Right Triangle Calculations Finding Sides And Trigonometric Ratios

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In a right triangle ABC with angle A = 90°, AD is perpendicular to BC. Given AD = 12 cm and BD = 9 cm, calculate: a) the lengths of the sides of the triangle; b) sin B, cos B, tan C, and cot C.

This article delves into the solution of a geometry problem involving a right triangle. We will explore how to calculate the lengths of the sides and various trigonometric ratios given specific information about the triangle. Let's consider the right triangle ABC, where angle A is 90 degrees, and AD is perpendicular to BC, with D lying on BC. We are given that AD = 12 cm and BD = 9 cm. Our goal is to calculate the lengths of the sides of the triangle and the values of sin B, cos B, tan C, and cot C.

Problem Statement

In right triangle ABC, with ∠A=90∘{ \angle A = 90^\circ }, AD is perpendicular to BC (AD⊥BC{ AD \perp BC }), and D lies on BC. Given AD = 12 cm and BD = 9 cm, calculate:

a) The lengths of the sides of the triangle. b) sin B, cos B, tan C, and cot C.

Solution

a) Calculating the lengths of the sides of the triangle

To begin, we focus on finding the lengths of the sides of the triangle ABC. This involves using the given information and applying geometric theorems, particularly those related to right triangles and similar triangles. The key here is to strategically use the provided lengths (AD and BD) and the right angle at A to deduce the remaining side lengths.

Firstly, consider the right triangle ADB. We are given AD = 12 cm and BD = 9 cm. To find the length of side AB, we can apply the Pythagorean theorem. The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. In triangle ADB, AB is the hypotenuse, so we have:

AB2=AD2+BD2{ AB^2 = AD^2 + BD^2 }

Substituting the given values:

AB2=122+92=144+81=225{ AB^2 = 12^2 + 9^2 = 144 + 81 = 225 }

Taking the square root of both sides:

AB=225=15 cm{ AB = \sqrt{225} = 15 \text{ cm} }

So, the length of side AB is 15 cm. This is a crucial first step, as it provides us with one side of the main triangle ABC and helps in further calculations.

Next, we need to find the length of DC. To do this, we can use the similarity of triangles. In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the triangle into two smaller triangles that are similar to the original triangle and to each other. Therefore, triangle ADB is similar to triangle CDA. This similarity implies that the ratios of corresponding sides are equal. Specifically, we can write:

BDAD=ADDC{ \frac{BD}{AD} = \frac{AD}{DC} }

Plugging in the given values:

912=12DC{ \frac{9}{12} = \frac{12}{DC} }

To solve for DC, we cross-multiply:

9⋅DC=12⋅12{ 9 \cdot DC = 12 \cdot 12 }

9DC=144{ 9DC = 144 }

Dividing both sides by 9:

DC=1449=16 cm{ DC = \frac{144}{9} = 16 \text{ cm} }

Thus, the length of DC is 16 cm. Now that we have BD and DC, we can find the length of BC by adding them:

BC=BD+DC=9+16=25 cm{ BC = BD + DC = 9 + 16 = 25 \text{ cm} }

Now we know the length of BC, which is the hypotenuse of triangle ABC. To find the length of AC, we can again use the Pythagorean theorem in triangle ABC:

AC2+AB2=BC2{ AC^2 + AB^2 = BC^2 }

We know AB = 15 cm and BC = 25 cm, so:

AC2+152=252{ AC^2 + 15^2 = 25^2 }

AC2+225=625{ AC^2 + 225 = 625 }

Subtracting 225 from both sides:

AC2=625−225=400{ AC^2 = 625 - 225 = 400 }

Taking the square root of both sides:

AC=400=20 cm{ AC = \sqrt{400} = 20 \text{ cm} }

Therefore, the length of side AC is 20 cm. We have now found all the sides of triangle ABC: AB = 15 cm, AC = 20 cm, and BC = 25 cm.

b) Calculating sin B, cos B, tan C, and cot C

Having determined the lengths of all sides of the triangle, we can now calculate the trigonometric ratios sin B, cos B, tan C, and cot C. These ratios relate the angles of a right triangle to the lengths of its sides. Understanding these relationships is crucial in trigonometry and has broad applications in various fields, including physics, engineering, and navigation.

Firstly, let's define the trigonometric ratios in the context of a right triangle. For an acute angle in a right triangle:

  • Sine (sin) is the ratio of the length of the opposite side to the length of the hypotenuse.
  • Cosine (cos) is the ratio of the length of the adjacent side to the length of the hypotenuse.
  • Tangent (tan) is the ratio of the length of the opposite side to the length of the adjacent side.
  • Cotangent (cot) is the ratio of the length of the adjacent side to the length of the opposite side. It is also the reciprocal of the tangent.

Now, let's calculate sin B. In triangle ABC, angle B is one of the acute angles. The side opposite to angle B is AC, and the hypotenuse is BC. Therefore:

sin⁡B=oppositehypotenuse=ACBC=2025=45{ \sin B = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AC}{BC} = \frac{20}{25} = \frac{4}{5} }

Thus, sin B = 4/5.

Next, let's calculate cos B. The side adjacent to angle B is AB, and the hypotenuse is BC. Therefore:

cos⁡B=adjacenthypotenuse=ABBC=1525=35{ \cos B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{BC} = \frac{15}{25} = \frac{3}{5} }

So, cos B = 3/5.

Now, let's move on to angle C. We need to find tan C and cot C. First, let's calculate tan C. In triangle ABC, the side opposite to angle C is AB, and the side adjacent to angle C is AC. Therefore:

tan⁡C=oppositeadjacent=ABAC=1520=34{ \tan C = \frac{\text{opposite}}{\text{adjacent}} = \frac{AB}{AC} = \frac{15}{20} = \frac{3}{4} }

Hence, tan C = 3/4.

Finally, let's calculate cot C. The cotangent is the reciprocal of the tangent, so:

cot⁡C=1tan⁡C=134=43{ \cot C = \frac{1}{\tan C} = \frac{1}{\frac{3}{4}} = \frac{4}{3} }

Alternatively, we can calculate cot C directly using the ratio of the adjacent side to the opposite side:

cot⁡C=adjacentopposite=ACAB=2015=43{ \cot C = \frac{\text{adjacent}}{\text{opposite}} = \frac{AC}{AB} = \frac{20}{15} = \frac{4}{3} }

Thus, cot C = 4/3.

Final Answers

a) The lengths of the sides of the triangle are: AB = 15 cm, AC = 20 cm, and BC = 25 cm. b) The trigonometric ratios are: sin B = 4/5, cos B = 3/5, tan C = 3/4, and cot C = 4/3.

This problem showcases the application of the Pythagorean theorem and the concept of similar triangles to find the sides of a right triangle. Additionally, it demonstrates how to calculate trigonometric ratios once the side lengths are known. These concepts are fundamental in geometry and trigonometry and are essential tools for solving a wide range of problems in mathematics and related fields. Understanding these principles allows for a deeper comprehension of geometric relationships and enhances problem-solving skills in various contexts.