Question

determining the number of possible triangles

how many distinct triangles can be formed for which ( mangle e = 64^circ ), ( g = 9 ), and ( e = 10 )?
(square) triangle(s)

how many distinct triangles can be formed for which ( mangle j = 129^circ ), ( k = 8 ), and ( j = 3 )?
(square) triangle(s)

Explanation:

First Problem: \( m\angle E = 64^\circ \), \( g = 9 \), \( e = 10 \)

Step1: Recall Law of Sines

The Law of Sines states that \(\frac{\sin E}{e}=\frac{\sin G}{g}\). We know \( E = 64^\circ \), \( e = 10 \), \( g = 9 \). Plugging in, we get \(\frac{\sin 64^\circ}{10}=\frac{\sin G}{9}\).

Step2: Solve for \(\sin G\)

\(\sin G=\frac{9\sin 64^\circ}{10}\). Calculate \(\sin 64^\circ\approx0.8988\), so \(\sin G\approx\frac{9\times0.8988}{10}\approx0.8089\).

Step3: Determine possible \( G \)

Since \(\sin G\approx0.8089\), \( G \) could be in the first quadrant (\( G_1\approx54^\circ \)) or second quadrant (\( G_2\approx180^\circ - 54^\circ = 126^\circ \)). Now check if \( G_2 + E \lt 180^\circ \): \( 126^\circ+64^\circ = 190^\circ\gt180^\circ \), so only \( G_1 \) is valid. Thus, 1 triangle.

Second Problem: \( m\angle J = 129^\circ \), \( k = 8 \), \( j = 37 \)

Step1: Recall Law of Sines

\(\frac{\sin J}{j}=\frac{\sin K}{k}\), so \(\frac{\sin 129^\circ}{37}=\frac{\sin K}{8}\).

Step2: Solve for \(\sin K\)

\(\sin K=\frac{8\sin 129^\circ}{37}\). \(\sin 129^\circ=\sin(180^\circ - 51^\circ)=\sin 51^\circ\approx0.7771\), so \(\sin K\approx\frac{8\times0.7771}{37}\approx0.168\).

Step3: Check validity

\( J = 129^\circ \) is obtuse. For a triangle, only one obtuse angle is allowed. Also, \( k = 8 \lt j = 37 \), so \( K \) must be acute (since side opposite smaller angle is smaller). But \(\sin K\approx0.168\) gives \( K\approx9.7^\circ \), and \( 129^\circ + 9.7^\circ\lt180^\circ \)? Wait, no: Wait, \( j = 37 \) is opposite \( J = 129^\circ \), \( k = 8 \) is opposite \( K \). Since \( j\gt k \), \( J\gt K \). But \( J = 129^\circ \) is obtuse, and \( K \) would be acute, but let's check the sum: \( 129^\circ + K + L = 180^\circ \), so \( K + L = 51^\circ \). But also, from Law of Sines, \(\frac{\sin K}{8}=\frac{\sin 129^\circ}{37}\), so \(\sin K=\frac{8\sin 129^\circ}{37}\approx\frac{8\times0.7771}{37}\approx0.168\), so \( K\approx9.7^\circ \), then \( L = 51^\circ - 9.7^\circ = 41.3^\circ \). Wait, but wait: Is \( j\) (37) longer than \( k\) (8)? Yes, so \( J\) (129°) is opposite \( j\) (37), \( K\) opposite \( k\) (8). Since \( j\gt k \), \( J\gt K \), which is true (129° > 9.7°). But also, check if \( k\sin J \lt j \)? Wait, no, the ambiguous case is for acute angles. For obtuse angle \( J \), if \( j\gt k \), then only one triangle. Wait, actually, when the given angle is obtuse (\( J = 129^\circ \)), we check if \( j\gt k \). Since \( j = 37 \), \( k = 8 \), \( 37\gt8 \), so only one triangle? Wait, no, wait: Wait, the length opposite the obtuse angle must be the longest side. Here, \( j = 37 \) is opposite \( J = 129^\circ \), and \( k = 8 \) is another side. Since \( 37\gt8 \), \( j \) is longer than \( k \), so \( J \) is the largest angle (obtuse), so only one triangle? Wait, no, wait, let's recalculate \(\sin K\): \(\sin K=\frac{8\sin 129^\circ}{37}\approx\frac{8\times0.7771}{37}\approx0.168\), so \( K\approx9.7^\circ \), then \( L = 180 - 129 - 9.7 = 41.3^\circ \), which is valid. But wait, is there another possible \( K \)? Since \(\sin K = 0.168\), the other angle would be \( 180 - 9.7 = 170.3^\circ \), but then \( J + K = 129 + 170.3 = 299.3\gt180 \), which is impossible. So only 1 triangle? Wait, no, wait, the first problem: when the angle is acute, we have to check the ambiguous case. The second problem: angle is obtuse, so if the side opposite the obtuse angle is longer than the other given side, then only one triangle. Since \( j = 37 \) (opposite \( J = 129^\circ \)) is longer than \( k = 8 \) (opposite \( K \)), so only one triangle. Wait, but wait, maybe I made a mistake. Wait, let's check the first problem again. First problem: \( E = 64^\circ \) (acute), \( e = 10 \), \( g = 9 \). So \( e\gt g \), so \( E\gt G \), so \( G \) must be acute, so only one triangle. Wait, no, earlier I thought there was a second angle, but \( 180 - 54 = 126 \), and \( 126 + 64 = 190\gt180 \), so invalid. So first problem: 1 triangle. Second problem: \( J = 129^\circ \) (obtuse), \( j = 37 \), \( k = 8 \). Since \( j\gt k \), the side opposite the obtuse angle is longer than the other side, so only one triangle. Wait, but wait, maybe the second problem has 0 triangles? Wait, no, let's check the Law of Sines again. \(\frac{\sin J}{j}=\frac{\sin K}{k}\), so \(\sin K=\frac{k\sin J}{j}=\frac{8\sin 129^\circ}{37}\). \(\sin 129^\circ\approx0.7771\), so \( 8\times0.7771\approx6.2168 \), divided by…

Answer:

First triangle: \(\boxed{1}\)
Second triangle: \(\boxed{1}\)