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Chapter 24- Solution Of Right Triangles | Exercise 24

Question 19

\text { If } \tan x^{\circ}=\frac{5}{12}, \tan y^{\circ}=\frac{3}{4} and AB = 48 m; find the length of CD.

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\text { Given } \tan x^{\circ}=\frac{5}{12}, \tan y^{\circ}=\frac{3}{4} and AB = 48 m;

Let length of BC = xm

From triangle ADC

\begin{aligned} \tan x^{\circ} &=\frac{D C}{A C} \\ \frac{5}{12} &=\frac{D C}{48+x} \\ 240+5 x &=12 C D.....(1) \end{aligned}

From triangle BDC

\begin{aligned} \tan y^{\circ} &=\frac{C D}{B C} \\ \frac{3}{4} &=\frac{C D}{x} \\ x &=\frac{4 C D}{3}.....(2) \end{aligned}

From (1)

\begin{aligned} 240+5\left(\frac{4 C D}{3}\right) &=12 C D \\ 240+\frac{20 C D}{3} &=12 C D \\ 720+20 C D &=36 C D \\ 16 C D &=720 \\ C D &=45 \end{aligned}

Length of CD is 45 m.

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Question 19

\text { If } \tan x^{\circ}=\frac{5}{12}, \tan y^{\circ}=\frac{3}{4} and AB = 48 m; find the length of CD.

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

\text { Given } \tan x^{\circ}=\frac{5}{12}, \tan y^{\circ}=\frac{3}{4} and AB = 48 m;

Let length of BC = xm

From triangle ADC

\begin{aligned} \tan x^{\circ} &=\frac{D C}{A C} \\ \frac{5}{12} &=\frac{D C}{48+x} \\ 240+5 x &=12 C D.....(1) \end{aligned}

From triangle BDC

\begin{aligned} \tan y^{\circ} &=\frac{C D}{B C} \\ \frac{3}{4} &=\frac{C D}{x} \\ x &=\frac{4 C D}{3}.....(2) \end{aligned}

From (1)

\begin{aligned} 240+5\left(\frac{4 C D}{3}\right) &=12 C D \\ 240+\frac{20 C D}{3} &=12 C D \\ 720+20 C D &=36 C D \\ 16 C D &=720 \\ C D &=45 \end{aligned}

Length of CD is 45 m.

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

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