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Introduction to Trigonometry
CH08 Question Bank
CBSE_CH08_QUESTION_BANK
Grade 10

Question:

[Case Study]

A carpenter uses a standard $30^\circ$-$60^\circ$-$90^\circ$ set-square (a common tool in a geometry box) to mark angles while cutting wood for a triangular shelf bracket.

(a) What is the value of $\sin 30^\circ + \sin 60^\circ$? [1 Mark]
(b) What is the value of $\cos 30^\circ \times \cos 60^\circ$? [1 Mark]
(c) The carpenter needs to check that $\tan 30^\circ \times \tan 60^\circ = 1$. Verify this. [1 Mark]
(d) Which pair of standard angles gives equal sine and cosine values? [1 Mark]

Step-by-Step Solution

Key Concept: Case study on trigonometry ratios and identities.
(a) What is the value of $\sin 30^\circ + \sin 60^\circ$? [1 Mark]
$\sin30^\circ+\sin60^\circ=\dfrac12+\dfrac{\sqrt3}{2}=\dfrac{1+\sqrt3}{2}$. [1.0 Mark]

(b) What is the value of $\cos 30^\circ \times \cos 60^\circ$? [1 Mark]
$\cos30^\circ\times\cos60^\circ=\dfrac{\sqrt3}{2}\times\dfrac12=\dfrac{\sqrt3}{4}$. [1.0 Mark]

(c) The carpenter needs to check that $\tan 30^\circ \times \tan 60^\circ = 1$. Verify this. [1 Mark]
$\tan30^\circ\times\tan60^\circ=\dfrac{1}{\sqrt3}\times\sqrt3=1$. Verified. [1.0 Mark]

(d) Which pair of standard angles gives equal sine and cosine values? [1 Mark]
At $45^\circ$, $\sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}$. [1.0 Mark]

Correct Answer: $\sin30^\circ+\sin60^\circ=\dfrac12+\dfrac{\sqrt3}{2}=\dfrac{1+\sqrt3}{2}$. [1.0 Mark] | $\cos30^\circ\times\cos60^\circ=\dfrac{\sqrt3}{2}\times\dfrac12=\dfrac{\sqrt3}{4}$. [1.0 Mark] | $\tan30^\circ\times\tan60^\circ=\dfrac{1}{\sqrt3}\times\sqrt3=1$. Verified. [1.0 Mark] | At $45^\circ$, $\sin45^\circ=\cos45^\circ=\dfrac{1}{\sqrt2}$. [1.0 Mark]
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