For solving this question, you need to apply trigonometric ratios and/or Pythagoras Theorem for each of the right triangles.
RIGHT TRIANGLE
A triangle is classified as a right triangle when it presents one of your angles equal to 90º. In this triangle from the trigonometric ratios or the Pythagoras Theorem ([tex]hypotenuse^2=(side_1)^2 + (side_2)^2[/tex]), it is possible finding angles or sides.
The main trigonometric ratios are presented below.
[tex]sin(x)=\frac{opposite\,side}{hypotenuse} \\ \\ cos(x)=\frac{adjacent\,side}{hypotenuse}\\ \\ tan(x)=\frac{sin(x)}{cos(x)}= \frac{opposite\,side}{hypotenuse}* \frac{hypotenuse}{adjacent\,side}= \frac{opposite\,side}{adjacent\,side}[/tex]
[tex]hypotenuse^2=(side_1)^2 + (side_2)^2\\ \\ 29^2=20^2+s^2\\ \\ 841=400+s^2\\ \\ s^2=441\\ \\ s=\sqrt{441}=21[/tex]
For angle D you will find:
[tex]sin(D)=\frac{opposite\,side}{hypotenuse}=\frac{20}{29} \\ \\ cos(D)=\frac{adjacent\,side}{hypotenuse}=\frac{21}{29} \\ \\ tan(D)=\frac{sin(D)}{cos(D)}= \frac{opposite\,side}{adjacent\,side}=\frac{20}{21}[/tex]
For angle E you will find:
[tex]sin(E)=\frac{opposite\,side}{hypotenuse}=\frac{21}{29} \\ \\ cos(E)=\frac{adjacent\,side}{hypotenuse}=\frac{20}{29} \\ \\ tan(E)=\frac{sin(E)}{cos(E)}= \frac{opposite\,side}{adjacent\,side}=\frac{21}{20}[/tex]
The question gives an angle (62°) and the adjacent side (25) from the angle 62° of the right triangle. Therefore, you can find x from the trigonometric ratio of tan (62°):
[tex]tan(62^{\circ \:})= \frac{opposite\,side}{adjacent\,side}=\frac{x}{25}\\ \\ \tan \left(62^{\circ \:}\right)=1.881 \\ \\ Then,\\ \\ 1.881=\frac{x}{25} \\ \\ x=25*1.881=47.025\\ \\ x=47.0[/tex]
The question gives an angle (26°) and the opposite side (11) from the angle 26° of the right triangle. Therefore, you can find x from the trigonometric ratio of sin (26°):
[tex]sin(26^{\circ \:})=\frac{opposite\,side}{hypotenuse}=\frac{11}{x} \\ \\ sin(26^{\circ \:})=0.438\\ \\ Then,\\ \\ 0.438=\frac{11}{x} \\ \\ 0.438x=11\\ \\ x=\frac{11}{0.438} \\ \\ x=25.1[/tex]
The question gives an angle (48°) and the hypotenuse (32) of the right triangle. Therefore, you can find x from the trigonometric ratio of sin (48°):
[tex]sin(48^{\circ \:})=\frac{opposite\,side}{hypotenuse}=\frac{x}{32} \\ \\ sin(48^{\circ \:})=0.743\\ \\ Then,\\ \\ 0.743=\frac{x}{32} \\ \\ x=32*0.743\\ \\ x=23.8[/tex]
The question gives an angle (12°) and the adjacent side (29) from the angle 12° of the right triangle. Therefore, you can find x from the trigonometric ratio of cos (12°):
[tex]cos(12^{\circ \:})=\frac{adjacent\,side}{hypotenuse}=\frac{29}{x} \\ \\ cos(12^{\circ \:})=0.978\\ \\ Then,\\ \\ 0.978=\frac{29}{x} \\ \\ 0.978x=29\\ \\ x=\frac{29}{0.978} \\ \\ x=29.6[/tex]
The question gives the adjacent side (14) from the angle x and the hypotenuse (15) of the right triangle. Therefore, you can find x from the trigonometric ratio of cos(x) :
[tex]cos(x)=\frac{adjacent\,side}{hypotenuse}=\frac{14}{15} \\ \\ cos(x)=0.933[/tex]
After that, you should calculate the arccos(x).
[tex]\arccos \left(\frac{14}{15}\right)=21.0^{\circ \:}\\ \\ Then,\\ \\ x=21.0^{\circ \:}[/tex]
The question gives the adjacent side (23) from angle x and the opposite side (19) from angle x of the right triangle. Therefore, you can find x from the trigonometric ratio of tan (x) :
[tex]tan(x)= \frac{opposite\,side}{adjacent\,side}=\frac{19}{23}\\ \\ \tan \left(x}\right)=0.826[/tex]
After that, you should calculate the arctan(x).
[tex]\arctan \left(\frac{19}{23}\right)=39.6^{\circ \:}\\ \\ Then,\\ \\ x=39.6^{\circ \:}[/tex]
The question gives the adjacent side (9) from angle x and the hypotenuse (17) of the right triangle. Therefore, you can find x from the trigonometric ratio of cos(x) :
[tex]cos(x)=\frac{adjacent\,side}{hypotenuse}=\frac{9}{17} \\ \\ cos(x)=0.529[/tex]
After that, you should calculate the arccos(x).
[tex]\arccos \left(\frac{9}{17}\right)=58.0^{\circ \:}\\ \\ Then,\\ \\ x=58.0^{\circ \:}[/tex]
The question gives the opposite side (43) from angle x and the hypotenuse (45) of the right triangle. Therefore, you can find x from the trigonometric ratio of sin(x) :
[tex]sin(x)=\frac{opposite\,side}{hypotenuse}=\frac{43}{45} \\ \\ sin(x)=0.956[/tex]
After that, you should calculate the arcsin(x).
[tex]\arcsin \left(\frac{43}{45}\right)=72.9^{\circ \:}\\ \\ Then,\\ \\ x=72.9^{\circ \:}[/tex]
Learn more about trigonometric ratios here:
brainly.com/question/11967894
