OK, let's try with no figure. We have an isosceles triangle sides s,s, and b.
Opposite b is angle t.
Draw the altitude h to bisect t. We have two right triangles, legs b/2 and h, hypotenuse s. The angle opposite b/2 is t/2 so
sin(t/2) = (b/2)/s = b/2s
So we arrived at the first part,
b = 2s sin(t/2)
The area of a triangle with sides s,s and included angle t is
A = (1/2) s² sin t
13/8 or 1 5/8
3/4 becomes 7/8 when add it is 13/8
Answer is C ig I’m sowwy if I’m wrong
Answer:
<h2>
AC = 36.01</h2>
Step-by-step explanation:
Given ΔABC and ΔADB, since both triangles are right angled triangles then the following are true.
From ΔADB, AB² = AD²+BD²
Given AB = 24 and AD = 16
BD² = AB² - AD²
BD² = 24²-16²
BD² = 576-256
BD² = 320
BD = 
BD = 17.9
from ΔABC, AC² = AB²+BC²
SInce AC = AD+DC and BC² = BD² + DC² (from ΔBDC )we will have;
(AD+DC)² = AB²+ (BD² + DC²)
Given AD = 16, AB = 24 and BD = 17.9, on substituting
(16+DC)² = 24²+17.9²+ DC²
256+32DC+DC² = 24²+17.9²+ DC²
256+32DC = 24²+17.9²
32DC = 24²+17.9² - 256
32DC = 640.41
DC = 
DC = 20.01
Remember that AC = AD+DC
AC = 16+20.01
AC = 36.01
Two is equivalent to twenty and 84