Question
If $\sin A = \frac { 3 } { 4 }$, calculate cos A and tan A.
✓
Answer
Given: A triangle ABC in which $\angle B = 90 ^ { \circ }$
$SinA=\frac34=\frac PH$
Let BC = 3k and AC = 4k where k is a positive integer.|
Using Pythagoras theorem,
$AB^2=AC^2-BC^2$
$A B = \sqrt { ( A C ) ^ { 2 } - ( B C ) ^ { 2 } } = \sqrt { ( 4 k ) ^ { 2 } - ( 3 k ) ^ { 2 } }$
$= \sqrt { 16 k ^ { 2 } - 9 k ^ { 2 } } =\sqrt{7k^2}= k \sqrt { 7 }$
Therefore,$$ $\cos A = \frac { B } { H } = \frac { A B } { A C } = \frac { k \sqrt { 7 } } { 4 k } = \frac { \sqrt { 7 } } { 4 }$
$\tan \mathrm { A } = \frac { \mathrm { P } } { \mathrm { B } } = \frac { \mathrm { BC } } { \mathrm { AB } } = \frac { 3 k } { k \sqrt { 7 } } = \frac { 3 } { \sqrt { 7 } }$
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