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In the diagram below, AACD is right-angled at C.I perpendicular to AD. Write down the ratio of: 4.1.1 cos D in AACD. 4.1.2 sin D in ACED.

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here's the solution for Question 4:

Question 4.1
In the diagram, △ACD\triangle ACD is right-angled at C, and CE⊥ADCE \perp AD.

4.1.1 Write down the ratio of cos⁡D\cos D in △ACD\triangle ACD.
In △ACD\triangle ACD, the angle at C is 90∘90^\circ.
The side adjacent to angle D is CD.
The hypotenuse is AD.

cos⁡D=adjacenthypotenuse=CDAD\cos D = \frac{\text{adjacent}}{\text{hypotenuse}} = \boxed{\frac{CD}{AD}}

4.1.2 Write down the ratio of sin⁡D\sin D in △CED\triangle CED.
In △CED\triangle CED, the angle at E is 90∘90^\circ.
The side opposite to angle D is CE.
The hypotenuse is CD.

sin⁡D=oppositehypotenuse=CECD\sin D = \frac{\text{opposite}}{\text{hypotenuse}} = \boxed{\frac{CE}{CD}}

Question 4.2
Given 5sin⁡θ−4=05 \sin \theta - 4 = 0 and 90∘<θ<270∘90^\circ < \theta < 270^\circ.

4.2.1 Determine with the aid of a diagram and without using a calculator the value of tan⁡θ\tan \theta.

  1. Solve for sin⁡θ\sin \theta.

    5sin⁡θ=45 \sin \theta = 4 sin⁡θ=45\sin \theta = \frac{4}{5}

  2. Determine the quadrant.

    Since sin⁡θ>0\sin \theta > 0 and 90∘<θ<270∘90^\circ < \theta < 270^\circ, θ\theta must be in the second quadrant.

  3. Draw a right-angled triangle in the second quadrant.

    Let y=4y=4 and r=5r=5.
    Use Pythagoras theorem to find xx: x2+y2=r2x^2 + y^2 = r^2.

    x2+42=52x^2 + 4^2 = 5^2 x2+16=25x^2 + 16 = 25 x2=9x^2 = 9 x=±3x = \pm 3

    Since θ\theta is in the second quadrant, xx is negative. So, x=−3x = -3.

  4. Calculate tan⁡θ\tan \theta.

    tan⁡θ=yx=4−3\tan \theta = \frac{y}{x} = \frac{4}{-3} tan⁡θ=−43\tan \theta = \boxed{-\frac{4}{3}}

    4.2.2 Determine with the aid of a diagram and without using a calculator the value of 2cos⁡2θ−12 \cos^2 \theta - 1.

  5. From the diagram in 4.2.1, x=−3x=-3, y=4y=4, r=5r=5.

    Calculate cos⁡θ\cos \theta.

    cos⁡θ=xr=−35\cos \theta = \frac{x}{r} = \frac{-3}{5}

  6. Substitute cos⁡θ\cos \theta into the expression.

    2cos⁡2θ−1=2(−35)2−12 \cos^2 \theta - 1 = 2 \left(-\frac{3}{5}\right)^2 - 1 =2(925)−1= 2 \left(\frac{9}{25}\right) - 1 =1825−1= \frac{18}{25} - 1 =1825−2525= \frac{18}{25} - \frac{25}{25} =−725= \boxed{-\frac{7}{25}}

    Question 4.3
    Determine, WITHOUT the use of a calculator, the value of: cos⁡45∘+5cosec⁡90∘×3tan⁡230∘sin⁡45∘\frac{\cos 45^\circ + 5 \operatorname{cosec} 90^\circ \times 3 \tan^2 30^\circ}{\sin 45^\circ}

  7. Recall the special angle values.

    cos⁡45∘=22\cos 45^\circ = \frac{\sqrt{2}}{2} cosec⁡90∘=1sin⁡90∘=11=1\operatorname{cosec} 90^\circ = \frac{1}{\sin 90^\circ} = \frac{1}{1} = 1 tan⁡30∘=13\tan 30^\circ = \frac{1}{\sqrt{3}} sin⁡45∘=22\sin 45^\circ = \frac{\sqrt{2}}{2}

  8. Substitute the values into the expression.

    22+5(1)×3(13)222\frac{\frac{\sqrt{2}}{2} + 5(1) \times 3 \left(\frac{1}{\sqrt{3}}\right)^2}{\frac{\sqrt{2}}{2}} =22+5×3×1322= \frac{\frac{\sqrt{2}}{2} + 5 \times 3 \times \frac{1}{3}}{\frac{\sqrt{2}}{2}} =22+522= \frac{\frac{\sqrt{2}}{2} + 5}{\frac{\sqrt{2}}{2}}

  9. Separate the terms in the numerator.

    =2222+522= \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} + \frac{5}{\frac{\sqrt{2}}{2}} =1+5×22= 1 + 5 \times \frac{2}{\sqrt{2}} =1+102= 1 + \frac{10}{\sqrt{2}}

  10. Rationalize the denominator.

    =1+102×22= 1 + \frac{10}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} =1+1022= 1 + \frac{10\sqrt{2}}{2} =1+52= 1 + 5\sqrt{2} =1+52= \boxed{1 + 5\sqrt{2}}

    Question 4.4
    Solve xx, WITHOUT the use of a calculator: 2sin⁡(x−25∘)=12 \sin(x - 25^\circ) = 1.

  11. Isolate the sine term.

    sin⁡(x−25∘)=12\sin(x - 25^\circ) = \frac{1}{2}

  12. Find the reference angle.

    The angle whose sine is 12\frac{1}{2} is 30∘30^\circ.

  13. Determine the quadrants where sine is positive (Quadrant I and Quadrant II).

    Case 1: Quadrant I

    x−25∘=30∘+k⋅360∘x - 25^\circ = 30^\circ + k \cdot 360^\circ x=30∘+25∘+k⋅360∘x = 30^\circ + 25^\circ + k \cdot 360^\circ x=55∘+k⋅360∘x = 55^\circ + k \cdot 360^\circ

    Case 2: Quadrant II

    x−25∘=(180∘−30∘)+k⋅360∘x - 25^\circ = (180^\circ - 30^\circ) + k \cdot 360^\circ x−25∘=150∘+k⋅360∘x - 25^\circ = 150^\circ + k \cdot 360^\circ x=150∘+25∘+k⋅360∘x = 150^\circ + 25^\circ + k \cdot 360^\circ x=175∘+k⋅360∘x = 175^\circ + k \cdot 360^\circ

    where kk is an integer.

    x=55∘+k⋅360∘ or x=175∘+k⋅360∘, where k∈Zx = \boxed{55^\circ + k \cdot 360^\circ \text{ or } x = 175^\circ + k \cdot 360^\circ, \text{ where } k \in \mathbb{Z}}

    Question 4.5
    Given that A=26.2∘A = 26.2^\circ and B=132.4∘B = 132.4^\circ. Calculate the value of cosec⁡A+cos⁡2B\operatorname{cosec} A + \cos 2B.

  14. Substitute the values of A and B into the expression.

    cosec⁡A+cos⁡2B=1sin⁡A+cos⁡(2B)\operatorname{cosec} A + \cos 2B = \frac{1}{\sin A} + \cos(2B) =1sin⁡26.2∘+cos⁡(2×132.4∘)= \frac{1}{\sin 26.2^\circ} + \cos(2 \times 132.4^\circ) =1sin⁡26.2∘+cos⁡(264.8∘)= \frac{1}{\sin 26.2^\circ} + \cos(264.8^\circ)

  15. Calculate the values using a calculator.

    sin⁡26.2∘≈0.4415\sin 26.2^\circ \approx 0.4415 cos⁡264.8∘≈−0.0906\cos 264.8^\circ \approx -0.0906 =10.4415+(−0.0906)= \frac{1}{0.4415} + (-0.0906) ≈2.2649−0.0906\approx 2.2649 - 0.0906 ≈2.1743\approx 2.1743 =2.17= \boxed{2.17}

    Question 4.6
    Simplify the following expression if sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1: cos⁡2x+tan⁡x⋅cot⁡x+sin⁡2x\cos^2 x + \tan x \cdot \cot x + \sin^2 x.

  16. Rearrange the terms.

    (cos⁡2x+sin⁡2x)+tan⁡x⋅cot⁡x(\cos^2 x + \sin^2 x) + \tan x \cdot \cot x

  17. Apply the given identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

    1+tan⁡x⋅cot⁡x1 + \tan x \cdot \cot x

  18. Use the identity cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x}.

    1+tan⁡x⋅1tan⁡x1 + \tan x \cdot \frac{1}{\tan x} =1+1= 1 + 1 =2= \boxed{2}

    Question 4.7
    A hiker climbs 250 m at a slope of 10∘10^\circ and thereafter completes 45 m at a slope of 12∘12^\circ. How high is he above his starting point?

  19. Calculate the vertical height gained in the first part of the climb.

    Let h1h_1 be the height.

    sin⁡10∘=h1250 m\sin 10^\circ = \frac{h_1}{250 \text{ m}} h1=250sin⁡10∘h_1 = 250 \sin 10^\circ h1≈250×0.1736h_1 \approx 250 \times 0.1736 h1≈43.40 mh_1 \approx 43.40 \text{ m}

  20. Calculate the vertical height gained in the second part of the climb.

    Let h2h_2 be the height.

    sin⁡12∘=h245 m\sin 12^\circ = \frac{h_2}{45 \text{ m}} h2=45sin⁡12∘h_2 = 45 \sin 12^\circ h2≈45×0.2079h_2 \approx 45 \times 0.2079 h2≈9.35 mh_2 \approx 9.35 \text{ m}

  21. Calculate the total height above the starting point.

    Total height=h1+h2\text{Total height} = h_1 + h_2 Total height≈43.40 m+9.35 m\text{Total height} \approx 43.40 \text{ m} + 9.35 \text{ m} Total height≈52.75 m\text{Total height} \approx 52.75 \text{ m} Total height=52.75 m\text{Total height} = \boxed{52.75 \text{ m}}
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