What is the area of a triangle with A=15°, B=113°, and b=7?
Answer:
C
Step-by-step explanation:
First, use the sine rule:
[tex]\dfrac{b}{\sin B}=\dfrac{a}{\sin A},\\ \\\dfrac{7}{\sin 113^{\circ}}=\dfrac{a}{\sin 15^{\circ}},\\ \\a=\dfrac{7\sin 15^{\circ}}{\sin 113^{\circ}}\approx 1.968\ un.[/tex]
Now, the sum of the measures of all interior angles of the triangle is equal to 180°,
[tex]\angle C=180^{\circ}-113^{\circ}-15^{\circ}=52^{\circ}.[/tex]
At last, the area of the triangle is
[tex]A=\dfrac{1}{2}ab\sin \angle C,\\ \\A=\dfrac{1}{2}\cdot 1.968\cdot 7\cdot \sin 52^{\circ}\approx 5.4\ un^2.[/tex]