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Chapter 24- Solution Of Right Triangles | Exercise 24

Question 1
  1. Find 'x' if:

(i)

(ii)

(iii)

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(i)

\begin{aligned} \sin 60^{\circ} &=\frac{20}{x} \\ \frac{\sqrt{3}}{2} &=\frac{20}{x} \\ x &=\frac{40}{\sqrt{3}} \end{aligned}

(ii)

\begin{aligned} \tan 30^{\circ} &=\frac{20}{x} \\ \frac{1}{\sqrt{3}} &=\frac{20}{x} \\ x &=20 \sqrt{3} \end{aligned}

(iii)

\begin{aligned} \sin 45^{\circ} &=\frac{20}{x} \\ \frac{1}{\sqrt{2}} &=\frac{20}{x} \\ x &=20 \sqrt{2} \end{aligned}

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Our top 5% students will be awarded a special scholarship to Lido.

subject-cta

Question 1

  1. Find 'x' if:

(i)

(ii)

(iii)

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

(i)

\begin{aligned} \sin 60^{\circ} &=\frac{20}{x} \\ \frac{\sqrt{3}}{2} &=\frac{20}{x} \\ x &=\frac{40}{\sqrt{3}} \end{aligned}

(ii)

\begin{aligned} \tan 30^{\circ} &=\frac{20}{x} \\ \frac{1}{\sqrt{3}} &=\frac{20}{x} \\ x &=20 \sqrt{3} \end{aligned}

(iii)

\begin{aligned} \sin 45^{\circ} &=\frac{20}{x} \\ \frac{1}{\sqrt{2}} &=\frac{20}{x} \\ x &=20 \sqrt{2} \end{aligned}

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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