If 5A=cosec (A-36^ ) and 5A is an acute angle, show that A=21^ . - Vidyadip

Question

If $5\text{A}=\text{cosec }(\text{A}-36^\circ)$ and $5\text{A}$ is an acute angle, show that $\text{A}=21^\circ.$

✓

Answer

Given: $\sec\text{5A}=\text{cosec}(\text{A}-36^\circ)$$\Rightarrow\text{cosec}(90^\circ-\text{5A})=\text{cosec}(\text{A}-36^\circ)$ $\big[\because\text{cosec}(90^\circ-\theta)=\sec\theta\big]$
$\Rightarrow90^\circ-\text{5A}=\text{A}-36^\circ$
$\Rightarrow\text{6A}=90^\circ+36^\circ$
$\Rightarrow\text{6A}=126^\circ$
$\Rightarrow\text{A}=21^\circ$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Trigonometric Identities→Trigonometric Identities — all question types→Maths STD 10 question papers→CBSE Board STD 10 question papers→CBSE Board question paper generator→

Similar questions

A shopkeeper buys a number of books for $₹ 1,800$ . If he had bought $15$ more books for the same amount, then each book would have cost him $₹ 20$ less. Find how many books he bought initially.

If A(-3, 5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.

Solve for x and y:
$\frac{5}{\text{x}}-\frac{3}{\text{y}}=1,$
$\frac{3}{\text{2x}}+\frac{2}{\text{3y}}=\text{5}$ $(\text{x}\neq0,\ \text{y}\neq0).$

Solve the following quadratic equations by factorization:
$\frac{2\text{x}}{\text{x}-4}+\frac{2\text{x}-5}{\text{x}-3}=\frac{25}{3}$

Solve the following system of equations graphically:
2x + 3y = 8,
x - 2y + 3 = 0

Find the roots of the following equation, if they exist, by applying the quadratic formula:
$4x^2 - 4a^2x + (a^4 - b^4) = 0$

Solve the following equations by using the method of completing the square:
$\text{x}^2-\big(\sqrt2+1\big)\text{x}+\sqrt2=0$

The sides of certain triangles are given below. Determine them are right triangles:
7cm, 24cm, 25cm.

In an A.P., if the $5^{\text {th }}$ and $12^{\text {th }}$ terms are 30 and 65 respectively, what is the sum of first 20 terms?

Prove that the points (2a, 4a), (2a, 6a) and $(2\text{a}+\sqrt{3}\text{a}, 5\text{a})$ are the vertices of an equilateral triangle.