We will have to use the Law of Sines, which states that:
![(\sin A)/(a)=(\sin B)/(b)=(\sin C)/(c)](https://img.qammunity.org/2023/formulas/mathematics/college/dwc66mjebk1jvhqbzm1audqtmju1mnx2km.png)
Procedure:
0. We have to remember that the sum of the interior angles of a right triangle adds up to 180°. Then:
![A+B+C=180](https://img.qammunity.org/2023/formulas/mathematics/college/6j3n21ol5wd272s82bkv7q3917lpom8agw.png)
As we are given two angles, B and C, we can isolate for A:
![A=180-B-C](https://img.qammunity.org/2023/formulas/mathematics/college/lgajwhkhxdz3pj6bi8594gyn8eiiwjhtg1.png)
Replacing the values we get:
![A=180-42-100](https://img.qammunity.org/2023/formulas/mathematics/college/489o7a1rz9dczf8i5h9vx9bol9gdpwg49h.png)
![A=38](https://img.qammunity.org/2023/formulas/mathematics/college/vajdzcpoku5x3xrhd5lxyzf9c6i4l2babp.png)
2. Now that we have the three angles, we can use the Law of Sines to get the sides:
![(\sin B)/(b)=(\sin C)/(c)](https://img.qammunity.org/2023/formulas/mathematics/college/d9hiajcvyuv8mnnmxn67ydhx6spp3x3xu8.png)
Isolating for c:
![c=(\sin C)/((\sin B)/(b))](https://img.qammunity.org/2023/formulas/mathematics/college/cpqzmvr2tzg4nns6u7666nv5d21rml9by8.png)
Replacing the values:
![c=(\sin 100)/((\sin 42)/(165))](https://img.qammunity.org/2023/formulas/mathematics/college/fqmt989wxl4d5oqq0qbjv2fvboioxbshbn.png)
![c\approx243](https://img.qammunity.org/2023/formulas/mathematics/college/re5mam3jutq2q6350mfuvdyl3seunjsk9a.png)
3. Finally, getting a also using the Law of Sines:
![a=(\sin A)/((\sin B)/(b))](https://img.qammunity.org/2023/formulas/mathematics/college/p6dztfh7w30h20y32xaw2vpfkpo0bd7bmv.png)
Replacing the values:
![a=(\sin 38)/((\sin 42)/(165))](https://img.qammunity.org/2023/formulas/mathematics/college/tgkslk4o9w794vd2yf82s27tcb0vz1nexk.png)
![a\approx152](https://img.qammunity.org/2023/formulas/mathematics/college/hm1wtjeksbdud96fme3gfz9acgtblayd81.png)
Answer: A.
• A = 38°
,
• a = 152m
,
• c = 243m