Question

Find the area of the given triangle to the nearest square unit. Angle a= 30 degrees, b=10, angle B=45 degrees

Answer 1

Answer:

$A=34\ units^2$

Step-by-step explanation:

Suppose we have a general triangle like the one shown in the figure.

We know the angle A, the angle B and the length b.

$A = 30\°\\\\B = 45\°\\\\b = 10$

By definition I know that the sum of the internal angles of a triangle is always equal to 180 °.

So

$A + B + C = 180\\\\30 + 45 + C = 180$

We solve the equation and thus we find the angle C.

$C = 180 - 30-45\\\\C = 105$

We already know the three triangle angles.

Now we use the sine theorem to calculate the sides c and a.

The  sine theorem says that:

$\frac{sin(A)}{a}=\frac{sin(B)}{b}=\frac{sin(C)}{c}$

Then

$\frac{sin(30)}{a}=\frac{sin(45)}{10}$

$\frac{sin(30)}{\frac{sin(45)}{10}}=a$

$a=7.071$

Also

$\frac{sin(105)}{c}=\frac{sin(45)}{10}$

$\frac{sin(105)}{\frac{sin(45)}{10}}=c$

$c=13.660$

Finally, we use the Heron formula to calculate the triangle area

$A=\sqrt{s(s-a)(s-b)(s-c)}$

Where s is:

$s=\frac{a+b+c}{2}$

Therefore

$s=\frac{7.071+10+13.660}{2}$

$s=15.37$

$A=\sqrt{15.37(15.37-7.071)(15.37-10)(15.37-13.66)}$

$A=34\ units^2$