can you help me to solve each triangle described below using ΔABC :
1)a = 10, A = 16°, B = 49°
2)a = 8, b = 7, A = 88°
A+B+C=180°, so C=115°
Now use law of sines:
b/sin49° = c/sin115° = 10/sin16°
#2:
sin88°/8 = sinB/7, so
B = 61° or 109°
But, since A=88°, B cannot be 109°, so there's no ambiguous case here.
Now you know A and B, so you can find C and c as above.
Sure! To solve each triangle, we can use the Law of Sines and the Law of Cosines. Let's start with the first triangle:
1) Triangle ABC where a = 10, A = 16°, and B = 49°:
Step 1: Since we know two angles, we can find angle C by using the Angles in a Triangle Property. The sum of angles in a triangle is always 180 degrees.
C = 180° - A - B
C = 180° - 16° - 49°
C = 115°
Step 2: Now, to find side b, we can use the Law of Sines:
sin A / a = sin B / b
sin 16° / 10 = sin 49° / b
b = (10 * sin 49°) / sin 16°
b ≈ 18.68 units
Step 3: To find side c, we can use the Law of Cosines:
c^2 = a^2 + b^2 - 2ab * cos C
c^2 = 10^2 + 18.68^2 - 2 * 10 * 18.68 * cos 115°
c ≈ 16.24 units
Thus, the sides of triangle ABC are approximately a = 10 units, b ≈ 18.68 units, and c ≈ 16.24 units.
Now let's move to the second triangle:
2) Triangle ABC where a = 8, b = 7, and A = 88°:
Step 1: Since we know two sides and an angle, we can find the remaining angle using the Law of Cosines:
cos A = (b^2 + c^2 - a^2) / (2bc)
cos 88° = (7^2 + c^2 - 8^2) / (2 * 7 * c)
c^2 = 7^2 + 8^2 - 2 * 7 * 8 * cos 88°
c ≈ 11.80 units
Step 2: Now, to find angle B, we can use the Law of Sines:
sin B / b = sin A / a
sin B / 7 = sin 88° / 8
sin B = (7 * sin 88°) / 8
B = arcsin[(7 * sin 88°) / 8]
B ≈ 73.78°
Step 3: To find angle C, we can use the Angles in a Triangle Property:
C = 180° - A - B
C = 180° - 88° - 73.78°
C ≈ 18.22°
Thus, the sides of triangle ABC are approximately a = 8 units, b = 7 units, and c ≈ 11.80 units, with angles A ≈ 88°, B ≈ 73.78°, and C ≈ 18.22°.