Question

Geometry: HELPPPP PLSS!!

Answer 1

A, C and G

Explanation:

`\boxed{\begin{minipage}{9.4 cm}\underline{Trigonometric ratios} \\\\$\sf \sin(\theta)=(O)/(H)\quad\cos(\theta)=(A)/(H)\quad\tan(\theta)=(O)/(A)$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf O$ is the side opposite the angle. \\\phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle. \\\phantom{ww}$\bullet$ $\sf H$ is the hypotenuse (the side opposite the right angle). \\\end{minipage}}`

From inspection of the given right triangle:

• θ = 25°
• O = a
• A = b
• H = c

Substitute the values into the trigonometric ratios:

`\implies \sf \sin(25^(\circ))=(a)/(c)`

`\implies \sf \cos(25^(\circ))=(b)/(c)`

`\implies \sf \tan(25^(\circ))=(a)/(b)`

`\boxed{\begin{minipage}{4 cm}\underline{Trigonometric values} \\\\$\sin(90^(\circ)- \theta)=\cos \theta$\\$\cos(90^(\circ)- \theta)=\sin \theta$\\$\tan(90^(\circ)- \theta)=\cot \theta$\\\end{minipage}}`

Using trigonometric values and the calculated trigonometric ratios:

`\begin{aligned} \sf \sin(90^(\circ)- 25^(\circ))&=\sf \cos (25^(\circ))\\\implies \sf \sin(65^(\circ)) &=\sf (b)/(c)\end{aligned}`

`\begin{aligned} \sf \cos(90^(\circ)- 25^(\circ))&= \sf \sin(25^(\circ))\\\implies \sf \cos(65^(\circ)) &=\sf (a)/(c)\end{aligned}`

`\begin{aligned}\sf \tan(90^(\circ)- 25^(\circ))&= \sf \cot(25^(\circ))\\\implies \sf \tan(65^(\circ)) &= \sf (1)/(\tan(25^(\circ)))\\\implies \sf \tan(65^(\circ)) & = \sf (b)/(a)\end{aligned}`

Therefore, the true equations are:

`\sf A: \quad \cos(25^(\circ))=(b)/(c)`

`\sf C: \quad \sin(65^(\circ))=(b)/(c)`

`\sf G: \quad \sin(25^(\circ))=\cos(65^(\circ))`