Question

In triangle ABC ,right angled at B, AB = 4 cm, and BC = 3 cm. Find the values of sin A,sin C, sec A, and sec C
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Answer 1

Step-by-step explanation:

AC^2=BC^2+AB^2

=3^2+4^2

=9+16

=25

squaring both sides

√AC^2=√5^2

square and root get cancelled

so,AC=5

sinA=opp/hyp

=3/5

sinc=4/5

secA=hyp/adj

=5/3

secC=5/4

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Answer 2

The values of sin A,sin C, sec A, and sec C are 3/5, 4/5, 5/4 and 5/3 respectively.

Step-by-step explanation:

Given information: In triangle ABC ,right angled at B, AB = 4 cm, and BC = 3 cm.

Using Pythagoras theorem,

$hypotenuse^2=base^2+perpendicular^2$

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

$AC^2=(4)^2+(3)^2$

$AC^2=25$

Taking square root on both sides.

$AC=5$

Using properties of trigonometry we get

$\sin A=\dfrac{Opposite}{Hypotenuse}=\dfrac{BC}{AC}=\dfrac{3}{5}$

$\sin C=\dfrac{Opposite}{Hypotenuse}=\dfrac{AB}{AC}=\dfrac{4}{5}$

$\sec A=\dfrac{Hypotenuse}{adjacent}=\dfrac{AC}{AB}=\dfrac{5}{4}$

$\sec C=\dfrac{Hypotenuse}{adjacent}=\dfrac{AC}{AC}=\dfrac{5}{3}$

Therefore, the values of sin A,sin C, sec A, and sec C are 3/5, 4/5, 5/4 and 5/3 respectively.

#Learn more

In triangle ABC, right-angled at B, if AB=5, BC=12 and AC=13, find all the six trigonometric ratios of angle A​.

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