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A right-angled triangle has sides a, b and c and the angle θ, as shown below. Write the following in terms of a, b and c: cos θ, tan θ, sin(90° - θ). If it is given that a = 5 and θ = 50°, calculate the numerical value of b. Hence, show that cos²θ + sin²θ = 1.

Mathematics

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Here's the solution for Question 4:

Question 4.1
From the given right-angled triangle:

  • Opposite side to θ\theta is aa.
  • Adjacent side to θ\theta is bb.
  • Hypotenuse is cc.

4.1.1 Write the following in terms of aa, bb, and cc:

4.1.1.1 cos⁡θ\cos \theta

cos⁡θ=adjacenthypotenuse=bc\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \boxed{\frac{b}{c}}

4.1.1.2 tan⁡θ\tan \theta

tan⁡θ=oppositeadjacent=ab\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \boxed{\frac{a}{b}}

4.1.1.3 sin⁡(90∘−θ)\sin(90^\circ - \theta)
The angle (90∘−θ)(90^\circ - \theta) is the angle at the top vertex.
The side opposite to (90∘−θ)(90^\circ - \theta) is bb.
The hypotenuse is cc.

sin⁡(90∘−θ)=oppositehypotenuse=bc\sin(90^\circ - \theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \boxed{\frac{b}{c}}

4.1.2 If it is given that a=5a=5 and θ=50∘\theta=50^\circ, calculate:

4.1.2.1 The numerical value of bb.

  1. Use the tangent ratio from 4.1.1.2.

    tan⁡θ=ab\tan \theta = \frac{a}{b}

  2. Substitute the given values a=5a=5 and θ=50∘\theta=50^\circ.

    tan⁡50∘=5b\tan 50^\circ = \frac{5}{b}

  3. Solve for bb.

    b=5tan⁡50∘b = \frac{5}{\tan 50^\circ} b≈51.19175b \approx \frac{5}{1.19175} b≈4.195b \approx \boxed{4.195}

    4.1.2.2 Hence, show that cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1.

  4. Calculate cc using sin⁡θ=ac\sin \theta = \frac{a}{c}.

    sin⁡50∘=5c\sin 50^\circ = \frac{5}{c} c=5sin⁡50∘c = \frac{5}{\sin 50^\circ} c≈50.76604c \approx \frac{5}{0.76604} c≈6.527c \approx 6.527

  5. Calculate sin⁡θ\sin \theta and cos⁡θ\cos \theta using the values of a,b,ca, b, c.

    sin⁡θ=ac=56.527≈0.7660\sin \theta = \frac{a}{c} = \frac{5}{6.527} \approx 0.7660 cos⁡θ=bc=4.1956.527≈0.6427\cos \theta = \frac{b}{c} = \frac{4.195}{6.527} \approx 0.6427

  6. Substitute these values into the identity.

    cos⁡2θ+sin⁡2θ=(0.6427)2+(0.7660)2\cos^2 \theta + \sin^2 \theta = (0.6427)^2 + (0.7660)^2 ≈0.41306+0.58676\approx 0.41306 + 0.58676 ≈0.99982\approx 0.99982

    Rounding to 1:

    cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = \boxed{1}

    (Note: The slight deviation from exactly 1 is due to rounding intermediate values. Using the exact values from the calculator for sin⁡50∘\sin 50^\circ and cos⁡50∘\cos 50^\circ would yield exactly 1.)

    Question 4.2
    Calculate the value of 2cosec⁡A+cos⁡3B2 \operatorname{cosec} A + \cos 3B, if A^=38.2∘\hat{A}=38.2^\circ and B^=146.4∘\hat{B}=146.4^\circ.

  7. Recall that cosec⁡A=1sin⁡A\operatorname{cosec} A = \frac{1}{\sin A}.

    2cosec⁡A+cos⁡3B=2(1sin⁡A)+cos⁡(3B)2 \operatorname{cosec} A + \cos 3B = 2 \left(\frac{1}{\sin A}\right) + \cos(3B)

  8. Substitute the given values for AA and BB.

    =2(1sin⁡38.2∘)+cos⁡(3×146.4∘)= 2 \left(\frac{1}{\sin 38.2^\circ}\right) + \cos(3 \times 146.4^\circ) =2(1sin⁡38.2∘)+cos⁡(439.2∘)= 2 \left(\frac{1}{\sin 38.2^\circ}\right) + \cos(439.2^\circ)

  9. Calculate the values.

    sin⁡38.2∘≈0.6184\sin 38.2^\circ \approx 0.6184 cos⁡(439.2∘)=cos⁡(439.2∘−360∘)=cos⁡(79.2∘)≈0.1874\cos(439.2^\circ) = \cos(439.2^\circ - 360^\circ) = \cos(79.2^\circ) \approx 0.1874 =2(10.6184)+0.1874= 2 \left(\frac{1}{0.6184}\right) + 0.1874 =2(1.61707)+0.1874= 2(1.61707) + 0.1874 =3.23414+0.1874= 3.23414 + 0.1874 =3.42= \boxed{3.42}

    Question 4.3
    Simplify sin⁡45∘⋅tan⁡260∘cos⁡45∘\frac{\sin 45^\circ \cdot \tan^2 60^\circ}{\cos 45^\circ} WITHOUT the use of a calculator.

  10. Recall the special angle values.

    sin⁡45∘=22\sin 45^\circ = \frac{\sqrt{2}}{2} tan⁡60∘=3\tan 60^\circ = \sqrt{3} cos⁡45∘=22\cos 45^\circ = \frac{\sqrt{2}}{2}

  11. Substitute the values into the expression.

    sin⁡45∘⋅tan⁡260∘cos⁡45∘=22⋅(3)222\frac{\sin 45^\circ \cdot \tan^2 60^\circ}{\cos 45^\circ} = \frac{\frac{\sqrt{2}}{2} \cdot (\sqrt{3})^2}{\frac{\sqrt{2}}{2}}

  12. Simplify the expression.

    =22⋅322= \frac{\frac{\sqrt{2}}{2} \cdot 3}{\frac{\sqrt{2}}{2}} =3= 3 =3= \boxed{3}

    Question 4.4
    Given that 5cos⁡θ−4=05 \cos \theta - 4 = 0, where 0∘<θ<90∘0^\circ < \theta < 90^\circ; and α+θ=90∘\alpha + \theta = 90^\circ and 0∘<α<90∘0^\circ < \alpha < 90^\circ, calculate the value of cot⁡α\cot \alpha.

  13. From 5cos⁡θ−4=05 \cos \theta - 4 = 0, find cos⁡θ\cos \theta.

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

  14. Use the identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find sin⁡θ\sin \theta.

    Since 0∘<θ<90∘0^\circ < \theta < 90^\circ, sin⁡θ\sin \theta is positive.

    sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta sin⁡2θ=1−(45)2\sin^2 \theta = 1 - \left(\frac{4}{5}\right)^2 sin⁡2θ=1−1625\sin^2 \theta = 1 - \frac{16}{25} sin⁡2θ=25−1625=925\sin^2 \theta = \frac{25 - 16}{25} = \frac{9}{25} sin⁡θ=925=35\sin \theta = \sqrt{\frac{9}{25}} = \frac{3}{5}

  15. Use the relationship α+θ=90∘\alpha + \theta = 90^\circ to express α\alpha.

    α=90∘−θ\alpha = 90^\circ - \theta

  16. Calculate cot⁡α\cot \alpha.

    cot⁡α=cot⁡(90∘−θ)\cot \alpha = \cot(90^\circ - \theta)

  17. Use the co-function identity cot⁡(90∘−θ)=tan⁡θ\cot(90^\circ - \theta) = \tan \theta.

    cot⁡α=tan⁡θ\cot \alpha = \tan \theta

  18. Calculate tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}.

    tan⁡θ=3545=34\tan \theta = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}

    Therefore,

    cot⁡α=34\cot \alpha = \boxed{\frac{3}{4}}
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