Class 10 NCERT解决方案-第8章三角学入门–练习8.1

问题1.在∆ ABC中，在B处成直角，AB = 24厘米，BC = 7厘米。决定 ：

(i)罪A，cos A(ii)罪C，cos C

解决方案：

Using Pythagoras theorem for ΔABC

AC² = AB² + BC²

= (24 cm)² + (7 cm)²

= (576 + 49) cm²

= 625 cm²

∴AC = 25 cm       

(i) sin A = opp/hyp

    sin A = 7/25

    cos A = adj/hyp = 24/25

    cos A = 24/25

(ii) sin C = opp/hyp

     sin C = 24/25

     cos C = adj/hyp

     cos C = 7/25

问题2。在图8.13中，找到tan P – cotR。

解决方案：

Applying Pythagoras theorem for ΔPQR, we obtain

PR² = PQ² + QR²

(13 cm)² = (12 cm)² + QR²

169 cm² = 144 cm² + QR²

25 cm² = QR²

QR = 5 cm

tan P = opp/adj

tan P = 5/12

cot R = adj/opp

cot R = 5/12

tan P – cot R = 5/12 – 5/12 = 0

问题3.如果sin A = 3/4，则计算cos A和tanA。

解决方案：

Using sin²A + cos²A = 1

(3/4)² + cos²A = 1

cos²A = 1 – (3/4)² = 1 – 9/16

cos A = 7^1/2/4

tan A = sin A/cos A

tan A = (3/4)/(7^1/2/4)

tan A = 3/7^1/2

问题4：给定15 cot A =8。求正A和sec A

解决方案：

Given, 15 cot A = 8

               cot A = 8/15

tan A = 1/cot A

tan A = 15/8

Using, 1 + tan²A = sec²A

           1 + (15/8)² = sec²A

           289/64 = sec²A

           sec A = 17/8 

We know, cos²A = 1/sec²A

                cos²A = 64/289

               sin²A = 1 – cos2A

               sin²A = 225/289

              sin A = 15/17

问题5：给定秒θ = 13/12，计算所有其他三角比。

解决方案。

Using Pythagoras theorem,

sin θ = 5/13

cos θ = 12/13

tan θ = 5/12

cosec θ = 13/5

cot θ = 12/5

问题6：如果∠A和∠B是锐角，使得cos A = cos B，则证明∠A=∠B。

解决方案：

让我们考虑其中CD⊥AB的ΔABC。

假设cos A = cos B

AD / AC = BD / BC…(1)

我们需要证明∠A=∠B。为了证明这一点，我们需要将AC扩展到P，使得BC = CP。

From equation (1), we obtain

AD/BD = AC/BC

AD/BD = AC/CP    (BC = CP by construction)

By using the converse of B.P.T (Basic Proportionality Theorem),

CD||BP

∠ACD = ∠CPB (Corresponding angles) … (3)

And, ∠BCD = ∠CBP (Alternate interior angles) … (4)

By construction, we have BC = CP.

∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)

From equations (3), (4), and (5), we obtain

∠ACD = ∠BCD … (6)

In ΔCAD and ΔCBD,

∠ACD = ∠BCD (Using equation (6))

∠CDA = ∠CDB (Both 90°)

Therefore, the remaining angles should be equal.

∴∠CAD = ∠CBD

⇒ ∠A = ∠B

问题7：如果cotθ = 7/8，则求值

(i)(((1 + sinθ )*(1 –sinθ))/(1 +cosθ)*(1-cosθ )))

(ii)婴儿床^2θ

解决方案：

(i) Using (a + b) * (a – b) = a² – b² in numerator and denominator

We get

(1 – sin²θ)/(1 – cos²θ)

Using sin²θ + cos²θ = 1

We get

cos²θ/sin²θ = cot²θ

Now

cot²θ = (7/8)2 = 49/64

(ii) cot²θ = (7/8)2 = 49/64

问题8.如果3 cot A = 4，请检查(1 – tan ² A)/(1 + tan ² A)= cos ² A – sin ² A

解决方案。

We know that, tanA = sinA / cosA   ….(1)

Using (1) on L.H.S

= (1 – sin²A/cos²A)/(1 + sin²A/cos²A)

which on rearranging becomes

= (cos²A – sin²A)/(cos²A + sin²A)

Using the identity,

cos²A + sin²A = 1

LHS becomes

= (cos²A – sin²A)

This is equal to RHS.

LHS = RHS (for every value of cot A)

Hence, Proved.

问题9：在ΔABC中，与B成直角。如果tan A = 1 /(3 ^1/2 )，则求出

(i)罪A cos C +罪A A sin C

(ii)cos A cos C − sin A sin C

解决方案：

Using Pythagoras theorem

(AB)² + (BC)² = (AC)²

(31/2)² + (1)² = (AC)²

which gives

AC = 2 cm

Using formulas

sin A = 1/2

sin C = 3^1/2/2

cos A = 3^1/2/2

cos C = 1/2

Now, (i) sin A cos C + cos A sin C

Substituting the values

= (1/2) * (1/2) + (3^1/2/2) * (3^1/2/2)

= 1/4 + 3/4

= 1

Now, (ii) cos A cos C − sin A sin C

Substituting the values

= (3^1/2/2) * (1/2) – (1/2) * (3^1/2/2)

= 3^1/2/4 – 3^1/2/4

= 0

问题10：在ΔPQR中，与Q成直角，PR + QR = 25 cm，PQ = 5 cm。确定sin P，cos P和tan P的值。

解决方案：

Given that, PR + QR = 25

PQ = 5

Let PR be x cm.

Therefore, QR = 25 − x cm

Applying Pythagoras theorem in ΔPQR, we obtain

PR² = PQ² + QR²

x² = (5)² + (25 − x)²

x² = 25 + 625 + x² − 50x

50x = 650

x = 13

Therefore, PR = 13 cm

QR = (25 − 13) cm = 12 cm

Now,

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

问题11。说明以下是对还是错。证明你的答案。

(i)tan P的值始终小于1。

(ii)对于某个角度A值，sec A = 12/5。

(iii)cos A是角度A的余割线的缩写。

(iv)婴儿床A是婴儿床和A的乘积

(v)对于某个角度θ，sinθ= 4/3

解决方案：

(i) Consider a ΔPQR, right-angled at Q as shown below.

Here tan P = 12/5 which is surely greater than 1.

Therefore, the statement is false.

(ii) Consider ΔABC with AB = 5 cm, AC = 12 cm and BC = x cm

Using Pythagoras theorem in ΔABC

(AB)² + (BC)2 = (AC)2

52 + x2 = 122

x = (144 – 25)1/2

x = (119)1/2

x = 10.9 cm

AB < BC < AC

So this triangle is valid,

Therefore, given statement is true.

(iii) Abbreviation used for cosecant A is cosec A. And cos A is the abbreviation used for cosine A.

Hence, the given statement is false.

(iv) cot A is not the product of cot and A. It is the cotangent of ∠A.

Hence, the given statement is false.

(v) sin θ = 4/3 

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ is not possible.

Hence, the given statement is false