Question

Area of triangle- please help :)

Answer 1

I will help you as soon as possible

Answer 2

Answer:

$\approx 45.1\:\mathrm{w/\: some \: MOE}$

Step-by-step explanation:

The area of $ABCD$ is comprised of two triangles, so we can find the area of both triangles and add them to get the total area of

Since the sum of the interior angles of a triangle add up to $180^{\circ}$, we have:

$\angle C=180-44-38=\fbox{ 98^{\circ} }$.

The Law of Sines is given as:

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ for any triangle.

Therefore, we have the proportion:

$\frac{\sin 98^{\circ}}{10}=\frac{\sin 44^{\circ}}{BC},\\BC=\frac{10\cdot \sin 44^{\circ}}{\sin 98^{\circ}},\\BC\approx 7.01$.

Now that we have two sides of a triangle and the angle between them, we can use $[\triangle]=\frac{1}{2}ab\sin C$  to find the area of $\triangle CBD$:

$[CBD]=\frac{1}{2}\cdot 10\cdot 7.01 \cdot \sin 38^{\circ},\\\\\left [CBD]\right\approx 21.58$.

Since $\triangle ABD$ is a right triangle, we can use basic trig rules for a right triangle to solve for the sides:

$\sin 55^{\circ}=\frac{AB}{10},\\AB=10\sin 55^{\circ},\\AB\approx 8.19$,

$\sin 35^{\circ}=\frac{AD}{10},\\AD=10\sin 35^{\circ},\\AD\approx 5.74$.

Therefore, the area of $\triangle ABD$ is:

$\frac{1}{2}\cdot 8.19\cdot 5.74\approx 23.49$.

The area of $ABCD$ is then

$21.58+23.49=45.07=\fbox{ 45.1 }$.

$\bigstar$ Note that intermediate results were rounded during this calculation. Therefore, there is a small margin of error when rounding to final answer to one decimal place. I recommend plugging in values for yourself using all equations I wrote above, then rounding only your final answer to one decimal place as requested by the problem.