Find the area of the triangle with the given measurements. round the solution to the nearest hundredth if necessary. a = 46°, b = 27 ft, c = 14 ft

We will use the law of cosines
side a² = b² + c² -2bc • cos(A)
side a² = 729 + 196 -2*27*14 * cos (46)
side a² = 925 -(756 * 0.69466)
side a² = 399.83704
side a = 19.995925585
We could round that to 20
a = 20 b = 27 c =14

We can calculate a triangle's area when we know all 3 sides by using Heron's Formula
area = square root (s • (s - a) • (s - b) • (s - c))

where s is the semi-perimeter
semi-perimeter = (side a + side b + side c) ÷ 2
s = (20 + 27 + 14) / 2
s = 30.5
Now we use Heron's Formula
area = square root (s • (s - a) • (s - b) • (s - c))
area = square root (30.5 • (30.5 - 20) • (30.5 - 27) • (30.5 - 14))
area = square root (30.5 • (10.5) • (3.5) • (16.5))
area = square root (18494.4375)
area = 135.9942553934  which rounds to
136 square feet
Source:
http://www.1728.org/triang.htm

 

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12