Question

Which trigonometric ratios are correct for triangle ABC? Select three options.

Answer 1

Answer:

The correct ratios are

$\sin(C)=\dfrac{\sqrt{3}}{2}, \tan(C)=\sqrt{3},\sin B=\dfrac{1}{2}$

Step-by-step explanation:

Consider the below figure attached with this question.

In triangle ABC, ∠A=90°, ∠B=30° and ∠C=60°, AC=9 and BC=18.

Using Pythagoras theorem we get

$hypotenuse^2=perpendicular^2+base^2$

$BC^2=AB^2+AC^2$

$(18)^2=AB^2+9^2$

$324-81=AB^2$

Taking square root on both sides.

$\sqrt{243}=AB$

$9\sqrt{3}=AB$

Trigonometric ratios are

$\sin (C)=\dfrac{opposite}{hypotenuse}=\dfrac{AB}{BC}=\dfrac{9\sqrt{3}}{18}=\dfrac{\sqrt{3}}{2}$

$\cos (B)=\dfrac{base}{hypotenuse}=\dfrac{AB}{BC}=\dfrac{9\sqrt{3}}{18}=\dfrac{\sqrt{3}}{2}$

$\tan (C)=\dfrac{opposite}{base}=\dfrac{AB}{AC}=\dfrac{9\sqrt{3}}{9}=\sqrt{3}$

$\sin (B)=\dfrac{opposite}{hypotenuse}=\dfrac{AC}{BC}=\dfrac{9}{18}=\dfrac{1}{2}$

$\tan (B)=\dfrac{opposite}{base}=\dfrac{AC}{AB}=\dfrac{9}{9\sqrt{3}}=\dfrac{1}{\sqrt{3}}$

Therefore, the correct ratios are

$\sin(C)=\dfrac{\sqrt{3}}{2}, \tan(C)=\sqrt{3},\sin B=\dfrac{1}{2}$

Answer 2

Answer:

A)

C)

D)

Step-by-step explanation:

sorry i couldn't type in actual written out answers, wouldn't copy and paste.

just took test, these r right.