Question

solve the following plz it's urgent
plz​

Answer 1

3rd answer is 3/5

4th answer is 8/17

plz. check my bi.o here and support me

Answer 2

$\huge\fcolorbox{red}{lime}{Solution}$

$\sf \sin \theta =\dfrac{4}{5}$

We know that:-

$\boxed {cos^{2}\theta=1-sin^{2}\theta }$

$\begin{gathered}cos^{2}\theta = 1-\big(\frac{4}{5}\big)^{2}\\=1-\frac{16}{25}\\=\frac{25-16}{25}\\=\frac{9}{25}\end{gathered}$

$\begin{gathered}\implies cos\theta =\sqrt{\frac{9}{25}}\\=\frac{3}{5}\end{gathered}$

$\cos\theta = \dfrac{3}{5}$

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cos θ = 15/17 .. .(i) [Given]

In right angled ∆ABC, ∠C = θ.

Let the common multiple be k.

∴ BC = 15k and AC = 17k

$Now, AC^2 = AB^2 + BC^2$

[Pythagoras theorem]

$(17k)^2 = AB^2 + (15K)^2$

$289k^2 = AB^2 + 225^2$

$AB^2 = 289k^2 – 225k^2 = 64k^2$

$AB = \sqrt{(64k^2) }$

[Taking square root of both sides]

AB= 8k

$\sf \sin \theta=\dfrac{8k}{17k} \\ \sf =\dfrac{8}{17}$