find the area of a triangle when a=22 degree, b=105 degree, and b=14 30.4 units^2 32.1 units^2 31.2 units^2 34.8 units^2

C = 180 - (22 + 105) = 180 - 127 = 53 degrees
Area of triangle = 1/2 ab sin C
But a / sin A = b / sin B or a = b sin A / sin B = (14 x sin 22) / sin 105 = 5.429
Therefore, Area of triangle = 1/2 x 5.429 x 14 x sin 53 = 30.4 units^2

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eudora eudora · Answer 2 · 2019-05-16T06:59:35-04:00

Answer:

Option A. 30.4 unit²

Step-by-step explanation:

In the given triangle ABC, side AC = 14 units, ∠ BAC = 22° and ∠ ABC = 105°.

We have to find the area of ΔABC.

Since area of an obtuse triangle = \(\frac{1}{2}(Base)(Height)\)

Here Base = x

Height = h

To find the measure of Base "x".

We will apply the sine rule in ΔABC

\(\frac{sin22}{x}=\frac{sin105}{y14}\)

\(\frac{0.375}{x}=\frac{0.966}{14}\)

We cross multiply on both the sides

14×0.375 = 0.966×(x)

5.25 = 0.966x

x = \(\frac{5.25}{0.966}\)

x = 5.43 unit

For the measurement of Height "h"

\(cos(15+22)=\frac{Base}{Hypotenuse}\)

\(cos37=\frac{h}{14}\)

\(0.978=\frac{h}{14}\)

h = 0.978×14 = 11.18 unit

Now we put the values of base and height in the formula

Area of Δ ABC = \(\frac{1}{2}(5.43)(11.18)\)

= \(\frac{60.71}{2}\)

= 30.35 ≈ 30.4 unit²