Day 93 – Application of trigonometric ratios

Introduction If a person wants to climb to a point higher, he has to use using a ladder. Hour interest is to find the angle at which the ladder should be inclined so that a person can climb without the risk of falling back. All these are possible due to the concept trigonometric ratios. In this lesson, we are going use trigonometric ratios to solve a right when we know the length of one side and the size of one angle.

Vocabulary Cosine This is the ratio of the adjacent side to the hypotenuse in a right triangle with reference to a specific angle Sine This is the ratio of the opposite side to the hypotenuse in a right triangle with reference to a specific angle Tangent This is the ratio of the opposite side to the adjacent side in a right triangle with reference to a specific angle

Consider ∆𝐴𝐵𝐶 below sin 𝜃= 𝑦 𝑧 ⟹𝑦=𝑧𝑠𝑖𝑛 𝜃 ⇒𝑧= 𝑦 sin 𝜃 tan 𝜃= 𝑦 𝑥 ⟹𝑦=𝑥𝑡𝑎𝑛 𝜃 ⇒𝑥= 𝑦 tan 𝜃 cos 𝜃= 𝑥 𝑧 ⟹𝑥=𝑧 cos 𝜃 ⇒𝑧= 𝑥 cos 𝜃 𝐴 𝐵 𝐶 𝜃 𝑧 𝑦 𝑥

Example 1 Find the length of side AB Solution Let the length of AB be 𝑥, cos 60 ° = 𝑥 24 ⟹𝑥=24 cos 60 𝑥=24× 1 2 𝑥=12 𝑖𝑛 𝐴 𝐵 𝐶 60 ° 24 𝑖𝑛

Example 2 Find the length of side XY 4 𝑖𝑛 𝑄 X Y Z 45 °

Solution tan 45 ° = 4 𝑄𝑌 𝑄𝑌=4× tan 45 ° 𝑄𝑌=4×1 =4 𝑖𝑛 Since XZ=ZY and ZQ is shared by two triangles, then ∆𝑄𝑋𝑍≅∆𝑄𝑌𝑍 and therefore XQ=QY= 4 𝑖𝑛 XY = XQ+QY = 4+4 = 8 𝑖𝑛

DETERMINING TRIGONOMETRIC INVERSES Sometimes we might be required to find the acute angles in a right triangle. This will always be possible when we know the length of any two of the sides. Consider the following illustration. sin 𝜃 = 12 13 =0.9231 𝜃 12 𝑖𝑛 5 𝑖𝑛 13 𝑖𝑛

Using a calculator follow the following steps Key in 0.9231 Key in the shift function. Key in sin Screen displays 67.38357305 Thus, 𝜃=𝑠𝑖𝑛 −1 0.9231= 67.38 °

homework 1. Find the length of JL J K L M 3 𝑖𝑛 70 °

Answers to homework 1. 8.77 𝑖𝑛

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