Question

In the figure one side of the triangle is 10 centimetres and angles on that side are 60° and 45°. Calculate the area of the triangle. 

 

Hint:

For an ASA (Angle-Side-Angle) triangle, a standard method to find the area is to drop a perpendicular to the known side. This divides the problem into two solvable right-angled triangles, allowing you to find the height and then the area.

Answer 1

We are given a triangle with one side (the base) and the two adjacent angles (ASA case). We need to calculate the triangle's area.

The area of a triangle is (1)/(2) × base × height. We will take the given side of 10 cm as the base. We need to find the corresponding height by dropping a perpendicular from the opposite vertex. This will create two right-angled triangles.

Let the base be b = 10 cm. Let the height (altitude) be h. This altitude divides the base into two segments, x₁ and x₂, such that x₁ + x₂ = 10.
The altitude creates two right-angled triangles with angles 60° and 45°.
In the first right-angled triangle (with the 60° angle):
(60^) = (h)/(x₁) √(3) = (h)/(x₁) x₁ = h√(3) In the second right-angled triangle (with the 45° angle):
(45^) = (h)/(x₂) 1 = (h)/(x₂) x₂ = h We know that x₁ + x₂ = 10. Substitute the expressions for x₁ and x₂:
h√(3) + h = 10 Factor out h:
h ( 1√(3) + 1 ) = 10 h ( 1 + √(3)√(3) ) = 10 Solve for h:
h = 10√(3)1 + √(3) To rationalize the denominator, multiply by the conjugate (√(3) - 1):
h = 10√(3)(√(3) - 1)( √(3) + 1)(√(3) - 1) = 10(3 - √(3))3 - 1 = 10(3 - √(3))2 = 5(3 - √(3)) Now, calculate the area:
Area = (1)/(2) × base × height = (1)/(2) × 10 × h = 5h Area = 5 × 5(3 - √(3)) = 25(3 - √(3)) The area of the triangle is 25(3 - √(3)) cm².