Question

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find the value of x in the triangle shown below.
x = \\(\square\\)°
(the triangle has two sides of length 12, one side of length 10.4, one angle of 62°, and the angle at the vertex between the two sides of length 12 is x°)

Explanation:

Step1: Identify triangle type

This is an isosceles triangle with two sides of length 12, so the angles opposite these sides are equal. Let the unknown angle at the top right be $\theta$.

Step2: Set up sine law

Use the Law of Sines: $\frac{\sin 62^\circ}{12} = \frac{\sin \theta}{10.4}$
Rearrange to solve for $\sin \theta$:
$\sin \theta = \frac{10.4 \times \sin 62^\circ}{12}$
Calculate $\sin 62^\circ \approx 0.8829$, so:
$\sin \theta = \frac{10.4 \times 0.8829}{12} \approx \frac{9.182}{12} \approx 0.7652$

Step3: Find angle $\theta$

Take inverse sine: $\theta \approx \arcsin(0.7652) \approx 49.9^\circ \approx 50^\circ$

Step4: Calculate $x$ using angle sum

The sum of angles in a triangle is $180^\circ$, so:
$x = 180^\circ - 62^\circ - 50^\circ$

Answer:

$58^\circ$