In a ΔABC right angled at B, AB = 8 unit and AC = 10 unit. What is the value of \(\sin^2\theta-\cos^2\theta\) where \(\theta\) is ∠ACB?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{7}{25}\)
In a right triangle, the hypotenuse is the side opposite the right angle, here AC. The third side comes from Pythagoras, and then the sine and cosine of θ = ∠ACB are read off: sine = opposite / hypotenuse, cosine = adjacent / hypotenuse.
\(BC=\sqrt{AC^2-AB^2}=\sqrt{100-64}=6\)
Side opposite θ is AB, so \(\displaystyle\sin\theta=\frac{8}{10}\)
Side next to θ is BC, so \(\displaystyle\cos\theta=\frac{6}{10}\)
\(\displaystyle\sin^2\theta-\cos^2\theta=\frac{64}{100}-\frac{36}{100}=\frac{28}{100}=\frac{7}{25}\)