Geometry: Solving Right Triangles
This is a free lesson from our course in Geometry
Right triangles are triangles in which one of the interior angles is 90. A 90 angle is called a right angle. Right triangles are also called right-angled triangles. In these the other two interior angles are complementary, i.e. their sum equals 90. Right triangles have special properties which help to conceptualize easily.
In a right triangle, the side opposite of the right angle is called the hypotenuse. The sides adjacent to the right angle are called the legs. When using the Pythagorean Theorem, the hypotenuse or its length is often labeled with a lower case c. The legs (or their lengths) generally are labeled as a, b. Either of the legs can be considered a base and the other leg would be height or altitude, as the right angle makes them perpendicular. When lengths of both the legs are known, then by setting one of these sides as the base (b) and the other as the height (h), the area of the right triangle can be calculated using general formula i.e. Area = (1/2) b* h or (h = a, in the  figure). Right Triangle
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<h2> Solving Right Triangles - Watch video</h2> <p> Pythagorean theorem, find, missing, formula, problems, sides, video, Pythagorean, solving right triangle, geometry, length, hypotenuse, Pythagorean theorem problems, example, solutions, right triangle, practice questions, quizzes</p> <p> If the lengths of the legs of a right triangle are a, b then the length of its hypotenuse c can be calculated by plugging in the given values of a, b in the formula</p>
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Pythagorean Theorem Converse
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(Continued from above) Important Properties 
Right triangles can be neither equilateral, obtuse nor acute triangles. 
Isosceles right triangles have two 45 angles as well as the 90 angle. 
All isosceles right triangles are similar, as corresponding angles in isosceles right triangles are equal.
While solving the word problems related to a right triangle (where length of two sides are known), remember the following: 
The diagonal across a rectangle creates two congruent right triangles. 
The angle between two walls, and the angle between the floor and a wall, is usually a right angle.
Sine, Cosine, and Tangent for Right Triangles
Sine, Cosine, and Tangent are the important and useful functions of an angle for right triangle calculations. For an angle , the sine function is (written as sin ), the cosine function (written as cos ), and the tangent function (written as tan ). For a known value of angle ; sin , cos , and tan can be found in the tables or calculator. For an angle between listed values, sine, cosine, or tangent of the angle can be worked from the values in the table or vice-versa. These three functions are related to right triangles in the following ways: 
the sine of a non-right angle equals the length of the leg opposite that angle divided by the length of the hypotenuse. 
the cosine of a non-right angle equals the length of the leg adjacent to it divided by the length of the hypotenuse. 
the tangent of a non-right angle equals the length of the leg opposite that angle divided by the length of the leg adjacent to it.
Note: for values of where cos 0 , tan = sin /cos  
In a right triangle with the two non-right angles 1and 2, and the side lengths a, b, c as shown in the figure below:
The functions of angle 1 are:

Similarly the functions of angle 2:
Right triangle
Special Right Triangles
A special right triangle is a right triangle whose sides are in a particular ratio. Pythagorean Theorem also can be used to solve these, but using these relationships you can save on some calculations.
As discussed earlier, there may be three important special triangles types: 3 - 4 - 5 triangles, 45- 45- 90 triangles and 30- 60 - 90 triangles.
A 3 - 4 - 5 triangle is right triangle whose lengths are in the ratio of 3 : 4 : 5. If two sides lengths of a right triangle you are given, you may check the ratio of lengths whether it confirms to 3 : 4 : 5 ratios.
A 45- 45- 90 triangle is a special right triangle whose angles are 45, 45 and 90. The lengths of the sides in this triangle are in the ratio of 1 : 1 : 2.
A 30- 60- 90 triangle is a special triangle whose angles are 30, 60and 90. The lengths of the sides in this triangle are in the ratio of 1 : 3 : 2.
Look at some of the examples to illustrate and develop skills to help doing Geometry home work and sharpening the skills:
Example 1: Given- in a right triangle length of a leg is 8 cm and a hypotenuse length is 17 cm. Find out the length of the other leg?
Step 1: Say leg a = 8 cm, and hypotenuse c = 17 cm.
Step 2: Use the formula c2 = a2 + b2, and plug in values of c and a i.e. 172 = 82 + b2.
Step 3: On simplification it gives, b2 = 172 - 82 = 225.
Therefore, b =15
The length of the other leg is 15 cm, as the final answer.
Example 2: Given- find the length of two sides, if sum of the sides of a right triangle is 23 in and the hypotenuse is 17 in.
Step 1: Say a and b are the lengths of two sides. Thus a + b = 23
Step 2: Solve it for b, you get b = 23 a. With this you know all the sides i.e. a, (23 - a) and c, which equals to 17.
Step 3: Use the formula c2 = a2 + b2, and plug in above known values of sides.
          172 = a2 + (23 a)2 (by substitution)
          289 = a2 + (529 - 46a + a2). Simplify it, 
         2a2 - 46a + 240 = 0
          a2 - 23a + 120 = 0 = a2 - 8a -15a + 120
          Simplifying it gives, a = 8 in or a = 15 in
Any of the above solutions can be said to be correct, since a and b values were not specified. Making a choice of a = 8, the other side will be 15.
The length of one side will be 15 in and other 8 in, as the final answer.
Example 3: Given- find sin() and cos() in the right triangle in the figure below:
Step 1: Find the hypotenuse c of the right triangle using Pythagorean Theorem.
            C = (122 + 162) = 400 = 20
Step 2: sin () in a right triangle is given by: sin () =16 / 20 = 0.8
Step 3: cos() in a right triangle is given by: cos() = 12 / 20 = 0.6
            sin () = 0.8 and cos() = 0.6. as the final answer.
Right Triangle-Pythagorean Theorem
Example 4: Given- solve the right triangle ABC given that c = 10 cm and b = 8 cm.
Step 1: Find out the third side a, using Pythagorean Theorem and formula
        c2 = a2 + b2
      c2 = a2 + b2 i.e. 152 = a2 + 122
        a2 = 152 - 122, on simplification it gives a = 9 cm
Step 2: Find out the value of angle A. cos A = 12/15 = 0.8
Step 3: Check up in Trigonometric Table, the value of angle for which cos is nearest to 0.8.     For 37, it is 0.7986 (it is very close to 0.8). Therefore, angle A is 37.
Step 4: Angles A, B are 37 and 90 respectively. You know sum of three angles of a triangle equals 180.
Therefore, angle C = (180 - 90 -37) = 53
Side a = 9 cm, and Angles A, B, and C are respectively 37, 90 and 53, as the final answer.
Example 5: Given- in the figure below, two triangles have common side 'a', find tan .
Step 1: Find a in the right triangle using right triangle i.e. tan (44) = a / 12
            a = 12 * tan (44)
Step 2: Look at the left triangle now: tan () = a / 8
Step 3: Substitute value of a from step 1 to
Step 2: tan () = 12 * tan (44) / 8.
            Simplification gives,
Step 4: tan () = 1.5 * tan (44) i.e. 1.5 * (0.9657)
            [take this value from Trigonometric Table]
            = 1.45
tan () equals to 1.45, as the final answer.
Example 6: Given- Two lines tangent to a circle at points P and Q have a point of intersection A. The size of angle PAQ is equal to degrees and the length of the radius of the circle is equal to r. Find the distance from point A to the center of the circle O in terms of and r.
Step 1: A line through the center o of the circle meeting at point on tangent to the circle is perpendicular to the tangent line; hence make the right angle at P and Q (Figure given).
Step 2: Triangles PAO and QAO are congruent, tan(/2) is given by tan (/2) = r /q .
It may be re-written as q = r / tan (/2)
Step 3: Since QAO is right triangle, c = (r2 + q2)
Step 4: Plug in q from Step 2, it gives: c = [r2 + (r/ tan (/2)2]
Step 5: Applying factoring r2 gives, c = [r2 (1+ 1/ tan (/2)2]
Step 6: The simplification gives; c = r [1+ 1/ tan (/2)2]
The distance from A to center O is r sqrt [1+ 1/ tan (/2)2], as the final answer.
Example 7: A triangle PQR shaped object (shown below), has an area of 24.97 in2. Side PQ has a length of 10 in and side PR has a length of 8 in and angle QPR is obtuse. Find angle QPR, and length of side QR to the nearest whole number.
Step 1: Say the size of angle QPR = .
Step 2: Use the formulas for the area triangle i.e.
           Area = 24.97 = (1/2) (PR)(PQ) sin() and solve for sin()
           sin() = 49.94 / (8*10) = 0.624.
Step 3: Pick up the solution that gives obtuse i.e.
           = - arcsin(0.624)
Step 4: Convert to degrees to obtain, which gives (approximately) = 141
Step 5: Use the cosine rule to calculate the length of side QR
            QR2 = QP2 + PR2 - 2(QP)(PR)cos()
            = 100 + 64 - 2(10)(8)cos(141), which gives QR = 17 in.
The length of the side QR is 7 in, the final answer.
Example 8: Jamie decided to purchase a 40 inches TV, for which height of the screen is 24 inches. Find out how much horizontal space length will it require to fix this rectangular TV on the wall?
You know that TV and monitor screens are measured across the diagonal,
Step 1: The diagonal of rectangular TV is 40 in (given)
Step 2: The height and width are like two legs of a right triangle, and diagonal as the hypotenuse. In this, hypotenuse and one leg is given and the other leg measure will be the horizontal space length required on the wall.
Step 3: Use the formula: c2 = a2 + b2. Given, a = 24, b =? , c = 40. Plug in the known values gives,
Step 4: 402 = 242 + b2. Therefore, b2 = 402 - 242 = 1600 - 576 = 1024
Step 5: Taking square root of both sides gives, b = 32 in.
The screen is 32 in wide and will occupy this length on the wall, as the final answer.
The video above will explain more in detail about Solving Right Triangles, and how to apply the concepts in solving real-world problems. This is explained with the help of several examples and done watching video. This helps you to deal with solving problems and help doing the Geometry home work.
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