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Solution of Triangles

Sine Rule, Cosine Rule, Tangent Rule or Napier’s Analogy, Projection Rule, Trigonometric Ratios of Half Angles, Area of a Triangle, m-n Theorem, Apollonius Theorem

In any ∆ABC, the lengths of the sides opposite to angles A, B and C are denoted by a, b and c respectively. This means, BC = a; AC = b and AB = c.

Sine Rule

In any ∆ABC:

asinA=bsinB=csinC=k, where k is a constant

Cosine Rule

In any ∆ABC:

cosA=b2+c2-a22bc cosB=c2+a2-b22ca cosC=a2+b2-c22ab

Tangent Rule or Napier’s Analogy

In any ∆ABC:

tanA-B2=a-ba+bcotC2  tanB-C2=b-cb+ccotA2 tanC-A2=c-ac+acotB2

Projection Rule

In any ∆ABC: a=bcosC+ccosB b=ccosA+acosC c=acosB+bcosA

Trigonometric Ratios of Half Angles

In any ∆ABC:

sinA2=s-bs-cbc, sinB2=s-cs-aca, sinC2=s-as-bab cosA2=ss-abc, cosB2=ss-bca, cosC2=ss-cab

tanA2=s-bs-css-a, tanB2=s-cs-ass-b, tanC2=s-as-bss-c

Area of a Triangle

If the area of the triangle is denoted by ∆, then

∆=12bcsinA=12casinB=12absinC=abc4R=rs=ss-as-bs-c

Here, R is the radius of circumcircle of ∆ABC; r is the radius of inscribed circle of ∆ABC and 2s = a + b + c.

m-n Theorem

Let D be a point on side BC of ∆ABC such that it divides the side BC in the ratio m : n.

1. m+ncotθ=mcotα-ncotβ​       2. m+ncotθ=ncotB-mcotC

Apollonius Theorem

In ∆ABC, if AD is the median through A, then AB2+AC2=2AD2+BD2. Reality checkQuestion:1

What is the perimeter of the triangle whose lengths of sides are three consecutive natural numbers and the largest angle is twice the smallest angle?

A)

13 units

B)

14 units

C)

15 units

D)

18 units

Let the sides of the triangle be x, x + 1 and x + 2, where x is a natural number. Suppose ∠C = α is the smallest angle.

∴ ∠A = 2α

Applying sine rule in ΔABC, we have

…(1)

Also,

…(2)

From (1) and (2), we have

x = 4    (x is a natural number)

∴  x + 1 = 5 and x + 2 = 6

Perimeter of the triangle = x + (x + 1) + (x + 2) = 15 units Hence, the correct answer is option C.

Question:2

The sides of a triangle are in arithmetic progression (A.P.). If the smallest angle of the triangle is θ and its largest angle exceeds the smallest angle by α, then what is the value of  tanθ+α2 ?

A)

B)

C)

D)

Let the sides of ΔABC be a − d, a and a + d, where d > 0.

The greatest side, AB = a + d and the smallest side, BC = a − d.

So, ∠A is the smallest angle and ∠C is the largest angle.

Now, ∠A = θ

∠C = θ + α

∠B = 180o − (2θ + α)

Applying sine rule, we get:

The correct option is (C).

Question:3

Prove that for any triangle where A is the area and s is the semi-perimeter of the triangle.

Let a, b and c be the sides of the triangle.

A.M. ≥ G.M.

Question:4

A)

B)

C)

D)

None of these

Hence, the correct option is (C).

Question:5

In ΔABC, let BC = a, CA = b and AB = c. If (a + b + c) (a − b + c) = 48 sq.cm and ∠B = 45° , what is the area of ΔABC?

A)

6 sq.cm

B)

C)

12 sq.cm

D)

(a + b + c) (a − b + c) = [(a + c) + b] [(a + c) − b]

= (a + c)2 − b2

= a2 + c2 + 2ac − b2

= (a2 + c2 − b2) + 2ac

It is given that ∠B = 45°.

cos B =

⇒ a2 + c2 − b2 =

∴ (a + b + c) (a − b + c) = = ac

It is given that (a + b + c) (a − b + c) = 48sq. cm.

⇒ac = 48sq. cm

⇒ ac = 24sq. cm

∴ Area of ΔABC =

= 12 sq.cm

The correct answer is given by option C.

Question:6The angles of a triangle ABC are in AP and the ratio of b to c is 6 : 2. Find all the angles.It is given that ∠A, ∠B and ∠C are in AP. ∴ 2∠B = ∠A + ∠C In ∆ABC, ∠A + ∠B + ∠C = 180° ⇒ 3∠B = 180° ⇒ ∠B = 60° Now, bsinB=csinC              Sine rule⇒bc=sinBsinC⇒62=sin60°sinC ⇒62=32sinC⇒sinC=12⇒∠C=45°∴∠A+60°+45°=180°⇒∠A=75° Hence, the angles of ∆ABC are 75°, 60° and 45°.Question:7If in triangle ABC, b+c15=c+a16=a+b17, then prove that cosA6=cosB11=cosC14.Let: b+c15=c+a16=a+b17=k⇒b+c=15k, c+a=16k, a+b=17k∴ b+c+ c+a+a+b=15k+16k+17k⇒2a+b+c=48k⇒a+b+c=24k ∴ a=9k, b=8k, c=7k So, cosA=b2+c2-a22bc=64k2+49k2-81k22×8k×7k=27cosB=c2+a2-b22ac=49k2+81k2-64k22×9k×7k=1121cosC=a2+b2-c22ab=81k2+64k2-49k22×9k×8k=23 ∴ cosA : cosB : cosC =6 : 11 : 14Question:8Find the area of the ΔABC, if a = 3, b = 32 and c = 5.We have a = 3, b = 32, c = 5 cosB=a2+c2-b22ac         =9+25-182×3×5         =1630         =815sinB=16115 Area of ∆ABC=12acsinB                      =12×3×5×16115                      =1612 square units

In any ∆ABC, the lengths of the sides opposite to angles A, B and C are denoted by a, b and c respectively. This means, BC = a; AC = b and AB = c.

Sine Rule

In any ∆ABC:

asinA=bsinB=csinC=k, where k is a constant

Cosine Rule

In any ∆ABC:

cosA=b2+c2-a22bc cosB=c2+a2-b22ca cosC=a2+b2-c22ab

Tangent Rule or Napier’s Analogy

In any ∆ABC:

tanA-B2=a-ba+bcotC2  tanB-C2=b-cb+ccotA2 tanC-A2=c-ac+acotB2…

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