Find the best tutors and institutes for Class 10 Tuition

Find Best Class 10 Tuition

Please select a Category.

Please select a Locality.

No matching category found.

No matching Locality found.

Outside India?

Application Of Trigonometry

Shuvham Singhal
29/07/2017 0 0

Solve the triangle shown below:

Applications of Right Triangle Trigonometry Example 1.svg


We need to find the lengths of all sides and the measures of all angles. In this triangle, two of the three sides are given. We can find the length of the third side using the Pythagorean Theorem:

8 2 + b 2 = {\displaystyle 8^{2}+b^{2}=\,\!} 8^{2}+b^{2}=\,\! 10 2 {\displaystyle 10^{2}\,\!} 10^{2}\,\!
64 + b 2 = {\displaystyle 64+b^{2}=\,\!} 64+b^{2}=\,\! 100 {\displaystyle 100\,\!} 100\,\!
b 2 = {\displaystyle b^{2}=\,\!} b^{2}=\,\! 36 {\displaystyle 36\,\!} 36\,\!
b = {\displaystyle b=\,\!} b=\,\! ± 6 ⇒ b = 6 {\displaystyle \pm 6\Rightarrow b=6\,\!} \pm 6\Rightarrow b=6\,\!

(You may have also recognized the "Pythagorean Triple", 6, 8, 10, instead of carrying out the Pythagorean Theorem.)

You can also find the third side using a trigonometric ratio. Notice that the missing side, b, is adjacent to angle A, and the hypotenuse is given. Therefore we can use the cosine function to find the length of b:

cos ? ( 53.13 ? ) = {\displaystyle \cos(53.13^{\circ })=\,\!} \cos(53.13^{\circ })=\,\! adjacent side hypotenuse = b 10 {\displaystyle {\frac {\text{adjacent side}}{\text{hypotenuse}}}={\frac {b}{10}}} {\frac {{\text{adjacent side}}}{{\text{hypotenuse}}}}={\frac {b}{10}}
0.6 = {\displaystyle 0.6=\,\!} 0.6=\,\! b 10 {\displaystyle {\frac {b}{10}}} {\frac {b}{10}}
b = {\displaystyle b=\,\!} b=\,\! 0.6 ( 10 ) = 6 {\displaystyle 0.6(10)=6\,\!} 0.6(10)=6\,\!

We could also use the tangent function, as the opposite side was given. It may seem confusing that you can find the missing side in more than one way. The point is, however, not to create confusion, but to show that you must look at what information is missing, and choose a strategy. Overall, when you need to identify one side of the triangle, you can either use the Pythagorean Theorem, or you can use a trig ratio.

To solve the above triangle, we also have to identify the measures of all three angles. Two angles are given: 90 degrees and 53.13 degrees. We can find the third angle using the triangle angle sum:

180 − 90 − 53.13 = 36.87°

Solve the triangle shown below.

Applications of Right Triangle Trigonometry Example 2.svg

Solution: In this triangle, we need to find the lengths of two sides. We can find the length of one side using a trig ratio. Then we can find the length of the third side either using a trig ratio, or the Pythagorean Theorem.

We are given the measure of angle A, and the length of the side adjacent to angle A. If we want to find the length of the hypotenuse, c, we can use the cosine ratio:

cos ? ( 40 ? ) = {\displaystyle \cos(40^{\circ })=\,\!} \cos(40^{\circ })=\,\! adjacent hypotenuse = 6 c {\displaystyle {\frac {\text{adjacent}}{\text{hypotenuse}}}={\frac {6}{c}}} {\frac {{\text{adjacent}}}{{\text{hypotenuse}}}}={\frac {6}{c}}
cos ? ( 40 ? ) = {\displaystyle \cos(40^{\circ })=\,\!} \cos(40^{\circ })=\,\! 6 c {\displaystyle {\frac {6}{c}}} {\frac {6}{c}}
c cos ? ( 40 ? ) = {\displaystyle c\cos(40^{\circ })=\,\!} c\cos(40^{\circ })=\,\! 6 {\displaystyle 6\,\!} 6\,\!
c = {\displaystyle c=\,\!} c=\,\! 6 cos ? ( 40 ? ) ≈ 7.83 {\displaystyle {\frac {6}{\cos(40^{\circ })}}\approx 7.83} {\frac {6}{\cos(40^{\circ })}}\approx 7.83

If we want to find the length of the other leg of the triangle, we can use the tangent ratio. (Why is this a better idea than to use the sine?)

tan ? ( 40 ? ) = {\displaystyle \tan(40^{\circ })=\,\!} \tan(40^{\circ })=\,\! opposite adjacent = a 6 {\displaystyle {\frac {\text{opposite}}{\text{adjacent}}}={\frac {a}{6}}} {\frac {{\text{opposite}}}{{\text{adjacent}}}}={\frac {a}{6}}
tan ? ( 40 ? ) = {\displaystyle \tan(40^{\circ })=\,\!} \tan(40^{\circ })=\,\! a c {\displaystyle {\frac {a}{c}}} {\frac {a}{c}}
a = {\displaystyle a=\,\!} a=\,\! 6 tan ? ( 40 ? ) ≈ 5.03 {\displaystyle 6\tan(40^{\circ })\approx 5.03\,\!} 6\tan(40^{\circ })\approx 5.03\,\!

Now we know the lengths of all three sides of this triangle. In the review questions, you will verify the values of c and a using the Pythagorean Theorem. Here, to finish solving the triangle, we only need to find the measure of angle B:

180 − 90 − 40 = 50°
0 Dislike
Follow 0

Please Enter a comment


Other Lessons for You

Private, Public and Joint Sector Enterprises-Class 11 NCERT, Class 9 ICSE,Commercial Studies
CBSE Std XIth Business Studies Classification of Commercial Organisations based on the ownership-

Rajiv Vadera | 01 Jul

0 0

Conservation of energy for a falling body
Law of conservation of energy - Energy can neither be created nor be destroyed; it can only be transformed from one form to another. Consider a body of mass m placed at A. h = AB is the height of...

Soumi Roy | 02/12/2018

5 0

Financial Accounting
Accounting is the art of recording, classifying, summarising in a significant manner regarding money, transaction & events which are at least in part financial interpreting the result thereof. (AICPA...

Rajeshwari M. | 23/11/2018

2 0

Area Formulas
Areas square = a2 rectangle = ab parallelogram = bh trapezoid = h/2 (b1 + b2) circle = pi r 2 ellipse = pi r1 r2 triangle = (1/2) b h equilateral triangle = (1/4)(3) a2 triangle...

Rajiv Singh | 11/09/2018

5 1

Precautions to Protect Magnets from Losing Their Magnetic Properties
Precautions to Protect Magnets from Losing Their Magnetic Properties Never drop magnets from heights. Never heat a magnet. Do not hammer a magnet. Certain items such as CD's, DVD's, debit cards,...

Rajiv Singh | 11/09/2018

1 1

Find Best Class 10 Tuition ?

Find Now » is India's largest network of most trusted tutors and institutes. Over 25 lakh students rely on, to fulfill their learning requirements across 1,000+ categories. Using, parents, and students can compare multiple Tutors and Institutes and choose the one that best suits their requirements. More than 6.5 lakh verified Tutors and Institutes are helping millions of students every day and growing their tutoring business on Whether you are looking for a tutor to learn mathematics, a German language trainer to brush up your German language skills or an institute to upgrade your IT skills, we have got the best selection of Tutors and Training Institutes for you. Read more