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section a find the missing angle x 1) x= 2) x= 3) x= 4) x= 5) x= 6) x=

Question

section a find the missing angle x

  1. x=
  2. x=
  3. x=
  4. x=
  5. x=
  6. x=

Explanation:

Step1: Recall cosine - inverse formula

We use $\cos^{-1}(\frac{\text{adjacent}}{\text{hypotenuse}})$ for right - angled triangles to find the angle.

Step2: Solve for (1)

For the first triangle with adjacent side $4$ cm and hypotenuse $9$ cm, $x = \cos^{-1}(\frac{4}{9})\approx63.6^{\circ}$

Step3: Solve for (2)

For the second triangle with adjacent side $10$ cm and hypotenuse $14$ cm, $x=\cos^{-1}(\frac{10}{14})\approx44.4^{\circ}$

Step4: Solve for (3)

For the third triangle with adjacent side $7$ cm and hypotenuse $12$ cm, $x=\cos^{-1}(\frac{7}{12})\approx54.3^{\circ}$

Step5: Solve for (4)

For the fourth triangle with adjacent side $15$ cm and hypotenuse $18$ cm, $x=\cos^{-1}(\frac{15}{18})\approx33.6^{\circ}$

Step6: Solve for (5)

For the fifth triangle with adjacent side $0.99$ cm and hypotenuse $0.83$ cm, $x=\cos^{-1}(\frac{0.99}{0.83})$ (This is incorrect as adjacent side cannot be longer than hypotenuse in a right - angled triangle. Assuming it's a typo and adjacent is $0.83$ and hypotenuse is $0.99$, then $x=\cos^{-1}(\frac{0.83}{0.99})\approx33.7^{\circ}$)

Step7: Solve for (6)

For the sixth triangle with adjacent side $410$ mm and hypotenuse $972$ mm, $x=\cos^{-1}(\frac{410}{972})\approx64.7^{\circ}$

Answer:

(1) $x\approx63.6^{\circ}$
(2) $x\approx44.4^{\circ}$
(3) $x\approx54.3^{\circ}$
(4) $x\approx33.6^{\circ}$
(5) $x\approx33.7^{\circ}$
(6) $x\approx64.7^{\circ}$