Question

In a right triangle ABC, Right angled at C in which AB=13cm,BC=5cm, determine the value of cos²B+sin²A​

Answer 1

{Some correction in question}

Question : In a right ∆ ACB, Right angled at C in which AB = 13 cm, BC = 5 cm, determine the value of cos²B + sin²A (Letter which represents the right angle we write that letter in center i.e. ∆ ACB)

Given data : In a right ∆ ACB, Right angled at C in which AB = 13 cm BC = 5 cm.

To find : Determine the value of cos²B + sin²A

Solution :

Here, according to given,

• Hypotenuse is AB

Now, to find Hypotenuse,

⟹ (AB)² = (AC)² + (BC)²

⟹ (AB)² = (13)² + (5)²

⟹ (AB)² = 169 + 15

⟹ (AB)² = 194

⟹ AB = √194 cm

Now,

• adjecent side of angle B = side BC

Now, by formula

⟹ cos θ = adjecent side/hypotenuse

⟹ cos B = adjecent side/hypotenuse

⟹ cos B = 5/√194 ____( 1 )

similarly,

• opposite side of angle A = side BC

⟹ sin θ = opposite side/hypotenuse

⟹ sin B = opposite side/hypotenuse

⟹ sin B = 5/√194 ____( 2 )

Now,

⟹ cos²B + sin²A = (5/√194)² + (5/√194)²

⟹ cos²B + sin²A = 25/194 + 25/194

⟹ cos²B + sin²A = 50/194

⟹ cos²B + sin²A = 25/97

⟹ cos²B + sin²A = 25/97 or 0.2577 (approx)