Question

what is the value of x

Answer 1

$\text{In}~ \triangle ABC,\\\\~~~~~~~~\sin 45^{\circ} =\dfrac{BC}{AC} \\\\\implies \dfrac 1{\sqrt 2} = \dfrac{BC}{6\sqrt 2}\\\\\\ \implies BC= 6\\\\\text{Now, In}~ \triangle BCD\\\\~~~~~~\cos 60^{\circ} = \dfrac{BD}{BC}\\\\\ \implies \dfrac 12 = \dfrac{x}{6}\\\\\\\implies x = \dfrac 62\\\\\\\implies x = 3$

Answer 2

Answer:

x = 3

Step-by-step explanation:

using the sine ratio in right triangle ABC and the exact value

sin45° = $\frac{1}{\sqrt{2} }$ , then

sin45° = $\frac{opposite}{hypotenuse}$ = $\frac{BC}{AC}$ = $\frac{BC}{6\sqrt{2} }$ = $\frac{1}{\sqrt{2} }$ ( cross- multiply )

BC × $\sqrt{2}$ = 6$\sqrt{2}$ ( divide both sides by $\sqrt{2}$ )

BC = 6

using the cosine ratio in right triangle BCD and the exact value

cos60° = $\frac{1}{2}$ , then

cos60° = $\frac{adjacent}{hypotenuse}$ = $\frac{BD}{BC}$ = $\frac{x}{6}$ = $\frac{1}{2}$ ( cross- multiply )

2x = 6 ( divide both sides by 2 )

x = 3