Question

I WILL GIVE THE BRAINLIEST

Answer 1

Answer:

$a=6$, $b=6\sqrt{2}$, $c=2\sqrt{3}$, and $d=6$

Step-by-step explanation:

Looking at the left triangle, we can solve for $a$ and $c$.

a:

$a$ is to the opposite side of 60° angle, also we know the hypotenuse, $4\sqrt{3}$. The ratio that relates opposite with hypotenuse is SINE. Thus we can write:

$sin(A)=\frac{opposite}{hypotenuse}\\sin(60)=\frac{a}{4\sqrt{3}}$

Cross multiplying and solving for $a$:

$sin(60)=\frac{a}{4\sqrt{3} }\\a=sin(60)*4\sqrt{3}\\a=\frac{\sqrt{3}}{2}*4\sqrt{3}\\a=\frac{12}{2}=6$

( we know $sin(60)=\frac{\sqrt{3}}{2}$ and also $\sqrt{a}*\sqrt{a}=a$ )

c:

$c$ is to the adjacent side of 60° angle, also we know the hypotenuse, $4\sqrt{3}$. The ratio that relates adjacent with hypotenuse is COS. Thus we can write:

$cos(A)=\frac{adjacent}{hypotenuse}\\cos(60)=\frac{c}{4\sqrt{3}}$

Cross multiplying and solving for $c$:

$cos(60)=\frac{c}{4\sqrt{3}}\\c=cos(60)*4\sqrt{3}\\c=\frac{1}{2}*4\sqrt{3}\\c=2\sqrt{3}$

( we know $cos(60)=\frac{1}{2}$ )

Looking at the triangle to the right, we can solve for $b$ and $d$.

b:

$a$ is to the opposite side of 45° angle. We have figured out that $a=6$. Also we know that $b$ is the hypotenuse.The ratio that relates opposite with hypotenuse is SINE. Thus we can write:

$sin(A)=\frac{opposite}{hypotenuse}\\sin(45)=\frac{6}{b}\\b=\frac{6}{sin(45)}\\b=\frac{6}{\frac{1}{\sqrt{2}}}\\b=6*\frac{\sqrt{2}}{1}\\b=6\sqrt{2}$

( we know $sin(45)=\frac{1}{\sqrt{2}}$ )

d:

To solve for $d$, we can use the pythagorean theorem. Given by:

$a^2+b^2=c^2$

Where,

• $a$ and $b$ are two legs of the right triangle, and
• $c$ is the hypotenuse (side opposite 90 degree angle)

In the triangle on the right, $b$ is the hypotenuse and $a$ and $d$ are the two legs. Using pythagorean theorem and solving for $d$, we get:

$d^2+a^2=b^2\\d^2+(6)^2=(6\sqrt{2})^2\\d^2+36=72\\d^2=72-36\\d^2=36\\d=6$

( we know that $\sqrt{a}*\sqrt{a}=a$ )

Looking at the answers, 2nd answer choice is right.