Question

find m∠c
a
88°
11 cm
b
c
27 cm
a) 20.1° b) 24°
c) 30° d) 28°

Explanation:

Step1: Identify the Law to Use

We have a triangle with side \( AB = 11 \) cm, side \( BC = 27 \) cm, and angle \( \angle A = 88^\circ \). 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. Here, side \( AB \) is opposite angle \( C \), side \( BC \) is opposite angle \( A \). So \( AB = c = 11 \), \( BC = a = 27 \), \( \angle A = 88^\circ \), and we need to find \( \angle C \).

Step2: Apply the Law of Sines

Using the Law of Sines: \( \frac{\sin C}{AB}=\frac{\sin A}{BC} \). Substitute the known values: \( \frac{\sin C}{11}=\frac{\sin 88^\circ}{27} \).

Step3: Solve for \( \sin C \)

Multiply both sides by 11: \( \sin C=\frac{11\times\sin 88^\circ}{27} \). Calculate \( \sin 88^\circ\approx\sin(90^\circ - 2^\circ)=\cos 2^\circ\approx0.9994 \). Then \( \sin C=\frac{11\times0.9994}{27}\approx\frac{10.9934}{27}\approx0.4072 \).

Step4: Find \( \angle C \)

Take the inverse sine: \( \angle C=\sin^{-1}(0.4072)\approx24.0^\circ \)? Wait, no, wait, let's recalculate. Wait, \( 11\times\sin 88^\circ = 11\times0.999390827\approx10.9933 \). Then \( 10.9933\div27\approx0.40716 \). \( \sin^{-1}(0.40716)\approx24^\circ \)? Wait, but let's check again. Wait, maybe I mixed up the sides. Wait, in triangle \( ABC \), angle at \( A \) is \( 88^\circ \), side \( AB = 11 \) (between \( A \) and \( B \)), side \( BC = 27 \) (between \( B \) and \( C \)). So side opposite angle \( A \) is \( BC = 27 \), side opposite angle \( C \) is \( AB = 11 \). So Law of Sines: \( \frac{\sin C}{AB}=\frac{\sin A}{BC} \). So \( \sin C=\frac{AB\times\sin A}{BC}=\frac{11\times\sin 88^\circ}{27} \). Let's compute \( \sin 88^\circ\approx0.9994 \). So \( 11\times0.9994 = 10.9934 \). \( 10.9934\div27\approx0.4072 \). Then \( \sin^{-1}(0.4072)\approx24^\circ \)? Wait, but the option A is \( 20.1^\circ \), maybe I made a mistake. Wait, wait, maybe the triangle is labeled differently. Wait, maybe \( AB = 11 \), \( AC \) is another side, and \( BC = 27 \). Wait, no, the diagram: \( A \) and \( B \) are vertical, \( AB = 11 \), \( BC = 27 \), angle at \( A \) is \( 88^\circ \). Wait, maybe I used the wrong sides. Wait, let's check the Law of Sines again. Let's denote: \( \angle A = 88^\circ \), side \( AB = c = 11 \) (opposite \( \angle C \)), side \( AC = b \) (opposite \( \angle B \)), side \( BC = a = 27 \) (opposite \( \angle A \)). So Law of Sines: \( \frac{a}{\sin A}=\frac{c}{\sin C} \). So \( \sin C=\frac{c\sin A}{a}=\frac{11\sin 88^\circ}{27} \). Let's calculate \( \sin 88^\circ = 0.999390827 \). So \( 11\times0.999390827 = 10.9933 \). \( 10.9933\div27 = 0.40716 \). Then \( \sin^{-1}(0.40716) \approx 24^\circ \)? But the option A is \( 20.1^\circ \). Wait, maybe I messed up the sides. Wait, maybe \( AB = 11 \) is adjacent to angle \( A \), and we should use the Law of Cosines? Wait, no, Law of Sines is for when we have a side - angle pair. Wait, let's check the calculation of \( \sin^{-1}(0.4072) \). Using a calculator, \( \sin^{-1}(0.4072)\approx24^\circ \)? Wait, no, let's use a calculator: \( \sin(20.1^\circ)\approx\sin(20^\circ + 0.1^\circ)=\sin 20^\circ\cos 0.1^\circ+\cos 20^\circ\sin 0.1^\circ\approx0.3420\times0.9998 + 0.9397\times0.0017\approx0.3419+0.0016\approx0.3435 \). \( \sin(24^\circ)\approx0.4067 \). Ah! There we go. \( \sin(24^\circ)\approx0.4067 \), which is very close to our calculated \( 0.4072 \). So \( \angle C\approx24^\circ \)? Wait, but the option A is \( 20.1^\circ \), option B is \( 24^\circ \). Wait, maybe my initial assumption of the sides was wrong. Wait, maybe \( AB = 11 \) is not opposite angle \( C \). Wait, let's re - examine the triangle. Point \( A \), \( B \), \( C \): \( AB = 11 \) (vertical side), \( BC = 27 \) (hypotenuse - like side), angle at \( A \) is \( 88^\circ \). So angle at \( A \) is between \( AB \) and \( AC \). So side \( AB = 11 \), side \( AC \) is horizontal, side \( BC = 27 \). So angle at \( A \): between \( AB \) (vertical) and \( AC \) (horizontal), so \( AB = 11 \), \( AC = b \), \( BC = 27 \). Then angle at \( C \) is between \( AC \) and \( BC \). So using Law of Sines: \( \frac{AB}{\sin C}=\frac{BC}{\sin A} \). So \( \sin C=\frac{AB\sin A}{BC}=\frac{11\sin 88^\circ}{27} \). As we calculated, \( \sin C\approx0.4072 \), so \( C=\sin^{-1}(0.4072)\approx24^\circ \). Wait, but the option A is \( 20.1^\circ \). Wait, maybe I made a mistake in the angle. Wait, \( \sin 88^\circ\) is approximately \( 0.9994 \), so \( 11\times0.9994 = 10.9934 \), divided by 27 is approximately \( 0.4072 \). The inverse sine of \( 0.4072 \) is approximately \( 24^\circ \) (since \( \sin 24^\circ\approx0.4067 \), which is very close). So the answer should be B) \( 24^\circ \)? Wait, but let's check with a calculator. Let's compute \( \sin^{-1}(11\sin 88^\circ/27) \). \( 11\sin 88^\circ = 11\times0.999390827 = 10.9932991 \). Divide by 27: \( 10.9932991\div27\approx0.40716 \). \( \sin^{-1}(0.40716)\approx24.0^\circ \). So the correct option is B.

Answer:

B) \( 24^\circ \)