22}) = 115^{\circ}$

Now let's solve for B:

$cos(B) = \frac{12^2-(5)^2+9^2}{2\cdot12\cdot9} = \frac{200}{216} = 0.\overline{925}$

$B = cos^{-1}(0.\overline{925}) = 22^{\circ}$

We can solve for C the same way, but since we now have A and B, we can use our knowledge that all interior angles of a triangle must add up to 180 to find C.

$C = 180^{\circ} - 115^{\circ} - 22^{\circ} = 43^{\circ}$

Solve the right triangle.

Pcq1

C=90°

B=45°

a=5

c= $\sqrt{50}$

A=45°

b=5

A=135°

b=5

A=135°

b=2.07

A=45°

b= $\sqrt{50}$

None of these answers are correct.

Explanation

Pcq1

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°.

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

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

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

$A=45^{\circ}$

Solve the right triangle.

Pcq1

C=90°

B=45°

a=5

c= $\sqrt{50}$

A=45°

b=5

A=135°

b=5

A=135°

b=2.07

A=45°

b= $\sqrt{50}$

None of these answers are correct.

Explanation

Pcq1

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°.

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

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

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

$A=45^{\circ}$

Which of the following are the missing sides of the triangle?

$\text{I. }4.55$

$\text{II. }5.49$

$\text{III. }1.72$

$\text{IV. }14.56$

$\text{I and III}$

$\text{II and IV}$

$\text{I and II}$

None of the above

Explanation

In order to solve this problem, we need to find

Since all the angles of a triangle add to $180^{\circ}$ , we can easily find it:

$180^{\circ}-65^{\circ}-20^{\circ}=95^{\circ}$

We can now use the Law of Sines to find the missing sides:

$\frac{a}{\sin 65}=\frac{5}{\sin 95}$

$a=\frac{5\times \sin 65}{\sin 95}=4.55$ which is I.

$\frac{b}{\sin 20}=\frac{5}{\sin 95}$

$b=\frac{5\times \sin 20}{\sin 95}=1.72$ which is III.

Our answers are then I and III.

Which of the following are the missing sides of the triangle?

$\text{I. }4.55$

$\text{II. }5.49$

$\text{III. }1.72$

$\text{IV. }14.56$

$\text{I and III}$

$\text{II and IV}$

$\text{I and II}$

None of the above

Explanation

In order to solve this problem, we need to find

Since all the angles of a triangle add to $180^{\circ}$ , we can easily find it:

$180^{\circ}-65^{\circ}-20^{\circ}=95^{\circ}$

We can now use the Law of Sines to find the missing sides:

$\frac{a}{\sin 65}=\frac{5}{\sin 95}$

$a=\frac{5\times \sin 65}{\sin 95}=4.55$ which is I.

$\frac{b}{\sin 20}=\frac{5}{\sin 95}$

$b=\frac{5\times \sin 20}{\sin 95}=1.72$ which is III.

Our answers are then I and III.

Solve the triangle

Law_of_cosines__sss_

$A = 115^{\circ}, B = 22^{\circ}, C=43^{\circ}$

$A = 100^{\circ}, B = 31^{\circ}, C=49^{\circ}$

$A = 95^{\circ}, B = 23^{\circ}, C=62^{\circ}$

None of the other answers

$A = 99^{\circ}, B = 21^{\circ}, C=54^{\circ}$

Explanation

Since we are given all 3 sides, we can use the Law of Cosines in the angle form:

$cos(A) = \frac{-a^2+b^2+c^2}{2bc}$

$cos(B) = \frac{a^2-b^2+c^2}{2ac}$

$cos(C) = \frac{a^2+b^2-c^2}{2ab}$

Let's start by finding angle A:

$cos(A) = \frac{-(12)^2+5^2+9^2}{2\cdot5\cdot9} = \frac{-38}{90} = -0.4\overline{22}$

$A = cos^{-1}(-0.4\overline{22}) = 115^{\circ}$

Now let's solve for B:

$cos(B) = \frac{12^2-(5)^2+9^2}{2\cdot12\cdot9} = \frac{200}{216} = 0.\overline{925}$

$B = cos^{-1}(0.\overline{925}) = 22^{\circ}$

We can solve for C the same way, but since we now have A and B, we can use our knowledge that all interior angles of a triangle must add up to 180 to find C.

$C = 180^{\circ} - 115^{\circ} - 22^{\circ} = 43^{\circ}$

Which of the following are the missing sides of the triangle?

$\text{I. }4.55$

$\text{II. }5.49$

$\text{III. }1.72$

$\text{IV. }14.56$

$\text{I and III}$

$\text{II and IV}$

$\text{I and II}$

None of the above