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12. solve for x (right triangle with hypotenuse 16√10, angle 30° at the…

Question

  1. solve for x

(right triangle with hypotenuse 16√10, angle 30° at the base, and vertical side x)

Explanation:

Step1: Identify trigonometric ratio

In a right triangle, $\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$. Here, $\theta = 30^\circ$, opposite side is $x$, hypotenuse is $16\sqrt{10}$.

Step2: Apply sine formula

$\sin(30^\circ)=\frac{x}{16\sqrt{10}}$. Since $\sin(30^\circ)=\frac{1}{2}$, substitute:
$\frac{1}{2}=\frac{x}{16\sqrt{10}}$

Step3: Solve for x

Multiply both sides by $16\sqrt{10}$:
$x = \frac{1}{2} \times 16\sqrt{10}=8\sqrt{10}$

Answer:

$8\sqrt{10}$