Question

★ Trigonometry ★

Consider ∆ ACB , right-angled at C , in which AB = 29 units , BC = 21 units and $\sf{ \angle ABC \:=\: \theta }$ (see attachment). Determine the values of :

• $\sf{ cos² \: \theta \:+\: sin² \: \theta }$
• $\sf{ cos² \: \theta \:-\: sin² \: \theta }$
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Answer 1

Answer:

• 1

• 41/841 or 0.0487

Step-by-step explanation:

In ∆ABC,

By Pythagoras Theorem,

AB² = AC² + BC²

AC² = AB² - BC²

AC² = 29² - 21²

AC² = 841 - 441

AC² = 400

AC = √400 = 20

cos∅ = Base/hypotenuse = BC/AB = 21/29

sin∅ = perpendicular/hypotenuse = AC/AB = 20/29

cos²∅ = (21/29)² = 441/841

sin²∅ = (20/29)² = 400/841

cos²∅+sin²∅ = 441/841 + 400/841 = 841/841 = 1

cos²∅-sin²∅ = 441/841 - 400/841 = 41/841 = 0.0487