Question

Please
The diagram below represents the arm of a crane RST:

Answer 1

Answer:

a) Distance between RT = 10.59 m.

b) α = 128.71°, β = 87.42°

Step-by-step explanation:

a) In the given diagram, let us draw an imaginary line joining RT.

Now, we have a right angled triangle, RTN, where the perpendicular = 10 m and base = 3.5 m.

Using Pythogoras Theorem, we have,

RT² = TN² + RN²

⇒ RT² = (10)² + (3.5)²

⇒ RT² = 100 + 12.25

⇒ RT = √112.25

⇒ RT = 10.59 m.

b) In the right angled Δ RTN, let ∠RTN = ∅

then, tan ∅ = $\frac{RN}{TN}$

⇒ tan ∅ = 0.35

⇒ ∅ = tan⁻¹ (0.35)

⇒ ∅ = 19.29°

Thus, in the Δ RTN,

∠TRN = 180° - (90 + 19.29)°     [∵ sum of 3 angles of a Δ = 180°]

⇒ ∠TRN = 70.71°

Consider Δ TSR, we have,

$cos \beta =\frac{TS^{2} + SR^{2} - TR^{2}}{2 (TS * SR)}$

⇒ $cos \beta = \frac{81+36-112.14}{108}$

⇒ $cos \beta = \frac{4.86}{108}$

⇒ $cos \beta = 0.045$

⇒ β = cos⁻¹ (0.045)

⇒ β = 87.42°

Similarly,

cos ∠TRS = $\frac{SR^{2} + TR^{2}-TS^{2}}{2 * SR*TR}$

⇒ cos ∠TRS = $\frac{36+112.14-81}{127.08}$

⇒ cos ∠TRS = $\frac{67.14}{127.08}$

⇒ cos ∠TRS = 0.53

⇒ ∠TRS = cos⁻¹ (0.53)

⇒ ∠TRS = 58°

Now, we have,

α = ∠TRN + ∠TRS

⇒ α = (70.71 + 58)°

⇒ α = 128.71°

Answer 2

Answer:

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