In the given right angle triangle- if Sin - -3-5- then find the value of 3 tan- nbsp- 5-1-4-3-

The correct option is C 4 In the given figure.Given, sin θ= 3/5 = AB/AC AC^2 = AB^2 + BC^2 5^2 = 3^2 + BC^2 BC = 4 Hence, 3 tanα nbsp;= nbsp;3×BC/AB= 4/3 ...

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Byju's Answer

Standard IX

Mathematics

Right Angled Triangle

In the given ...

Question

In the given right angle triangle, if $S i n θ = \frac{3}{5}$ , then find the value of $3 t a n α$ .

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Solution

The correct option is A
4

In the given figure.Given,
$s i n θ = \frac{3}{5} = \frac{A B}{A C}$
${A C}^{2} = {A B}^{2} + {B C}^{2}$
$5^{2} = 3^{2} + {B C}^{2}$
BC = 4

Hence,
$3 t a n α = 3 \times \frac{B C}{A B} = \frac{4}{3}$
$3 t a n α$ = 4

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In the given right angle triangle, if Sinθ =35, then find the value of 3tanα.

The correct option is A 4 In the given figure.Given, sin θ=35=ABAC AC2=AB2+BC2 52=32+BC2 BC = 4 Hence, 3tanα=3×BCAB=43 3tanα = 4

In the given right angle triangle, if $S i n θ = \frac{3}{5}$ , then find the value of $3 t a n α$ .

The correct option is A
4

In the given figure.Given,
$s i n θ = \frac{3}{5} = \frac{A B}{A C}$
${A C}^{2} = {A B}^{2} + {B C}^{2}$
$5^{2} = 3^{2} + {B C}^{2}$
BC = 4

Hence,
$3 t a n α = 3 \times \frac{B C}{A B} = \frac{4}{3}$
$3 t a n α$ = 4