class 10 maths Introduction to Trigonometry questions with solutions

Q 1 :

The government plans to build ramp-like walls as boundary protection around a sensitive area. These walls are being constructed to reinforce the region’s boundary. The government has specified that the walls should have a height of 6 m above ground level with an inclination of approximately $30 ° .$

Based on the given information, answer the questions given below:

(i) Find the distance of DF.

12 m
11 m
13 m
14 m

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(1)

$I n △ D E F, w e h a v e s i n 30 ° = \frac{E D}{D F} \Rightarrow \frac{1}{2} = \frac{6}{D F} [∵ E D = 6 m] \Rightarrow D F = 12 m$

Q 2 :

The government plans to build ramp-like walls as boundary protection around a sensitive area. These walls are being constructed to reinforce the region’s boundary. The government has specified that the walls should have a height of 6 m above ground level with an inclination of approximately $30 degree.$

Based on the given information, answer the questions given below:

(ii) If angle is increased to $60 °$ , then what should be the value of DE to keep the distance between E and F same as before?

14 m
18 m
16 m
11 m

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(2)

$L e t D' b e t h e n e w p o s i t i o n o f D w h e n ∠ E F D = 60 ° . I n △ D E F, w e h a v e t a n 30 ° = \frac{E D}{E F} \Rightarrow \frac{1}{\sqrt{3}} = \frac{6}{E F} [∵ E D = 6 m] \Rightarrow E F = 6 \sqrt{3} m In △ D' EF, we have \tan 60 ° = \frac{D' E}{EF} \Rightarrow \sqrt{3} = \frac{D' E}{6 \sqrt{3}} \Rightarrow D' E = 6 \sqrt{3} \times \sqrt{3} = 18 m$

This should be the new height of the wall to keep the distance EF same as before and to increase the angle to $60 ° .$

Q 3 :

The government plans to build ramp-like walls as boundary protection around a sensitive area. These walls are being constructed to reinforce the region’s boundary. The government has specified that the walls should have a height of 6 m above ground level with an inclination of approximately $30 degree.$

Based on the given information, answer the questions given below:

(iii) (a) In the figure, what is the value of sin D · sin E?

$\frac{\sqrt{3}}{2}$
$\frac{\sqrt{- 3}}{2}$
$\frac{\sqrt{3}}{1}$
$\frac{\sqrt{3}}{4}$

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(1)

$s i n D = \frac{E F}{D F} = \frac{6 \sqrt{3}}{12} = \frac{\sqrt{3}}{2} a n d s i n E = 1 (as ∠ E = 90 °) N o w, s i n D \cdot s i n E = (\frac{\sqrt{3}}{2}) \cdot 1 = \frac{\sqrt{3}}{2}$

Q 4 :

The government plans to build ramp-like walls as boundary protection around a sensitive area. These walls are being constructed to reinforce the region’s boundary. The government has specified that the walls should have a height of 6 m above ground level with an inclination of approximately $30 degree.$

Based on the given information, answer the questions given below:

(iii) (b) In the figure, if DF + FE = 25 m and DE = 5 m, then what is the value of FE?

11 m
10 m
12 m
14 m

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(3)

$L e t F E = x m T h e n D F = (25 - x) m a n d D E = 5 m U \sin g P y t h a g o r a s t h e o r e m : D F^{2} = F E^{2} + D E^{2} \Rightarrow (25 - x)^{2} = x^{2} + 5^{2} 625 + x^{2} - 50 x = x^{2} + 25 600 = 50 x \Rightarrow x = 12 T h u s, t h e v a l u e o f F E i s 12 m$

Q 5 :

Riya and her sister visited their friend’s farmhouse. Upon arrival, Riya noticed that the roof of her friend’s house was shaped like a trapezium, with the dimensions indicated in the figure.

On the basis of above information, answer the following questions:

(i) What is the height of roof i.e., DE?

1 m
5 m
3 m
2 m

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(3)

$G i v e n, A B C D i s t r a p e z i u m w i t h A B ∥ C D . C D = 8 m, ∠ D A E = 30 ° a n d A D = B C = 6 m W e h a v e t o f i n d t h e h e i g h t o f r o o f i . e ., h e i g h t o f t r a p e z i u m (D E o r C F) . I n △ A E D, r i g h t a n g l e d a t E, w e h a v e i n ∠ D A E = \frac{D E}{A D} \Rightarrow s i n 30 ° = \frac{D E}{6} [∵ ∠ D A E = 30 ° and A D = 6 m] \Rightarrow \frac{1}{2} = \frac{D E}{6} \Rightarrow D E = 3 m Hence, height of roof is 3 m .$

Q 6 :

Riya and her sister visited their friend’s farmhouse. Upon arrival, Riya noticed that the roof of her friend’s house was shaped like a trapezium, with the dimensions indicated in the figure.

On the basis of above information, answer the following questions:

(ii) Find the measurement of ∠B.

$30 °$
$20 °$
$40 °$
$25 °$

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(1)

We have to find ∠B.

Since DE and CF are distances between two parallel lines AB and CD,

$D E = C F I n △ A D E a n d △ B C F : ∠ E = ∠ F = 90 ° A D = B C [F r o m F i g . 8.39] D E = C F [F r o m F i g . 8.39] ∴ △ A D E ≅ △ B C F [B y R H S C o n g r u e n c y] \Rightarrow ∠ A = ∠ B = 30 °$

Q 7 :

Riya and her sister visited their friend’s farmhouse. Upon arrival, Riya noticed that the roof of her friend’s house was shaped like a trapezium, with the dimensions indicated in the figure.

On the basis of above information, answer the following questions:

(iii) (a) What is the length of larger side AB of roof?

$(U s e \sqrt 3 = 1.73)$

17.35 m
15.5 m
13.5 m
18.38 m

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(4)

We have to find the length of larger side of roof i.e., AB.

$I n △ A E D, r i g h t a n g l e d a t E, w e h a v e c o s ∠ D A E = \frac{A E}{A D} \Rightarrow c o s 30 ° = \frac{A E}{6} [∵ ∠ D A E = 30 ° and A D = 6 m] \Rightarrow \frac{\sqrt{3}}{2} = \frac{A E}{6} \Rightarrow A E = 3 \sqrt{3} m ∵ c o s 30 \circ = 32 A l s o, △ A D E ≅ △ B C F \Rightarrow A E = B F = 3 \sqrt{3} m Now, AB = AE + EF + BF = 3 \sqrt{3} + 8 + 3 \sqrt{3} [∵ DEFC is a rectangle \Rightarrow CD = EF = 8 m] = 6 \sqrt{3} + 8 = 6 \times 1.73 + 8 = 18.38 m$

Hence, length of larger side of roof is 18.38 m

Q 8 :

Riya and her sister visited their friend’s farmhouse. Upon arrival, Riya noticed that the roof of her friend’s house was shaped like a trapezium, with the dimensions indicated in the figure.

On the basis of above information, answer the following questions:

(iii) (b) $F i n d t h e v a l u e o f \frac{{s i n}^{2} A + {s i n}^{2} D}{s i n A \sec D}$

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5
4

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(1)

We have,

$\frac{{s i n}^{2} A + {s i n}^{2} D}{s i n A \sec D} = \frac{{s i n}^{2} 30 ° + {s i n}^{2} 60 °}{s i n 30 ° s e c 60 °} [∵ ∠ A = 30 ° \Rightarrow ∠ D = 60 °] = \frac{{(\frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}{\frac{1}{2} \times 2} = \frac{\frac{1}{4} + \frac{3}{4}}{1} = \frac{1}{1} = 1$

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