Trigonometric ratios: finding an angle questions with answers

Questions

1. Question 1Opposite and adjacent: tan⁻¹

   In the diagram, $\angle XYZ = 90^\circ$. Find $x$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: Label the two known sides from the angle x°From the angle x°: the side $5$ cm is the opposite and the side $12$ cm is the adjacent.
   2. Step 2: Choose the ratio that uses those two sides (SOH CAH TOA) and write it down$\tan x^\circ = \dfrac{5}{12}$
   3. Step 3: Two sides known: use the inverse to find the angle$x = \tan^{-1}\left(\dfrac{5}{12}\right)$$= 22.6198\ldots$$= 22.6^\circ$ (1 d.p.)
2. Question 2With the hypotenuse: sin⁻¹ or cos⁻¹

   In the diagram, $\angle KLM = 90^\circ$. Find $x$, correct to 1 decimal place.

   Show the worked solution

   1. Step 1: Label the two known sides from the angle x°From the angle x°: the side $20$ m is the hypotenuse and the side $15$ m is the adjacent.
   2. Step 2: Choose the ratio that uses those two sides (SOH CAH TOA) and write it down$\cos x^\circ = \dfrac{15}{20}$
   3. Step 3: Two sides known: use the inverse to find the angle$x = \cos^{-1}\left(\dfrac{15}{20}\right)$$= 41.4096\ldots$$= 41.4^\circ$ (1 d.p.)
3. Question 3A ratio as an exact fraction

   In the diagram, $\angle PQR = 90^\circ$, $PQ = 16$ cm and $QR = 30$ cm. Write down the value of $\tan \angle PRQ$, giving your answer as a fraction in its simplest form.

   Show the worked solution

   1. Step 1: Label the two sides that tan uses, from ∠PRQFrom $\angle PRQ$: $PQ$ is the opposite and $QR$ is the adjacent.
   2. Step 2: tan = opposite ÷ adjacent: write the fraction, then divide the top and bottom by their HCF$\tan \angle PRQ = \dfrac{16}{30}$$= \dfrac{8}{15}$
4. Question 4A ratio as a fraction, then a point dividing a side

   In the diagram, $\angle EFG = 90^\circ$, $EG = 25$ cm and $\sin \angle FEG = \dfrac{4}{5}$. The point $H$ lies on $FG$ such that $FH : HG = 1 : 2$. Find the length of $FH$.

   Show the worked solution

   1. Step 1: The ratio tells you the sides in proportion: sin uses FG and EG$\dfrac{FG}{EG} = \dfrac{4}{5}$ (from the sin ratio)$FG = \dfrac{4}{5} \times 25$$= 20$ cm
   2. Step 2: The ratio cuts FG into equal parts: FH is the first number of parts out of the total$FH = \dfrac{1}{3} \times 20$$= 6.6666\ldots$$= 6.67$ cm (3 s.f.)