Question

given 15 cot A=8 find sin A and sec A​

Answer 1

Given

• 15 cot A = 8

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To Find

• sin A
• sec A

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Solution

⇒ 15 cot A = 8

⇒ cot A = $\dfrac{8}{15}$

Also,

⇒ cot A = $\dfrac{1}{tan \ A}$

⇒ cot A = $\dfrac{1}{\frac{CB}{AB}}$

⇒ cot A = $\bf \dfrac{AB}{CB}$

Hence, let's assume side AB to be '8x' and side CB to be '15x' since they are in the ratio of 8:15.

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Considering, ABC to be a right angled triangle, let's find it's hypotenuse.

⇒ AC² = AB² + BC²           (Pythagoras Theorem)

⇒ AC² = (8x)² + (15x)²

⇒ AC² = 64x² + 225x²

⇒ AC² = 289x²

⇒ AC = √289x²

⇒ AC = 17x

∴ AC is 17x.

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Now,

⇒ sin A = $\dfrac{BC}{AC}$

⇒ sin A = $\dfrac{15x}{17x}$

⇒ sin A = $\bf \dfrac{15}{17}$

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Also,

⇒ sec A = $\dfrac{1}{cos \ A}$

⇒ sec A = $\dfrac{1}{\frac{AB}{AC}}$

⇒ sec A = $\dfrac{AC}{AB}$

⇒ sec A = $\dfrac{17x}{8x}$

⇒ sec A = $\bf \dfrac{8}{5}$

∴ sin A = $\bf \dfrac{15}{17}$ and sec A = $\bf \dfrac{8}{5}$.

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