Question

3 for $\angle p$, create trig ratios for sin, cos, and tan:
drag & drop the answer
30
18
24
note: use ctrl+d to drag the option via keyboard
(right triangle with right angle at q, sides: rq=18, qp=24, rp=30; angles at r, q, p)
$\sin(p)=\frac{\square}{\square}$ $\cos(p)=\frac{\square}{\square}$
$\tan(p)=\frac{\square}{\square}$

Explanation:

Step1: Recall trigonometric ratios

In a right triangle, for an acute angle \( \theta \):
\( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \), \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \), \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \).

For \( \angle P \) in right triangle \( \triangle PQR \) (right-angled at \( Q \)):

• Opposite side to \( \angle P \): \( QR = 18 \)
• Adjacent side to \( \angle P \): \( QP = 24 \)
• Hypotenuse: \( PR = 30 \)

Step2: Calculate \( \sin(P) \)

\( \sin(P) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{QR}{PR} = \frac{18}{30} \)

Step3: Calculate \( \cos(P) \)

\( \cos(P) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{QP}{PR} = \frac{24}{30} \)

Step4: Calculate \( \tan(P) \)

\( \tan(P) = \frac{\text{opposite}}{\text{adjacent}} = \frac{QR}{QP} = \frac{18}{24} \)

Answer:

\( \sin(P) = \boldsymbol{\frac{18}{30}} \), \( \cos(P) = \boldsymbol{\frac{24}{30}} \), \( \tan(P) = \boldsymbol{\frac{18}{24}} \)