Question

find m∠a
7 m b
119°
c a
32 m
a) 11° b) 14.2°
c) 8° d) 7.8°

Explanation:

Step1: Identify the Law to Use

We have a triangle with two sides and the included angle? Wait, no, here we have side \( BC = 7\) m, side \( AC = 32\) m, and angle at \( B = 119^\circ\). Wait, actually, to find angle \( A\), we can use the Law of Sines. The Law of Sines states that \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\), where \(a, b, c\) are the lengths of the sides opposite angles \(A, B, C\) respectively.

First, let's label the triangle properly. In \(\triangle ABC\), side opposite angle \(A\) is \(BC = 7\) m (let's call this \(a = 7\)), side opposite angle \(B\) is \(AC = 32\) m (let's call this \(b = 32\)), and angle \(B = 119^\circ\). We need to find angle \(A\).

Step2: Apply the Law of Sines

Using the Law of Sines: \(\frac{a}{\sin A}=\frac{b}{\sin B}\)

Substitute the known values: \(\frac{7}{\sin A}=\frac{32}{\sin 119^\circ}\)

First, calculate \(\sin 119^\circ\). \(\sin 119^\circ=\sin(180^\circ - 61^\circ)=\sin 61^\circ\approx0.8746\)

So, \(\frac{7}{\sin A}=\frac{32}{0.8746}\)

Solve for \(\sin A\): \(\sin A=\frac{7\times0.8746}{32}\)

Calculate the numerator: \(7\times0.8746 = 6.1222\)

Then, \(\sin A=\frac{6.1222}{32}\approx0.1913\)

Now, find the angle whose sine is approximately \(0.1913\). Using the inverse sine function: \(A=\sin^{-1}(0.1913)\approx11^\circ\) (since \(\sin 11^\circ\approx0.1908\), which is close to \(0.1913\))

Answer:

A) \(11^\circ\)