Question

If sinA = 3/4 calculate cos A and tan A.

Answer 1

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Answer 2

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$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\put(7.7,2.9){\large{A}}\put(7.7,1){\large{B}}\put(10.6,1){\large{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\put(10.5,1){\line(-4,3){2.5}}\put(9,0.7){\sf{\large{3k}}}\put(9.4,1.9){\sf{\large{4k}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

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$\rm\:sinA=\dfrac{p}{h}=\dfrac{BC}{AC}=\dfrac{3}{4}$

$\rm\:Let\:BC=3k$

$\rm\:and\:AC=4k$

Now,

$\rm\:In\:right\:{\triangle\:ABC}$

$\rm\:AB=\sqrt{{(AC)}^{2}-{(BC)}^{2}}$

$\rm\:AB=\sqrt{{(4k)}^{2}-{(3k)}^{2}}$

$\rm\:AB=\sqrt{{16k}^{2}-{9k}^{2}}$

$\rm\:AB=\sqrt{{7k}^{2}}$

$\rm\:AB=\sqrt{7}\:k$

Then,

$\rm\:cosA=\dfrac{b}{h}=\dfrac{AB}{AC}=\dfrac{\sqrt{7}\cancel{k}}{4\cancel{k}}$

$\rm\:cosA=\dfrac{\sqrt{7}}{4}$

$\rm\:tanA=\dfrac{p}{b}=\dfrac{BC}{AB}=\dfrac{3\cancel{k}}{\sqrt{7}\cancel{k}}$

$\rm\:tanA=\dfrac{3}{\sqrt{7}}$