Question

find $m\angle b$
a ----c (11 yd, angle at c is 98°), a to b is 19 yd, triangle abc
a) 36.2° b) 31°
c) 34° d) 35°

Explanation:

Step1: Identify Law of Sines

In triangle \(ABC\), we know side \(AC = 11\) yd, side \(AB=19\) yd, and angle \(C = 98^\circ\). We use the Law of Sines: \(\frac{\sin A}{BC}=\frac{\sin B}{AC}=\frac{\sin C}{AB}\). Wait, actually, let's label correctly: \(AC = 11\) (side \(b\) if \(B\) is opposite? Wait, no, let's define: in \(\triangle ABC\), angle \(C = 98^\circ\), side \(AC = 11\) (side opposite \(B\)), side \(AB = 19\) (side opposite \(C\))? Wait, no, \(AB\) is from \(A\) to \(B\), length 19, \(AC\) is from \(A\) to \(C\), length 11, angle at \(C\) is \(98^\circ\). So using Law of Sines: \(\frac{\sin B}{AC}=\frac{\sin C}{AB}\). So \(\sin B=\frac{AC \cdot \sin C}{AB}\).

Step2: Substitute Values

\(AC = 11\), \(AB = 19\), \(\angle C = 98^\circ\). So \(\sin B=\frac{11 \cdot \sin 98^\circ}{19}\). Calculate \(\sin 98^\circ\approx\sin(90^\circ + 8^\circ)=\cos 8^\circ\approx0.9903\). Then \(11 \cdot 0.9903\approx10.8933\). Then \(\sin B\approx\frac{10.8933}{19}\approx0.5733\). Then \(B=\arcsin(0.5733)\approx34.9^\circ\approx35^\circ\)? Wait, no, wait, maybe I mixed up sides. Wait, \(AC\) is 11, \(BC\) is? Wait, no, the triangle: \(A\) to \(C\) is 11, \(A\) to \(B\) is 19, angle at \(C\) is \(98^\circ\). So side \(AC = 11\) (length between \(A\) and \(C\)), side \(AB = 19\) (length between \(A\) and \(B\)), angle at \(C\) is \(98^\circ\). So by Law of Sines: \(\frac{\sin B}{AC}=\frac{\sin C}{AB}\). So \(\sin B=\frac{AC \cdot \sin C}{AB}=\frac{11 \cdot \sin 98^\circ}{19}\). Let's compute \(\sin 98^\circ\approx0.990268\). Then \(11 \times 0.990268\approx10.8929\). Then \(10.8929 / 19\approx0.5733\). Then \(\arcsin(0.5733)\approx34.9^\circ\approx35^\circ\), which is option D? Wait, but let's check again. Wait, maybe I had the sides wrong. Wait, \(AC = 11\), \(BC\) is another side, \(AB = 19\). Wait, angle at \(C\) is \(98^\circ\), so side opposite angle \(B\) is \(AC = 11\), side opposite angle \(C\) is \(AB = 19\). So Law of Sines: \(\frac{\sin B}{11}=\frac{\sin 98^\circ}{19}\). So \(\sin B=\frac{11 \sin 98^\circ}{19}\approx\frac{11 \times 0.9903}{19}\approx\frac{10.8933}{19}\approx0.5733\). Then \(B\approx\arcsin(0.5733)\approx34.9^\circ\approx35^\circ\), which is option D. Wait, but let's check the options: D is \(35^\circ\).

Answer:

D) \(35^\circ\)