Suppose triangle ABC is isosceles, with the two equal sides being 10 cm in length and the equal angles being 40 degrees. What is the height and what is the area of the triangle?

Question

Aaliyah Anderson · Accepted Answer

#h = 10 sin 40, A = 50 sin 80# #B = C = 40º# #M = (B + C)/2# #Delta AMC# is a right triangle #(x, h, 10)# = (basis, height, hypotenuse) #sin 40º = h/10# #cos 40º = x/10# Area#(Delta ABC) = 1/2 cdot (2x) cdot h = 100 sin 40 cos 40# We use: #sin 2x = 2 sin x cos x#

James Atkins · Answer

undefined To find the height and the area of the isosceles triangle ABC with two equal sides of length 10 cm and equal angles of 40 degrees:

1. First, divide the triangle into two congruent right triangles by drawing a perpendicular line from the vertex opposite the base to the midpoint of the base.

2. Since the triangle is isosceles, the perpendicular bisects the base, so each half of the base is 5 cm.

3. In each right triangle, the angle opposite the base (which is half of the 40-degree angle) is 20 degrees.

4. Using trigonometric ratios, we can find the height of each right triangle. Let's denote the height as $ h $.

$$ \tan(20^\circ) = \frac{h}{5} $$

$$ h = 5 \tan(20^\circ) $$

5. Calculate the value of $ h $:

$$ h \approx 5 \times 0.36397 $$

$$ h \approx 1.81985 $$

$$ h \approx 1.82 \, \text{cm} $$

6. Since the height of the entire triangle is the sum of the heights of the two right triangles, the height of the entire triangle is $ 2 \times 1.82 = 3.64 $ cm.

7. To find the area of the triangle, use the formula:

$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$

$$ \text{Area} = \frac{1}{2} \times 10 \times 3.64 $$

$$ \text{Area} = 18.2 \, \text{cm}^2 $$

Therefore, the height of the triangle is approximately 3.64 cm and the area of the triangle is approximately 18.2 square centimeters.