area of triangle XYZ

given:side z=11 side x=7 and side y=13
what are the angles and area please?
please explain

If you are given all 3 sides, you must use the cosine law.

lets find ange Ø, where Ø is opposite the 13
(I always find the largest angle, because the calculator will give us the correct angle if the angle is obtuse)

13^2 = 11^2 + 7^2 - 2(11)(7)cosØ
154cosØ = 121 + 49 - 169
cosØ = 1/154 = .00649
Ø = 89.63°

You now could use the Sine Law to find a second angle
The third angle is then trivial.

could you elaborate?im kind of new to this.

To find the area of triangle XYZ, we can use Heron's formula, which states that the area of a triangle can be calculated using its side lengths.

First, let's find the semi-perimeter of the triangle, which is half the sum of the side lengths:

S = (11 + 7 + 13) / 2 = 31 / 2 = 15.5

Next, we can use Heron's formula:

Area = √[s(s - a)(s - b)(s - c)]

where s is the semi-perimeter of the triangle, and a, b, and c are the side lengths.

Area = √[15.5(15.5 - 11)(15.5 - 7)(15.5 - 13)]
= √[15.5 * 4.5 * 8.5 * 2.5]
= √(892.5)
≈ 29.88

Therefore, the area of triangle XYZ is approximately 29.88 square units.

Now, let's find the angles of triangle XYZ using the Law of Cosines. The Law of Cosines states that for any triangle with side lengths a, b, and c, and opposite angles A, B, and C respectively, the following relationship holds:

c^2 = a^2 + b^2 - 2ab * cos(C)

In our case, side z (11) would be considered c, side x (7) would be considered a, and side y (13) would be considered b.

To find angle X, we will use the Law of Cosines:

cos(X) = (a^2 + b^2 - c^2) / (2ab)
= (7^2 + 13^2 - 11^2) / (2 * 7 * 13)
= (49 + 169 - 121) / (182)
= 97 / 182
≈ 0.533

Now we can find the measure of angle X by taking the inverse cosine (arccos) of 0.533:

X = arccos(0.533)
≈ 57.35°

Similarly, using the Law of Cosines, we can find the measures of angles Y and Z as well.

Angle Y: cos(Y) = (b^2 + c^2 - a^2) / (2bc)
≈ 0.605
Y ≈ 52.8°

Angle Z: cos(Z) = (a^2 + c^2 - b^2) / (2ac)
≈ 0.852
Z ≈ 31.4°

Therefore, the angles of triangle XYZ are approximately X ≈ 57.35°, Y ≈ 52.8°, and Z ≈ 31.4°, and the area is approximately 29.88 square units.