Question

question 4 of 26
given the triangle below, what is ( m angle b ), rounded to the nearest tenth?
triangle diagram: vertices b (top), a (bottom-left), c (bottom-right); ( ab = 10 ), ( ac = 14 ), ( angle c = 36^circ ); note: triangle not drawn to scale
a. ( 55.4^circ )
b. ( 62.1^circ )
c. ( 60.3^circ )
d. ( 57.6^circ )

Explanation:

Step1: Identify the Law to Use

We have a triangle with two sides and a non - included angle? Wait, no, we have side \(AB = 10\), side \(AC=14\), and angle \(C = 36^{\circ}\). We can use the Law of Sines, which 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.

In triangle \(ABC\), side opposite angle \(C\) is \(AB = 10\) (let's call it \(c = 10\)), side opposite angle \(B\) is \(AC = 14\) (let's call it \(b = 14\)), and angle \(C=36^{\circ}\). Let angle \(B = \beta\).

From the Law of Sines: \(\frac{\sin\beta}{b}=\frac{\sin C}{c}\)

Step2: Substitute the Values

Substitute \(b = 14\), \(c = 10\), and \(C = 36^{\circ}\) into the formula:

\(\sin\beta=\frac{b\sin C}{c}=\frac{14\times\sin(36^{\circ})}{10}\)

First, calculate \(\sin(36^{\circ})\approx0.5878\)

Then, \(14\times0.5878 = 8.2292\)

\(\sin\beta=\frac{8.2292}{10}=0.82292\)

Step3: Find the Angle

Now, we need to find the angle \(\beta\) such that \(\sin\beta = 0.82292\). We use the inverse sine function:

\(\beta=\sin^{- 1}(0.82292)\)

Using a calculator, \(\sin^{-1}(0.82292)\approx55.4^{\circ}\) (Wait, no, wait. Wait, maybe I mixed up the sides. Wait, side \(AB = 10\) is opposite angle \(C\), side \(AC = 14\) is opposite angle \(B\), and side \(BC\) is opposite angle \(A\). Wait, let's re - check.

Wait, angle \(C\) is at vertex \(C\), so side opposite angle \(C\) is \(AB\) (length 10), side opposite angle \(B\) is \(AC\) (length 14), side opposite angle \(A\) is \(BC\).

Law of Sines: \(\frac{AB}{\sin C}=\frac{AC}{\sin B}\)

So \(\frac{10}{\sin36^{\circ}}=\frac{14}{\sin B}\)

Cross - multiply: \(10\sin B=14\sin36^{\circ}\)

\(\sin B=\frac{14\sin36^{\circ}}{10}\)

\(\sin36^{\circ}\approx0.5878\)

\(14\times0.5878 = 8.2292\)

\(\sin B=\frac{8.2292}{10}=0.82292\)

\(B=\sin^{-1}(0.82292)\approx55.4^{\circ}\)? Wait, but let's check the triangle angle sum. Wait, maybe I made a mistake in side labeling. Wait, maybe the side \(AB = 10\) is adjacent to angle \(A\) and \(B\), and \(AC = 14\) is adjacent to angle \(A\) and \(C\). Wait, maybe it's a triangle with \(AC = 14\) (side \(b\)), \(AB = 10\) (side \(c\)), angle \(C = 36^{\circ}\). Then by Law of Sines, \(\frac{c}{\sin C}=\frac{b}{\sin B}\), so \(\sin B=\frac{b\sin C}{c}=\frac{14\sin36^{\circ}}{10}\approx\frac{14\times0.5878}{10}\approx0.8229\), so \(B=\sin^{-1}(0.8229)\approx55.4^{\circ}\) or \(180 - 55.4=124.6^{\circ}\). But if \(B = 124.6^{\circ}\), then angle \(A=180 - 36 - 124.6 = 19.4^{\circ}\), and let's check the Law of Sines for angle \(A\). But the sum of angles in a triangle is \(180^{\circ}\). Let's see the lengths: \(AC = 14\) is longer than \(AB = 10\), so angle \(B\) should be larger than angle \(C\) (since larger side is opposite larger angle). Angle \(C = 36^{\circ}\), so angle \(B\) should be larger than \(36^{\circ}\). \(55.4^{\circ}\) is larger than \(36^{\circ}\), and \(124.6^{\circ}\) is also larger. But let's check the possible triangle. If we have \(AC = 14\), \(AB = 10\), angle \(C = 36^{\circ}\), using the Law of Sines, we have two possible triangles (ambiguous case). But in the options, \(55.4^{\circ}\) is option A. Wait, maybe my initial side labeling was wrong. Wait, maybe the side \(AB = 10\) is opposite angle \(C\), and side \(BC\) is opposite angle \(A\), and side \(AC = 14\) is opposite angle \(B\). Wait, no, let's re - draw mentally. Vertex \(A\), \(B\), \(C\) with \(A\) connected to \(C\) (length 14), \(A\) connected to \(B\) (length 10), and \(B\) connected to \(C\). So angle at \(C\) is \(36^{\circ}\), between \(AC\) and \(BC\). So side \(AB\) is opposite angle \(C\) (length 10), side \(AC\) is adjacent to angle \(C\) (length 14), side \(BC\) is adjacent to angle \(C\) (let's call it \(a\)). Then by Law of Sines, \(\frac{AB}{\sin C}=\frac{AC}{\sin B}\), so \(\sin B=\frac{AC\sin C}{AB}=\frac{14\sin36^{\circ}}{10}\approx0.8229\), so \(B=\sin^{-1}(0.8229)\approx55.4^{\circ}\) or \(124.6^{\circ}\). But if \(B = 124.6^{\circ}\), then angle \(A=180 - 36 - 124.6 = 19.4^{\circ}\), and then by Law of Sines, \(\frac{BC}{\sin A}=\frac{AB}{\sin C}\), \(BC=\frac{AB\sin A}{\sin C}=\frac{10\sin19.4^{\circ}}{\sin36^{\circ}}\approx\frac{10\times0.332}{0.5878}\approx5.65\). But if \(B = 55.4^{\circ}\), then angle \(A = 180-36 - 55.4=88.6^{\circ}\), and \(BC=\frac{AB\sin A}{\sin C}=\frac{10\sin88.6^{\circ}}{\sin36^{\circ}}\approx\frac{10\times0.999}{0.5878}\approx17.0\). Now, in the triangle, side \(AC = 14\), side \(AB = 10\). If \(B = 124.6^{\circ}\), side \(BC\approx5.65\), which is shorter than \(AB = 10\), but angle \(A\) would be \(19.4^{\circ}\), which is smaller than angle \(C = 36^{\circ}\), but side \(BC\) is opposite angle \(A\), so it should be shorter than side \(AB\) (opposite angle \(C\)), which is consistent. But the options have \(55.4^{\circ}\) as option A. Let's check the calculation again.

\(\sin36^{\circ}\approx0.587785\)

\(14\times0.587785 = 8.229\)

\(8.229\div10 = 0.8229\)

\(\sin^{-1}(0.8229)\approx55.4^{\circ}\) (using calculator: \(\sin(55.4^{\circ})\approx\sin(55 + 0.4^{\circ})=\sin55^{\circ}\cos0.4^{\circ}+\cos55^{\circ}\sin0.4^{\circ}\approx0.8192\times0.99997 + 0.5736\times0.00698\approx0.8192+0.0040\approx0.8232\), which is close to 0.8229)

So the measure of angle \(B\) is approximately \(55.4^{\circ}\)

Answer:

A. \(55.4^{\circ}\)