TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 10 Properties of Triangles Important Questions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Properties of Triangles Important Questions

I.

Question 1.
In ΔABC, If a=3,b=4 and sin A=3/4, find the angle B.
Solution :
By sine rule,\(\frac{a}{\sin A}=\frac{b}{\sin B}\)
⇒ sinB =\(\frac{b \sin A}{a}=\frac{4 \times 3 / 4}{3}\) = 1
⇒ sinB = 1 = B = 90°

Question 2.
If the lengths of the sides of a triangle are 3, 4, 5, find the circumradius of the triangle.
Solution:
Since 3²+4²= 5² the triangle is right angled and hypotenuse = 5 = circum diameter.
∴ Circum radius = \(\frac{5}{2}\) cm.

Question 3.
lf a=6,b=5,c=9, then find the angle A.
Solution:

Question 4.
In a Δ ABC, show that ∑(b +c)cosA = 2s
Solution:
LH.S. = ∑ (b + c) cos A
= (b + c) cos A+ (c + a) cos B + (a + b) cos C
= (bcosA+ acos B)+(ccosA + acosC) + (b cos C + c cos B)
= c+b+a
= a+b+c=2s=R.H.S.

Question 5.
In a Δ ABC, if
(a) (a+b+c)(b+c- a) = 3bc, findA.
Solution:
Given (a + b + c)(b + c-a) = 3bc
⇒ (2s) 2(s – a) = 3 bc

(b) If a = 4, b = 5, c = 7, find cos \(\frac{B}{2}\)
Solution:
2s = a+b+c = 4+5+7=16
⇒ s = 8
∴ s – b=8-5=3 and
\(\cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{a c}}=\sqrt{\frac{8 \times 3}{4 \times 7}}=\sqrt{\frac{6}{7}}\)

Question 6.
If ΔABC, find \(b \cos ^2 \frac{C}{2}+c \cos ^2 \frac{B}{2}\)
Solution:

Question 7.
If \(\tan \frac{\mathrm{A}}{2}=\frac{5}{6}\) and \(\tan \frac{\mathrm{C}}{2}=\frac{2}{5}\)
Solution:

⇒ 3s – 3b = s
⇒ 2s = 3b ⇒ a+b +c = 3b ⇒ a = 2b
⇒ a, b, c are in A.P.

Question 8.
If \(\cot \frac{A}{2}=\frac{b+c}{a}\),find angle B
Solution:
\(\cot \frac{A}{2}=\frac{b+c}{a}\)

Question 9.
If tan \(\left(\frac{\mathrm{C}-\mathrm{A}}{2}\right)=\mathrm{k} \cot \frac{\mathrm{B}}{2}\),find K
Solution:
Using Napiers rule,

Question 10
In ΔABC, Show that \(\frac{b^2-c^2}{a^2}=\frac{\sin (B-C)}{\sin (B+C)}\)
Solution:

Question 11.
Show that \((b-c)^2 \cos ^2 \frac{A}{2}+(b+c)^2 \sin ^2 \frac{A}{2}=a^2\)
Solution:
L.H.S= \((b-c)^2 \cos ^2 \frac{A}{2}+(b+c)^2 \sin ^2 \frac{A}{2}\)

Question 12.
If ΔABC, Prove that \(\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{r}\)
Solution:

Question 13.
Show that r r₁ r₂ r₃ = Δ²
Solution:
L.H.S. r. r₁ . r₂ . r₃

Question 14.
In an equilateral triangle, find the value of  \(\frac{\mathbf{r}}{\mathbf{R}}\)
Solution:

Question 15.
The perimeter of ΔABC is 12 cm and it’s in radius is 1 cm. Then find the area of the triangle.
Solution:
Given 2s = 12 ⇒ s = 6 cm. ; r = 1 cm
Area of Δ ABC,
Δ = rs = (1) (6) = 6sq.cm

Question 16.
Show that r r₁ = (s – b) (s – c).
Solution:
L.H.S. = r. r₁

Question 17.
In ΔABC, is = 6sq. cm and s = 1.5cm find ‘r’
Solution:
\(\mathrm{r}=\frac{\Delta}{\mathrm{s}}=\frac{6}{1.5}=\frac{6}{3 / 2}=4 \mathrm{~cm}\)

Question 18.
Show that \(r r_1 \cot \frac{A}{2}=\Delta\)
Solution:

II.

Question 1.
In a \(\Delta \mathrm{ABC}\),prove that \(\tan \left(\frac{B-C}{2}\right)=\frac{b-c}{b+c} \cot \frac{A}{2}\)
Solution:

Question 2.
Prove that cot A+cot B+ cot C \(=\frac{a^2+b^2+c^2}{4 \Delta}\)
Solution:

Question 3.
Show that r + r₃ + r₁ – r₂ = 4R cos B.
Solution:

Question 4.
In a ΔABC, if r₁= 8, r₂=12, r₃ = 24, find a, b, c.
Solution:

Question 5.
If the sides of a triangle are 13, 14, 15, then find the circum diameter.
Solution:
Let a= 13,b= 14,c= 15

Question 6.
Show that
a²cotA+b²cotB+c²cotC = \(\frac{\mathbf{a b c}}{\mathbf{R}}\)
Solution:

Question 7.
Prove that a(b cos C – c cos B) = b² – c².
Solution:

Question 8.
Show that \(\frac{c-b \cos A}{b-c \cos A}=\frac{\cos B}{\cos C}\)
Solution:
We have c = a cos B + b cos A and
b = a cos C + c cos A
\(\text { L.H.S. }=\frac{c-b \cos A}{b-c \cos A}\)

Question 9.
Show that b² sin 2C + c²sin 2B = 2bc sin A.
Solution:

Question 10.
Show that
\(a \cos ^2 \frac{A}{2}+b \cos ^2 \frac{B}{2}+c \cos ^2 \frac{C}{2}=s+\frac{\Delta}{R}\)
Solution:

Question 11.
In a ΔABC, if acosA=bcosB, prove that the triangle is either isosceles or right angled.
Solution:
a cos A = b cos B
⇒ 2R sin A cos A = 2R sin B cos B
⇒ sin 2A = sin 2B
⇒ sin 2A = sin (180° – 2B)
⇒ 2A = 2B or 2A = 180° – 2B
⇒ A=B or A=90°- B
⇒ A=B or A+B=90°
⇒  c = 90°
∴ The triangle is isosceles or right angled.

Question 12.
If \(\cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}\) = 3 : 5 : 7, Show that a : b : c =6 : 5 : 4.
Solution:

Thens – a = 3ks  – b = 5ks – c= 7k
∴ 3s – (a + b + c) = 15k
= 3s – 2s= 15k=s= 15k
∴ a = s – 3k = a = 12k
b = s – 5k=b= 10k
c = s – 7k c = 8k
∴ a : b : c = 12k : 10k : 8k = 6 : 5 : 4.

Question 13.
Prove that a³cos(B – C)+b³cos(C – A) + cos³ cos (A – B) = 3 abc.
Solution:
∑a³cos(B-C)
= ∑ a². a cos (B – C)
=∑ a² 2R sin A cos(B – C)
= ∑ a² 2sin(B + C) cos (B – C)
= \(R \Sigma a^2\left(2 \frac{b}{2 R} \cos B+2 \frac{c}{2 R} \cos C\right)\)
= ∑ a²(bcos B +ccosC)
= a² (bcos B + ccos C) + b²(ccos C + acosA) + c² (a cos A + b cos B)
= ab (a cos B+ b cos A) + bc (b cos C + c cos B) + ca(ccos A+acos C)
= abc + bca + cab
= 3 abc = R.H.S.

Question 14.
If P₁ P₂ p₃ are the altitudes of a ΔABC to the opposite sides show that

Solution:

Question 15.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 450 and from a point B Is 600, where B is a point at a distance 30 meters from the point A measured along the line AB which makes an angle 30° with AQ. Find the height of the tower.
Solution:

Question 16.
Two trees A and B are on the same side of a tiver. From a point C in the river, the distances of the trees A and B are 250 m and 300 m respectively. If the angle C is 450, find the distance between the trees (use \(\sqrt{2}\) = 1.414).
Solution:

Given AC = 300 m and BC = 250 m and in the Δ ABC, using cosine rule
AB² = AC² + BC² – 2AC.BC. cos 45°
= (300)² + (250)² -2 (300) (250) – \(\frac{1}{\sqrt{2}}\)
= 46450
∴ AB = 215.5 m (approximately)

Question 17.
Express \(\frac{a \cos A+b \cos B+c \cos C}{a+b+c}\) in terms of R and r.
Solution:

Question 18.
If a = 13, b = 14, c = 15, find r₁.
Solution:
\(\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\mathrm{s} \Rightarrow \mathrm{s}=\frac{13+14+15}{2}=\frac{42}{2}=21\)
s – a = 21- 13=8
s – b = 21 – 14 = 7 and
s – c = 21 – 15 = 6
Δ² = 21 x 8 x 7 6
Δ = 7 x 12 = 84sq. units
r₁ = \(\frac{\Delta}{s-a}=\frac{84}{8}\) = 10.5 units

Question 19.
If r r₂ = r₁ r₃, then find B.
Solution:
Given r r₂ = r₁ r₃

Question 20.
In a Δ ABC, show that the sides a, b, and c are in A.P. If and only If r₁, r₂, r₃ are in H.P.
Solution :
r₁, r₂, r₃ are in H.P.

⇔ s – a, s – b, s – c are in A.P
⇔ – a, – b, – c are in AP.
⇔ a, b, c are in A.P.

Question 21.
If A = 90°, show that 2 (r + R) = b + c
Solution:
L.H.S. = 2 (r + R)
= 2r + 2R
= 2(s – a) tan\(\frac{\mathrm{A}}{2}\) + 2R . 1
= 2 (s – a) tan 45° + 2R sinA (∵ A = 90°)
= (2s – 2a) + a
= 2s – a = a + b + c – a = b + c = R.H.S.

Question 22.
Show that

Solution:

Question 23.
Prove that Σ (r₁ + r) \(\tan \left(\frac{B-C}{2}\right)=0\)
Solution:

Question 24.
Show that \(\left(r_1+r_2\right) \sec ^2 \frac{C}{2}=\left(r_2+r_3\right) \sec ^2 \frac{A}{2} =\left(r_3+r_1\right) \sec ^2 \frac{B}{2}\)
Solution:

Hence the triangle is right angled at A.

Question 25.
Show that \(\frac{a b-r_1 r_2}{r_3}=\frac{b c-r_2 r_3}{r_1}=\frac{c a-r_3 r_1}{r_2}\)
Solution:

III.

Question 1.
If A, A₁, A₂, A₃ are the areas of in circle and exercise of a triangle respectively then prove that
\(\frac{1}{\sqrt{A_1}}+\frac{1}{\sqrt{A_2}}+\frac{1}{\sqrt{A_3}}=\frac{1}{\sqrt{A^2}}\)
Solution:
Let r₁, r₂, r₃ and r be the radii of excircies and incircie respectively whose areas are A₁, A₂, A₃ and A.

Question 2.
In a Δ ABC prove that
\(\frac{r_1}{b c}+\frac{r_2}{c a}+\frac{r_3}{a b}=\frac{1}{r}-\frac{1}{2 R}\)
Solution:

Question 3.
If (r₂ – r₁) (r₃ – r₃) = 2r₂r₃, show that A=90°
Solution:
Given (r₂ – r₁) (r₃ – r₁) = 2r₂r₃

Question 4.
Prove that \(\frac{r_1\left(r_2+r_3\right)}{\sqrt{r_1 r_2+r_2 r_3+r_3 r_1}}=a\)
Solution: