Given Triangle ABC: Right Triangles
1. If a = 6 and <B = 53º, find b, c, and <A
2. If c = 13 and b = 12, find a, <A, and <B
3. If <A = 20º and b = 10, find <B, a and c
4. If c = 30 and <A = 40º, find <B, a
These are right triangles, so angle C is 90
1. isnt CosineB=a/c so you can solve for c. Then, solve for b.
I will be happy to critique your thinking.
To solve these problems, we can use the trigonometric functions sine, cosine, and tangent.
1. Given that side a = 6 and angle B = 53º, we can find side b using the cosine function:
cos(B) = b / a
cos(53º) = b / 6
b = 6 * cos(53º)
Now we can find side c using the Pythagorean theorem:
c^2 = a^2 + b^2
c^2 = 6^2 + (6 * cos(53º))^2
c^2 = 36 + 36 * cos(53º)^2
c = sqrt(36 + 36 * cos(53º)^2)
Finally, to find angle A, we can use the fact that the sum of angles in a triangle is 180º:
A = 180º - B - 90º
A = 180º - 53º - 90º
2. Given side c = 13 and side b = 12, we can find side a using the Pythagorean theorem:
a^2 = c^2 - b^2
a^2 = 13^2 - 12^2
a = sqrt(13^2 - 12^2)
To find angle A, we can use the sine function:
sin(A) = a / c
sin(A) = sqrt(13^2 - 12^2) / 13
A = arcsin(sqrt(13^2 - 12^2) / 13)
Finally, to find angle B, we can use the fact that the sum of angles in a triangle is 180º:
B = 180º - A - 90º
3. Given angle A = 20º and side b = 10, we can find angle B using the fact that the sum of angles in a triangle is 180º:
B = 180º - A - 90º
To find side a, we can use the tangent function:
tan(A) = a / b
a = b * tan(A)
Finally, to find side c, we can use the Pythagorean theorem:
c^2 = a^2 + b^2
c^2 = (b * tan(A))^2 + b^2
c = sqrt((b * tan(A))^2 + b^2)
4. Given side c = 30 and angle A = 40º, we can find angle B using the fact that the sum of angles in a triangle is 180º:
B = 180º - A - 90º
To find side a, we can use the sine function:
sin(A) = a / c
a = c * sin(A)