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Two right-angled triangles, ABC and ADC, are shown - HSC - SSCE Mathematics Standard - Question 22 - 2019 - Paper 1

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Question 22

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Two right-angled triangles, ABC and ADC, are shown. Calculate the size of angle $\theta$, correct to the nearest minute.

Worked Solution & Example Answer:Two right-angled triangles, ABC and ADC, are shown - HSC - SSCE Mathematics Standard - Question 22 - 2019 - Paper 1

Step 1

Calculate AC

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Answer

Using the Pythagorean theorem:

AC2=AD2+DC2AC^2 = AD^2 + DC^2

Substituting the values:

AC2=(2.5)2+(6)2AC^2 = (2.5)^2 + (6)^2

Calculating:

AC2=6.25+36=42.25AC^2 = 6.25 + 36 = 42.25

Thus, taking the square root:

AC=42.25=6.5 cmAC = \sqrt{42.25} = 6.5 \text{ cm}

Step 2

Use the cosine ratio to find $\theta$

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Answer

Using the cosine ratio:

cos⁡(θ)=AdjacentHypotenuse=4.96.5\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4.9}{6.5}

Calculating:

cos⁡(θ)=0.754≈0.754\cos(\theta) = 0.754 \approx 0.754

Now, to find the angle θ\theta:

θ=cos⁡−1(0.754)\theta = \cos^{-1}(0.754)

Using a calculator, we find:

θ≈41.075∘\theta \approx 41.075^{\circ}

Step 3

Round to the nearest minute

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Answer

The decimal part, 0.075 times 60 gives approximately 4.5 minutes. Therefore, rounding gives:

θ≈41∘5′\theta \approx 41^{\circ} 5^{\prime}

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