Question

1. find the measure of angle a.

a triangle with vertices a, b, c. side ab is 25 cm, side bc is 16 cm, angle at b is 97°. find the measure of angle a.

Explanation:

Step1: Identify the Law to Use

We have a triangle with two sides and the included angle? Wait, no, here we have side \( AB = 25 \) cm, side \( BC = 16 \) cm, angle at \( B = 97^\circ \), and we need to find angle at \( A \). Wait, actually, to find angle \( A \), we can use the Law of Sines or Law of Cosines. Wait, first, let's recall the Law of Sines: \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \), and Law of Cosines: \( c^{2}=a^{2}+b^{2}-2ab\cos C \). Wait, in triangle \( ABC \), side opposite angle \( A \) is \( BC = 16 \) cm (let's call it \( a = 16 \)), side opposite angle \( B \) is \( AC \) (let's call it \( b \)), side opposite angle \( C \) is \( AB = 25 \) cm (let's call it \( c = 25 \)). Wait, angle at \( B \) is \( 97^\circ \), so we can use the Law of Sines: \( \frac{a}{\sin A}=\frac{c}{\sin C} \)? Wait, no, maybe I mixed up. Wait, let's label the triangle properly: \( AB = c = 25 \), \( BC = a = 16 \), \( AC = b \), angle at \( B \) is \( B = 97^\circ \). We need to find angle \( A \). So by Law of Sines: \( \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \). Wait, but we don't know side \( b \) (AC). Wait, maybe first use Law of Cosines to find side \( AC \), then use Law of Sines to find angle \( A \).

Step2: Find Side \( AC \) (let's call it \( b \)) using Law of Cosines

Law of Cosines: \( b^{2}=a^{2}+c^{2}-2ac\cos B \)
Here, \( a = 16 \), \( c = 25 \), \( B = 97^\circ \)
So \( b^{2}=16^{2}+25^{2}-2\times16\times25\times\cos(97^\circ) \)
Calculate \( 16^{2}=256 \), \( 25^{2}=625 \), \( 2\times16\times25 = 800 \)
\( \cos(97^\circ)\approx\cos(90^\circ + 7^\circ)=-\sin(7^\circ)\approx - 0.1219 \) (using calculator: \( \cos(97^\circ)\approx - 0.1218693434 \))
So \( b^{2}=256 + 625-800\times(-0.1218693434) \)
\( b^{2}=881 + 800\times0.1218693434 \)
\( 800\times0.1218693434\approx97.4955 \)
So \( b^{2}\approx881 + 97.4955 = 978.4955 \)
So \( b\approx\sqrt{978.4955}\approx31.28 \) cm

Step3: Use Law of Sines to Find Angle \( A \)

Now, by Law of Sines: \( \frac{a}{\sin A}=\frac{b}{\sin B} \)
We know \( a = 16 \), \( b\approx31.28 \), \( B = 97^\circ \), \( \sin(97^\circ)\approx\sin(90^\circ + 7^\circ)=\cos(7^\circ)\approx0.992546 \)
So \( \sin A=\frac{a\times\sin B}{b}=\frac{16\times0.992546}{31.28} \)
Calculate numerator: \( 16\times0.992546\approx15.8807 \)
Then \( \sin A\approx\frac{15.8807}{31.28}\approx0.5077 \)
Then \( A\approx\arcsin(0.5077)\approx30.5^\circ \) (since \( \sin(30^\circ)=0.5 \), \( \sin(31^\circ)\approx0.5150 \), so around \( 30.5^\circ \), approximately \( 31^\circ \) or more accurately, let's calculate:

\( \arcsin(0.5077)\approx30.5^\circ \) (using calculator: \( \arcsin(0.5077)\approx30.5^\circ \))

Wait, let's check the calculations again. Wait, maybe I made a mistake in labeling. Wait, in triangle \( ABC \), angle at \( B \) is \( 97^\circ \), side \( AB = 25 \), side \( BC = 16 \). So side \( AB \) is adjacent to angle \( B \) and angle \( A \), side \( BC \) is adjacent to angle \( B \) and angle \( C \). So using Law of Sines: \( \frac{BC}{\sin A}=\frac{AB}{\sin C} \), but we also know that \( A + B + C=180^\circ \), so \( C = 180^\circ - A - 97^\circ=83^\circ - A \). Wait, maybe that's more complicated. Alternatively, use Law of Sines directly: \( \frac{BC}{\sin A}=\frac{AB}{\sin C} \), but we don't know \( C \). Wait, no, the correct approach is:

In triangle \( ABC \):

- \( AB = c = 25 \)
- \( BC = a = 16 \)
- \( \angle B = 97^\circ \)
- Find \( \angle A \)

Using Law of Sines: \( \frac{a}{\sin A}=\frac{c}{\sin C} \), but we also know that \( A + B + C = 180^\circ \), so \( C = 180^\circ - A - 97^\circ=83^\circ - A \). This seems circular. So better to use Law of Cosines to find side \( AC \) (let's call it \( b \)) first:

\( b^{2}=a^{2}+c^{2}-2ac\cos B \)

\( a = 16 \), \( c = 25 \), \( B = 97^\circ \)

\( \cos(97^\circ)\approx - 0.1218693434 \)

\( b^{2}=16^{2}+25^{2}-2\times16\times25\times(-0.1218693434) \)

\( b^{2}=256 + 625 + 800\times0.1218693434 \)

\( 800\times0.1218693434 = 97.4954747 \)

\( b^{2}=256 + 625 + 97.4954747 = 978.4954747 \)

\( b=\sqrt{978.4954747}\approx31.28 \)

Now, using Law of Sines: \( \frac{a}{\sin A}=\frac{b}{\sin B} \)

\( \sin A=\frac{a\times\sin B}{b} \)

\( \sin B=\sin(97^\circ)\approx0.9925461516 \)

\( \sin A=\frac{16\times0.9925461516}{31.28} \)

\( 16\times0.9925461516\approx15.88073843 \)

\( \sin A\approx\frac{15.88073843}{31.28}\approx0.5077 \)

\( A=\arcsin(0.5077)\approx30.5^\circ \) (since \( \sin(30^\circ)=0.5 \), \( \sin(31^\circ)\approx0.5150 \), so \( 30.5^\circ \) is close, approximately \( 31^\circ \) or more accurately, using calculator, \( \arcsin(0.5077)\approx30.5^\circ \))

Step2 (Alternative Check):

Wait, maybe I made a mistake in labeling. Let's re - label:

- Vertex \( A \), \( B \), \( C \)
- Side \( AB = 25 \) (between \( A \) and \( B \))
- Side \( BC = 16 \) (between \( B \) and \( C \))
- Angle at \( B \): \( \angle ABC = 97^\circ \)
- We need to find \( \angle BAC=\angle A \)

So by Law of Sines:

\( \frac{BC}{\sin\angle A}=\frac{AB}{\sin\angle C} \)

And \( \angle A+\angle B+\angle C = 180^\circ\Rightarrow\angle C = 180^\circ - \angle A - 97^\circ=83^\circ - \angle A \)

Substitute into Law of Sines:

\( \frac{16}{\sin A}=\frac{25}{\sin(83^\circ - A)} \)

Using \( \sin(83^\circ - A)=\sin83^\circ\cos A-\cos83^\circ\sin A \)

\( \sin83^\circ\approx0.9925 \), \( \cos83^\circ\approx0.1219 \)

So:

\( 16(\sin83^\circ\cos A-\cos83^\circ\sin A)=25\sin A \)

\( 16\times0.9925\cos A-16\times0.1219\sin A = 25\sin A \)

\( 15.88\cos A-1.9504\sin A = 25\sin A \)

\( 15.88\cos A=25\sin A + 1.9504\sin A \)

\( 15.88\cos A=26.9504\sin A \)

Divide both sides by \( \cos A \):

\( 15.88 = 26.9504\tan A \)

\( \tan A=\frac{15.88}{26.9504}\approx0.589 \)

\( A=\arctan(0.589)\approx30.5^\circ \), which matches the previous result.

So the measure of angle \( A \) is approximately \( 31^\circ \) (or more precisely \( 30.5^\circ \))

Answer:

The measure of angle \( A \) is approximately \( \boldsymbol{31^\circ} \) (or more precisely \( 30.5^\circ \))