How to find an Angle in a right-angled Triangle ? Triangles are trilateral close polygons formed by the overlap of three lines. It is encountered a lot in casual life. It is one of the basic shapes of geometry. It has three sides, three angles, and three vertices. A Right Angled Triangle is one where one of the angles is always equal to 90°. Pythagoras Theorem is derived for right-angled triangles, Which states that the square of the hypotenuse ( the longest side ) is equal to the sum of the squares of base and plumb line. Given the length of at least two sides of a right-angled triangulum, we can find the measure of any slant of the right-angled triangle. For this, we use diverse trigonometric functions like sine, cosine, tangent, cotangent, secant, and cosecant. These help us to relate the angles of a right-angled triangulum with its sides.

Properties

• There is a right-angle vertex among the three vertices
• The side opposite to the right-angled vertex is called the hypotenuse.
• The length of the sides follows the Pythagoras theorem, which states

hypotenuse2 = base2 + altitude2

• The hypotenuse is the longest side of a right-angled triangle.
• The angles other than the right angle are acute angles since the value is less than 90o

Trigonometric functions

• cosθ: This gives the ratio of the base by the hypotenuse of a right-angled triangle.

cosθ = base / hypotenuse

• sinθ: This gives the ratio of altitude by the hypotenuse of a right-angled triangle.

sinθ = altitude / hypotenuse

• tanθ: It is the ratio of altitude by the base of a right-angled triangle.

tanθ = altitude / base

• cotθ: It is the inverse of tanθ
• secθ: It is the inverse of cosθ
• cosecθ: It is the inverse of sinθ

To find the angles of a right-angled triangle, we can take the trigonometric inverse of the ratio of given sides of the triangle. Example:

If sinθ = x, then we can write θ = sin-1x. This returns the angle for which the sine respect of the lean is adam. similarly, there exists cos-1θ, tan-1θ, cot-1θ, sec-1θ, and cosec-1θ

Sample Problems

Question 1. Given a right-angled triangle, with base equals 10cm and hypotenuse equals 20cm. Find the value of the base angle. Solution:

Given, Base = 10cm Hypotenuse = 20cm Let, the respect of the base slant be θ. We can write cosθ = base / hypotenuse = 10/20 = 1/2 θ = cos-1 ( 1/2 ) = 60o Thus, the value of base angle is 60o.

Question 2. Find the value of angles of a right angles triangle, given that one of the acute angles is twice the other. Solution:

Since we know the total of all the three angles in a triangle is 180o. Since one of the angles is 90o and one of the acute angles is doubly the early, we can consider them as θ and 2θ. so, we can write

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90o + θ + 2θ = 180o 3θ = 180o – 90o 3θ = 90o θ = 90o/3 = 30o 2θ = 2 × 30o = 60o So, the angles are 30o, 60o, and 90o.

Question 3. Find the value of the angle of elevation of a ladder of length 5m, given that base of the ladder is at a distance of 3m from the wall. Solution:

Since the ladder acts as a hypotenuse of a right angles triangle and basis distance equals 3m, we can write Hypotenuse = 5m Base = 3m Let the lean of elevation be θ. so, we can write cosθ = Base / Hypotenuse = 3/5 θ = cos-1 ( 3/5 ) θ = 53o thus, the rate of the angle of natural elevation is 53o .

Question 4. Find the value of hypotenuse, given the length of the altitude is 8m and the base angle equals 30o. Solution:

Given, the base angle is adequate to 30o and altitude equals 8m, we can apply the sine routine to find the duration of the hypotenuse. sin30o = altitude / hypotenuse hypotenuse = altitude / sin30o Since the measure of sin30o equals 1/2, we can write hypotenuse = altitude / ( 1/2 ) = 2 × altitude thus, hypotenuse = 2 × 8 = 16m Thus, the length of the hypotenuse is equal to 16m.

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