Trigonometric Ratios

Suppose a boy is standing in front of a lamp post at a certain distance. The height of the boy is 170 cm and the length of his shadow is 150 cm.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_5b13610f.png

You can see from the above figure that the boy and his shadow form a right-angled triangle as shown in the figure below.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_544e53c0.png

The ratio of the height of the boy to his shadow is 170:150 i.e., 17:15.

Is this ratio related to either of the angles of ΔABC?

We can also conclude the following:

cos Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_112616b2.gif , tan Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m717555f.gif

Also, note that

tan A = https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_3acc138c.gifand cot A =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_5b48b6ff.gif

Let us now solve some more examples based on trigonometric ratios.

Example 1: In a triangle ABC, right-angled at B, side AB = 40 cm and BC = 9 cm. Find the value of sin A, cos A, and tan A.

Solution:

It is given that AB = 40 cm and BC = 9 cm

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_6b0bf128.png

Using Pythagoras theorem in ΔABC, we obtain

(AC)2 = (AB)2 + (BC)2

(AC)2 = (40)2 + (9)2

(AC)2 = 1600 + 81

(AC)2 = 1681

(AC)2 = (41)2

AC = 41 cm

Now, sin Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_3c0cf6b6.gif

cos Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m663cc10c.gif

tan Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m1cf2fd98.gif

Example 2: From the given figure, find the values of cosec C and cot C, if AC = BC + 1.

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_64acc864.png

Solution:

Now, it is given that AB = 5 cm and

AC = BC + 1 … (1)

By Pythagoras theorem, we obtain

(AB)2 + (BC)2 = (AC)2

⇒ (AC)2 − (BC)2 = (AB)2

⇒ (BC + 1)2 – (BC)2 = (5)2 [Using (1)]

⇒ (BC)2 + 1 + 2BC – (BC)2 = 25

⇒ 2BC = 25 – 1

⇒ 2BC = 24

⇒ BC = 12 cm

∴ AC = 12 + 1 = 13 cm

Thus, cosec C =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_f27f207.gif

=

And, cot C =https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m7084a06e.gif

=https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m5aab2519.gif

Example 3: In a right-angled triangle ABC, which is right-angled at B, tan A https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m7993c13f.gif. Find the value of cos A and sec A.

Solution:

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m1d3892ea.png

It is given that tan Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_220db86f.gif

We know that tan Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m3c751f97.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m427cd988.gifhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_220db86f.gif

Let BC = 12k and AB = 5k

Using Pythagoras theorem in ΔABC, we obtain

(AC)2 = (AB)2 + (BC)2

= (5k)2 + (12k)2

= 25k2 + 144k2

(AC)2 = 169k2

AC = 13k

Now, cos Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_2e2ace9b.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m4979a9b4.gif

Sec Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_329ae5ba.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_m7913daa8.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/453/950/10.1.8.1.1_ok_SS_html_24c4de83.gif

Use Trigonometric Ratios In Solving Problems

If we know the value of a trigonometric ratio, then we can find the values of other trigonometric ratios and the value of any expression involving these trigonometric ratios.

Example1: If cot A https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m4f327529.gif, then find the value ofhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_5295a750.gif .

Solution:

It is given that cot A https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m75b8deaf.gif.

We know that

tan Ahttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m78cecd3d.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m2e69b953.gif

tan A = 3

Thenhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m63f682f3.gif,

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_2b366ac2.gif

Example 2: Find the value of (sin2 θ + cos2 θ), if sec θhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_11d3c1a3.gif .

Solution:

We have sec θ https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_400250c7.gif… (i)

We know that sec θ https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_7964033f.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m44d38068.png

Let ABC be a triangle in which ∠C = θ, therefore we have

sec θhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m64ee5023.gif

From equations (i) and (ii), we have

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m65eee47f.gif

Let us take AC = 25k and BC = 24k.

Using Pythagoras theorem, we have,

(AC)2 = (AB)2 + (BC)2

(25k)2 = (AB)2 + (24k)2

625k2 = (AB)2 + 576k2

AB2 = (625 − 576)k2

AB2 = 49k2

AB = 7k

∴ sin θhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_m1239917b.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_29b659a3.gif

cos θhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_6d28aab8.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_66ce5cb4.gif

Now, sin2 θ + cos2 θhttps://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_42331700.gif

https://img-nm.mnimgs.com/img/study_content/lp/1/10/9/135/178/454/500/10.1.8.1.2_html_4baf0317.gif

= 1

∴ sin2 θ + cos2 θ = 1