Question

ABCD is a quadrilateral
Work out angle x
​

Answer 1

First of all, we can work out the length of BD: using the pythagorean theorem, we have

$BD = \sqrt{6^2+8^2}=10$

Now, we can work on triangle BCD and use the sine theorem: we have

$\dfrac{BD}{\sin(\hat{C})} = \dfrac{CD}{\sin(x)}$

Plugging the values and solving for $\sin(x)$ we have

$\dfrac{10}{\sin(29)} = \dfrac{13}{\sin(x)} \iff \sin(x)=\dfrac{13\cdot \sin(29)}{10}\approx 0.63$

We deduce

$x=\arcsin(0.63)\approx 39$

Answer 2

Applying Pythagorean Theorem and the Law of Sines, the value of angle x in quadrilateral ABCD is: $\mathbf{x = 39^{\circ}}$

To work out angle x, we would need to first of all find the length of BD.

Apply the Pythagorean Theorem to find BD since triangle ABD is a right triangle.

Given:

AB = 8 cm

AD = 6 cm

BD = ? (hypotenuse)

Based on the Pythagorean Theorem, we would have the following:

$BD^2 = AB^2 + AD^2$

• Substitute

$BD^2 = 8^2 + 6^2\\\\BD^2 = 100\\\\BD = \sqrt{100} \\\\BD = 10 cm$

Next, apply the Law of Sines to work out angle x:

Base on the Law of Sines, we will have,

$\frac{sin(x)}{DC} = \frac{sin(C)}{BD}$

DC = 13 cm

BD = 10 cm

C = 29

• Plug in the values

$\frac{sin(x)}{13} = \frac{sin(29)}{10}$

• Cross multiply

$\frac{sin(x)}{13} = \frac{sin(29)}{10}\\\\sin(x) \times 10 = sin(29) \times 13$

• Divide both sides by 10

$sin(x) = \frac{sin(29) \times 13}{10} \\\\sin(x) = 0.63\\\\x = sin^{-1}(0.63)\\\\\mathbf{x = 39^{\circ}}$

Therefore, applying Pythagorean Theorem and the Law of Sines, the value of angle x in quadrilateral ABCD is: $\mathbf{x = 39^{\circ}}$

Learn more here:

https://brainly.com/question/3482956